Rd Sharma 2019 2020 for Class 8 Maths Chapter 6 - Algebraic Expressions And Identities

Share with your friends

Rd Sharma 2019 2020 Solutions for Class 8 Maths Chapter 6 Algebraic Expressions And Identities are provided here with simple step-by-step explanations. These solutions for Algebraic Expressions And Identities are extremely popular among Class 8 students for Maths Algebraic Expressions And Identities Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2019 2020 Book of Class 8 Maths Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2019 2020 Solutions. All Rd Sharma 2019 2020 Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

Page No 6.13:

Question 1:

Find each of the following product:
5x²× 4x³

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices. However, use of these laws are subject to their applicability in the given expressions.

In the present problem, to perform the multiplication, we can proceed as follows:

$5 x^{2} \times 4 x^{3} = (5 \times 4) \times (x^{2} \times x^{3})$
$= 20 x^{5}$ ( $∵$ $a^{m} \times a^{n} = a^{m + n}$ )

Thus, the answer is $20 x^{5}$ .

Page No 6.13:

Question 2:

Find each of the following product:
−3a² × 4b⁴

Answer:

To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a^{m} \times a^{n} = a^{m + n}$ , wherever applicable.

We have:

$- 3 a^{2} \times 4 b^{4} = (- 3 \times 4) \times (a^{2} \times b^{4}) = - 12 a^{2} b^{4}$

Thus, the answer is $- 12 a^{2} b^{4}$ .

Page No 6.13:

Question 3:

Find each of the following product:
(−5xy) × (−3x²yz)

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, $a^{m} \times a^{n} = a^{m + n}$ , wherever applicable.

We have:

$(- 5 x y) \times (- 3 x^{2} y z) = \{(- 5) \times (- 3)\} \times (x \times x^{2}) \times (y \times y) \times z = 15 \times (x^{1 + 2}) \times (y^{1 + 1}) \times z = 15 x^{3} y^{2} z$

Thus, the answer is $15 x^{3} y^{2} z$ .

Page No 6.13:

Question 4:

Find each of the following product:
$\frac{1}{4} x y \times \frac{2}{3} x^{2} y z^{2}$

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the the law of indices, that is, $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$\frac{1}{4} x y \times \frac{2}{3} x^{2} y z^{2} = (\frac{1}{4} \times \frac{2}{3}) \times (x \times x^{2}) \times (y \times y) \times z^{2} = (\frac{1}{4} \times \frac{2}{3}) \times (x^{1 + 2}) \times (y^{1 + 1}) \times z^{2} = \frac{1}{6} x^{3} y^{2} z^{2}$

Thus, the answer is $\frac{1}{6} x^{3} y^{2} z^{2}$ .

Page No 6.14:

Question 5:

Find each of the following product:
$(- \frac{7}{5} x y^{2} z) \times (\frac{13}{3} x^{2} y z^{2})$

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(- \frac{7}{5} x y^{2} z) \times (\frac{13}{3} x^{2} y z^{2}) = (- \frac{7}{5} \times \frac{13}{3}) \times (x \times x^{2}) \times (y^{2} \times y) \times (z \times z^{2}) = (- \frac{7}{5} \times \frac{13}{3}) \times (x^{1 + 2}) \times (y^{2 + 1}) \times (z^{1 + 2}) = - \frac{91}{15} x^{3} y^{3} x^{3}$

Thus, the answer is $- \frac{91}{15} x^{3} y^{3} x^{3}$ .

Page No 6.14:

Question 6:

Find each of the following product:
$(\frac{- 24}{25} x^{3} z) \times (- \frac{15}{16} x z^{2} y)$

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(- \frac{24}{25} x^{3} z) \times (- \frac{15}{16} x z^{2} y) = \{(- \frac{24}{25}) \times (- \frac{15}{16})\} \times (x^{3} \times x) \times (z \times z^{2}) \times y = \{(- \frac{24}{25}) \times (- \frac{15}{16})\} \times (x^{3 + 1}) \times (z^{1 + 2}) \times y$
$= \frac{9}{10} x^{4} y z^{3}$

Thus, the answer is $\frac{9}{10} x^{4} y z^{3}$ .

Page No 6.14:

Question 7:

Find each of the following product:
$(- \frac{1}{27} a^{2} b^{2}) \times (\frac{9}{2} a^{3} b^{2} c^{2})$

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:

$(- \frac{1}{27} a^{2} b^{2}) \times (\frac{9}{2} a^{3} b^{2} c^{2}) = (- \frac{1}{27} \times \frac{9}{2}) \times (a^{2} \times a^{3}) \times (b^{2} \times b^{2}) \times c^{2} = (- \frac{1}{27} \times \frac{9}{2}) \times (a^{2 + 3}) \times (b^{2 + 2}) \times c^{2} = - \frac{1}{6} a^{5} b^{4} c^{2}$

Thus, the answer is $- \frac{1}{6} a^{5} b^{4} c^{2}$ .

Page No 6.14:

Question 8:

Find each of the following product:
$(- 7 x y) \times (\frac{1}{4} x^{2} y z)$

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(- 7 x y) \times (\frac{1}{4} x^{2} y z) = (- 7 \times \frac{1}{4}) \times (x \times x^{2}) \times (y \times y) \times z = (- 7 \times \frac{1}{4}) \times (x^{1 + 2}) \times (y^{1 + 1}) \times z = - \frac{7}{4} x^{3} y^{2} z$

Thus, the answer is $- \frac{7}{4} x^{3} y^{2} z$ .

Page No 6.14:

Question 9:

Find each of the following product:
(7ab) × (−5ab²c) × (6abc²)

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(7 a b) \times (- 5 a b^{2} c) \times (6 a b c^{2}) = \{7 \times (- 5) \times 6\} \times (a \times a \times a) \times (b \times b^{2} \times b) \times (c \times c^{2}) = \{7 \times (- 5) \times 6\} \times (a^{1 + 1 + 1}) \times (b^{1 + 2 + 1}) \times (c^{1 + 2}) = - 210 a^{3} b^{4} c^{3}$

Thus, the answer is $- 210 a^{3} b^{4} c^{3}$ .

Page No 6.14:

Question 10:

Find each of the following product:
(−5a) × (−10a²) × (−2a³)

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:

$(- 5 a) \times (- 10 a^{2}) \times (- 2 a^{3}) = \{(- 5) \times (- 10) \times (- 2)\} \times (a \times a^{2} \times a^{3}) = \{(- 5) \times (- 10) \times (- 2)\} \times (a^{1 + 2 + 3}) = - 100 a^{6}$

Thus, the answer is $- 100 a^{6}$ .

Page No 6.14:

Question 11:

Find each of the following product:
(−4x²) × (−6xy²) × (−3yz²)

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(- 4 x^{2}) \times (- 6 x y^{2}) \times (- 3 y z^{2}) = \{(- 4) \times (- 6) \times (- 3)\} \times (x^{2} \times x) \times (y^{2} \times y) \times z^{2} = \{(- 4) \times (- 6) \times (- 3)\} \times (x^{2 + 1}) \times (y^{2 + 1}) \times z^{2} = - 72 x^{3} y^{3} z^{2}$

Thus, the answer is $- 72 x^{3} y^{3} z^{2}$ .

Page No 6.14:

Question 12:

Find each of the following product:
$(- \frac{2}{7} a^{4}) \times (- \frac{3}{4} a^{2} b) \times (- \frac{14}{5} b^{2})$

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(- \frac{2}{7} a^{4}) \times (- \frac{3}{4} a^{2} b) \times (- \frac{14}{5} b^{2}) = \{(- \frac{2}{7}) \times (- \frac{3}{4}) \times (- \frac{14}{5})\} \times (a^{4} \times a^{2}) \times (b \times b^{2}) = \{- (\frac{2}{7} \times \frac{3}{4} \times \frac{14}{5})\} \times a^{4 + 2} \times b^{1 + 2} = \{- (\frac{2}{7} \times \frac{3}{4_{2}} \times \frac{14^{2^{1}}}{5})\} \times a^{6} \times b^{3} = - \frac{3}{5} a^{6} b^{3}$

Thus, the answer is $- \frac{3}{5} a^{6} b^{3}$ .

Page No 6.14:

Question 13:

Find each of the following product:
$(\frac{7}{9} a b^{2}) \times (\frac{15}{7} a c^{2} b) \times (- \frac{3}{5} a^{2} c)$

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(\frac{7}{9} a b^{2}) \times (\frac{15}{7} a c^{2} b) \times (- \frac{3}{5} a^{2} c) = \{\frac{7}{9} \times \frac{15}{7} \times (- \frac{3}{5})\} \times (a \times a \times a^{2}) \times (b^{2} \times b) \times (c^{2} \times c) = \{\frac{7^{1}}{9_{3}} \times \frac{15^{3}}{7} \times (- \frac{3^{1}}{5})\} \times (a \times a \times a^{2}) \times (b^{2} \times b) \times (c^{2} \times c) = \{\frac{7^{1}}{9_{3}} \times \frac{15^{3^{1}}}{7} \times (- \frac{3^{1}}{5})\} \times (a^{1 + 1 + 2}) \times (b^{2 + 1}) \times (c^{2 + 1}) = - a^{4} b^{3} c^{3}$

Thus, the answer is $- a^{4} b^{3} c^{3}$ .

Page No 6.14:

Question 14:

Find each of the following product:
$(\frac{4}{3} u^{2} v w) \times (- 5 u v w^{2}) \times (\frac{1}{3} v^{2} w u)$

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(\frac{4}{3} u^{2} v w) \times (- 5 u v w^{2}) \times (\frac{1}{3} v^{2} w u) = \{\frac{4}{3} \times (- 5) \times \frac{1}{3}\} \times (u^{2} \times u \times u) \times (v \times v \times v^{2}) \times (w \times w^{2} \times w) = \{\frac{4}{3} \times (- 5) \times \frac{1}{3}\} \times (u^{2 + 1 + 1}) \times (v^{1 + 1 + 2}) \times (w^{1 + 2 + 1}) = - \frac{20}{9} u^{4} v^{4} w^{4}$

Thus, the answer is $- \frac{20}{9} u^{4} v^{4} w^{4}$ .

Page No 6.14:

Question 15:

Find each of the following product:
$(0.5 x) \times (\frac{1}{3} x y^{2} z^{4}) \times (24 x^{2} y z)$

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:
$(0.5 x) \times (\frac{1}{3} x y^{2} z^{4}) \times (24 x^{2} y z) = (0.5 \times \frac{1}{3} \times 24) \times (x \times x \times x^{2}) \times (y^{2} \times y) \times (z^{4} \times z) = (0.5 \times \frac{1}{3} \times 24) \times (x^{1 + 1 + 2}) \times (y^{2 + 1}) \times (z^{4 + 1}) = 4 x^{4} y^{3} z^{5}$

Thus, the answer is $4 x^{4} y^{3} z^{5}$ .

Page No 6.14:

Question 16:

Find each of the following product:
$(\frac{4}{3} p q^{2}) \times (- \frac{1}{4} p^{2} r) \times (16 p^{2} q^{2} r^{2})$

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:

$(\frac{4}{3} p q^{2}) \times (- \frac{1}{4} p^{2} r) \times (16 p^{2} q^{2} r^{2}) = \{\frac{4}{3} \times (- \frac{1}{4}) \times 16\} \times (p \times p^{2} \times p^{2}) \times (q^{2} \times q^{2}) \times (r \times r^{2}) = \{\frac{4}{3} \times (- \frac{1}{4}) \times 16\} \times (p^{1 + 2 + 2}) \times (q^{2 + 2}) \times (r^{1 + 2}) = - \frac{16}{3} p^{5} q^{4} r^{3}$

Thus, the answer is $- \frac{1}{3} p^{5} q^{4} r^{3}$ .

Page No 6.14:

Question 17:

Find each of the following product:
(2.3xy) × (0.1x) × (0.16)

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:
$(2.3 x y) \times (0.1 x) \times (0.16) = (2.3 \times 0.1 \times 0.16) \times (x \times x) \times y = (2.3 \times 0.1 \times 0.16) \times (x^{1 + 1}) \times y = 0.0368 x^{2} y$

Thus, the answer is $0.0368 x^{2} y$ .

Page No 6.14:

Question 18:

Express each of the following product as a monomials and verify the result in each case for x = 1:
(3x) × (4x) × (−5x)

Answer:

We have to find the product of the expression in order to express it as a monomial.

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:

$(3 x) \times (4 x) \times (- 5 x) = \{3 \times 4 \times (- 5)\} \times (x \times x \times x) = \{3 \times 4 \times (- 5)\} \times (x^{1 + 1 + 1}) = - 60 x^{3}$

Substituting x = 1 in LHS, we get:

$LHS = (3 x) \times (4 x) \times (- 5 x) = (3 \times 1) \times (4 \times 1) \times (- 5 \times 1) = - 60$

Putting x = 1 in RHS, we get:
$RHS = - 60 x^{3} = - 60 {(1)}^{3} = - 60 \times 1 = - 60$

$∵$ LHS = RHS for x = 1; therefore, the result is correct

Thus, the answer is $- 60 x^{3}$ .

Page No 6.14:

Question 19:

Express each of the following product as a monomials and verify the result in each case for x = 1:
(4x²) × (−3x) × $(\frac{4}{5} x^{3})$

Answer:

We have to find the product of the expression in order to express it as a monomial.

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(4 x^{2}) \times (- 3 x) \times (\frac{4}{5} x^{3}) = \{4 \times (- 3) \times \frac{4}{5}\} \times (x^{2} \times x \times x^{3}) = \{4 \times (- 3) \times \frac{4}{5}\} \times (x^{2 + 1 + 3}) = - \frac{48}{5} x^{6}$

$∴$ $(4 x^{2}) \times (- 3 x) \times (\frac{4}{5} x^{3}) = - \frac{48}{5} x^{6}$

Substituting x = 1 in LHS, we get:

$LHS = (4 x^{2}) \times (- 3 x) \times (\frac{4}{5} x^{3}) = (4 \times 1^{2}) \times (- 3 \times 1) \times (\frac{4}{5} \times 1^{3}) = 4 \times (- 3) \times \frac{4}{5} = - \frac{48}{5}$

Putting x = 1 in RHS, we get:

$RHS = - \frac{48}{5} x^{6} = - \frac{48}{5} \times 1^{6} = - \frac{48}{5}$

$∵$ LHS = RHS for x = 1; therefore, the result is correct

Thus, the answer is $- \frac{48}{5} x^{6}$ .

Page No 6.14:

Question 20:

Express each of the following product as a monomials and verify the result in each case for x = 1:
(5x⁴) × (x²)³ × (2x)²

Answer:

We have to find the product of the expression in order to express it as a monomial.

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n} and {(a^{m})}^{n} = a^{m n}$ .

We have:

$(5 x^{4}) \times {(x^{2})}^{3} \times {(2 x)}^{2} = (5 x^{4}) \times (x^{6}) \times (2^{2} \times x^{2}) = (5 \times 2^{2}) \times (x^{4} \times x^{6} \times x^{2}) = (5 \times 2^{2}) \times (x^{4 + 6 + 2}) = 20 x^{12}$
$∴$ $(5 x^{4}) \times {(x^{2})}^{3} \times {(2 x)}^{2} = 20 x^{12}$

Substituting x = 1 in LHS, we get:

$LHS = (5 x^{4}) \times {(x^{2})}^{3} \times {(2 x)}^{2} = (5 \times 1^{4}) \times {(1^{2})}^{3} \times {(2 \times 1)}^{2} = (5 \times 1) \times (1^{6}) \times {(2)}^{2} = 5 \times 1 \times 4 = 20$

Put x =1 in RHS, we get:

$RHS = 20 x^{12} = 20 \times {(1)}^{12} = 20 \times 1 = 20$

$∵$ LHS = RHS for x = 1; therefore, the result is correct.

Thus, the answer is $20 x^{12}$ .

Page No 6.14:

Question 21:

Express each of the following product as a monomials and verify the result in each case for x = 1:
(x²)³ × (2x) × (−4x) × (5)

Answer:

We have to find the product of the expression in order to express it as a monomial.

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n} and {(a^{m})}^{n} = a^{m n}$ .

We have:

${(x^{2})}^{3} \times (2 x) \times (- 4 x) \times 5 = (x^{6}) \times (2 x) \times (- 4 x) \times 5 = \{2 \times (- 4) \times 5\} \times (x^{6} \times x \times x) = \{2 \times (- 4) \times 5\} \times (x^{6 + 1 + 1}) = - 40 x^{8}$

$∴$ ${(x^{2})}^{3} \times (2 x) \times (- 4 x) \times 5 = - 40 x^{8}$

Substituting x = 1 in LHS, we get:

$LHS = {(x^{2})}^{3} \times (2 x) \times (- 4 x) \times 5 = {(1^{2})}^{3} \times (2 \times 1) \times (- 4 \times 1) \times 5 = 1^{6} \times 2 \times (- 4) \times 5 = 1 \times 2 \times (- 4) \times 5 = - 40$

Putting x = 1 in RHS, we get:

$RHS = - 40 x^{8} = - 40 {(1)}^{8} = - 40 \times 1 = - 40$

$∵$ LHS = RHS for x = 1; therefore, the result is correct

Thus, the answer is $- 40 x^{8}$ .

Page No 6.14:

Question 22:

Write down the product of −8x²y⁶ and −20xy. Verify the product for x = 2.5, y = 1.

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(- 8 x^{2} y^{6}) \times (- 20 x y) = \{(- 8) \times (- 20)\} \times (x^{2} \times x) \times (y^{6} \times y) = \{(- 8) \times (- 20)\} \times (x^{2 + 1}) \times (y^{6 + 1}) = - 160 x^{3} y^{7}$
$∴$ $(- 8 x^{2} y^{6}) \times (- 20 x y) = - 160 x^{3} y^{7}$

Substituting x = 2.5 and y = 1 in LHS, we get:

$LHS = (- 8 x^{2} y^{6}) \times (- 20 x y) = \{- 8 {(2.5)}^{2} {(1)}^{6}\} \times \{- 20 (2.5) (1)\} = \{- 8 (6.25) (1)\} \times \{- 20 (2.5) (1)\} = (- 50) \times (- 50) = 2500$

Substituting x = 2.5 and y = 1 in RHS, we get:

$RHS = - 160 x^{3} y^{7} = - 160 {(2.5)}^{3} {(1)}^{7} = - 160 (15.625) \times 1 = - 2500$

Because LHS is equal to RHS, the result is correct.

Thus, the answer is $- 160 x^{3} y^{7}$ .

Page No 6.14:

Question 23:

Evaluate (3.2x⁶y³) × (2.1x²y²) when x = 1 and y = 0.5

Ans

First multiply the expressions and then substitute the values for the variables.

To multiply algebric experssions use the commutative and the associative laws along with the law of indices, $a^{m} \times a^{n} = a^{m + n}$ .
We have,

$(3.2 x^{6} y^{3}) \times (2.1 x^{2} y^{2}) = (3.2 \times 2.1) \times (x^{6} \times x^{2}) \times (y^{3} \times y^{2}) = 6.72 x^{8} y^{5}$
Hence, $(3.2 x^{6} y^{3}) \times (2.1 x^{2} y^{2}) = 6.72 x^{8} y^{5}$

Now, substitute 1 for x and 0.5 for y in the result.

$6.72 x^{8} y^{5} = 6.72 {(1)}^{8} {(0.5)}^{5} = 6.72 \times 1 \times 0.03125 = 0.21$

Hence, the answer is $0.21$ .

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(3.2 x^{6} y^{3}) \times (2.1 x^{2} y^{2}) = (3.2 \times 2.1) \times (x^{6} \times x^{2}) \times (y^{3} \times y^{2}) = (3.2 \times 2.1) \times (x^{6 + 2}) \times (y^{3 + 2}) = 6.72 x^{8} y^{5}$

$∴$ $(3.2 x^{6} y^{3}) \times (2.1 x^{2} y^{2}) = 6.72 x^{8} y^{5}$

Substituting x = 1 and y = 0.5 in the result, we get:

$6.72 x^{8} y^{5} = 6.72 {(1)}^{8} {(0.5)}^{5} = 6.72 \times 1 \times 0.03125 = 0.21$

Thus, the answer is $0.21$ .

Page No 6.14:

Question 24:

Find the value of (5x⁶) × (−1.5x²y³) × (−12xy²) when x = 1, y = 0.5.

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(5 x^{6}) \times (- 1.5 x^{2} y^{3}) \times (- 12 x y^{2}) = \{5 \times (- 1.5) \times (- 12)\} \times (x^{6} \times x^{2} \times x) \times (y^{3} \times y^{2}) = \{5 \times (- 1.5) \times (- 12)\} \times (x^{6 + 2 + 1}) \times (y^{3 + 2}) = 90 x^{9} y^{5}$

$∴$ $(5 x^{6}) \times (- 1.5 x^{2} y^{3}) \times (- 12 x y^{2}) = 90 x^{9} y^{5}$

Substituting x = 1 and y = 0.5 in the result, we get:

$90 x^{9} y^{5} = 90 {(1)}^{9} {(0.5)}^{5} = 90 \times 1 \times 0.03125 = 2.8125$

Thus, the answer is 2.8125.

Page No 6.14:

Question 25:

Evaluate (2.3a⁵b²) × (1.2a²b²) when a = 1 and b = 0.5.

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(2.3 a^{5} b^{2}) \times (1.2 a^{2} b^{2}) = (2.3 \times 1.2) \times (a^{5} \times a^{2}) \times (b^{2} \times b^{2}) = (2.3 \times 1.2) \times (a^{5 + 2}) \times (b^{2 + 2}) = 2.76 a^{7} b^{4}$

$∴$ $(2.3 a^{5} b^{2}) \times (1.2 a^{2} b^{2}) = 2.76 a^{7} b^{4}$

Substituting a =1 and b = 0.5 in the result, we get:

$2.76 a^{7} b^{4} = 2.76 {(1)}^{7} {(0.5)}^{4} = 2.76 \times 1 \times 0.0625 = 0.1725$

Thus, the answer is $0.1725$ .

Page No 6.14:

Question 26:

Evaluate (−8x²y⁶) × (−20xy) for x = 2.5 and y = 1.

Answer:

Page No 6.14:

Question 27:

Express each of the following product as a monomials and verify the result for x = 1, y = 2:
(−xy³) × (yx³) × (xy)

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(- x y^{3}) \times (y x^{3}) \times (x y) = (- 1) \times (x \times x^{3} \times x) \times (y^{3} \times y \times y) = (- 1) \times (x^{1 + 3 + 1}) \times (y^{3 + 1 + 1}) = - x^{5} y^{5}$

To verify the result, we substitute x = 1 and y = 2 in LHS; we get:

$LHS = (- x y^{3}) \times (y x^{3}) \times (x y) = \{(- 1) \times 1 \times 2^{3}\} \times (2 \times 1^{3}) \times (1 \times 2) = \{(- 1) \times 1 \times 8\} \times (2 \times 1) \times 2 = (- 8) \times 2 \times 2 = - 32$

Substituting x = 1 and y = 2 in RHS, we get:

$RHS = - x^{5} y^{5} = (- 1) {(1)}^{5} {(2)}^{5} = (- 1) \times 1 \times 32 = - 32$

Because LHS is equal to RHS, the result is correct.

Thus, the answer is $- x^{5} y^{5}$ .

Page No 6.14:

Question 28:

Express each of the following product as a monomials and verify the result for x = 1, y = 2:
$(\frac{1}{8} x^{2} y^{4}) \times (\frac{1}{4} x^{4} y^{2}) \times (x y) \times 5$

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(\frac{1}{8} x^{2} y^{4}) \times (\frac{1}{4} x^{4} y^{2}) \times (x y) \times 5 = (\frac{1}{8} \times \frac{1}{4} \times 5) \times (x^{2} \times x^{4} \times x) \times (y^{4} \times y^{2} \times y) = (\frac{1}{8} \times \frac{1}{4} \times 5) \times (x^{2 + 4 + 1}) \times (y^{4 + 2 + 1}) = \frac{5}{32} x^{7} y^{7}$

To verify the result, we substitute x = 1 and y = 2 in LHS; we get:

$LHS = (\frac{1}{8} x^{2} y^{4}) \times (\frac{1}{4} x^{4} y^{2}) \times (x y) \times 5 = \{\frac{1}{8} \times {(1)}^{2} \times {(2)}^{4}\} \times \{\frac{1}{4} \times {(1)}^{4} \times {(2)}^{2}\} \times (1 \times 2) \times 5 = (\frac{1}{8} \times 1 \times 16) \times (\frac{1}{4} \times 1 \times 4) \times (1 \times 2) \times 5 = 2 \times 1 \times 2 \times 5 = 20$

Substituting x = 1 and y = 2 in RHS, we get:

$RHS = \frac{5}{32} x^{7} y^{7} = \frac{5}{32} {(1)}^{7} {(2)}^{7} = \frac{5}{32} \times 1 \times 128^{4} = 20$

Because LHS is equal to RHS, the result is correct.

Thus, the answer is $\frac{5}{32} x^{7} y^{7}$ .

Page No 6.14:

Question 29:

Express each of the following product as a monomials and verify the result for x = 1, y = 2:
$(\frac{2}{5} a^{2} b) \times (- 15 b^{2} a c) \times (- \frac{1}{2} c^{2})$

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(\frac{2}{5} a^{2} b) \times (- 15 b^{2} a c) \times (- \frac{1}{2} c^{2}) = \{\frac{2}{5} \times (- 15) \times (- \frac{1}{2})\} \times (a^{2} \times a) \times (b \times b^{2}) \times (c \times c^{2}) = \{\frac{2}{5} \times (- 15) \times (- \frac{1}{2})\} \times (a^{2 + 1}) \times (b^{1 + 2}) \times (c^{1 + 2}) = 3 a^{3} b^{3} c^{3}$

$∵$ The expression doesn't consist of the variables x and y.

$∴$ The result cannot be verified for x = 1 and y = 2

Thus, the answer is $3 a^{3} b^{3} c^{3}$ .

Page No 6.14:

Question 30:

Express each of the following product as a monomials and verify the result for x = 1, y = 2:
$(- \frac{4}{7} a^{2} b) \times (- \frac{2}{3} b^{2} c) \times (- \frac{7}{6} c^{2} a)$

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(- \frac{4}{7} a^{2} b) \times (- \frac{2}{3} b^{2} c) \times (- \frac{7}{6} c^{2} a) = \{(- \frac{4}{7}) \times (- \frac{2}{3}) \times (- \frac{7}{6})\} \times (a^{2} \times a) \times (b \times b^{2}) \times (c \times c^{2}) = \{(- \frac{4}{7}) \times (- \frac{2}{3}) \times (- \frac{7}{6})\} \times (a^{2 + 1}) \times (b^{1 + 2}) \times (c^{1 + 2}) = - \frac{4}{9} a^{3} b^{3} c^{3}$

$∵$ The expression doesn't consist of the variables x and y.

$∴$ The result cannot be verified for x = 1 and y = 2.

Thus, the answer is $- \frac{4}{9} a^{3} b^{3} c^{3}$ .

Page No 6.14:

Question 31:

Express each of the following product as a monomials and verify the result for x = 1, y = 2:
$(\frac{4}{9} a b c^{3}) \times (- \frac{27}{5} a^{3} b^{2}) \times (- 8 b^{3} c)$

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(\frac{4}{9} a b c^{3}) \times (- \frac{27}{5} a^{3} b^{2}) \times (- 8 b^{3} c) = \{(\frac{4}{9}) \times (- \frac{27}{5}) \times (- 8)\} \times (a \times a^{3}) \times (b \times b^{2} \times b^{3}) \times (c^{3} \times c) = \{(\frac{4}{9}) \times (- \frac{27}{5}) \times (- 8)\} \times (a^{1 + 3}) \times (b^{1 + 2 + 3}) \times (c^{3 + 1}) = \frac{96}{5} a^{4} b^{6} c^{4}$

Thus, the answer is $\frac{96}{5} a^{4} b^{6} c^{4}$ .

$∵$ The expression doesn't consist of the variables x and y.

$∴$ The result cannot be verified for x = 1 and y = 2

Page No 6.14:

Question 32:

Evaluate each of the following when x = 2, y = −1.
$(2 x y) \times (\frac{x^{2} y}{4}) \times (x^{2}) \times (y^{2})$

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(2 x y) \times (\frac{x^{2} y}{4}) \times (x^{2}) \times (y^{2}) = (2 \times \frac{1}{4}) \times (x \times x^{2} \times x^{2}) \times (y \times y \times y^{2}) = (2 \times \frac{1}{4}) \times (x^{1 + 2 + 2}) \times (y^{1 + 1 + 2}) = \frac{1}{2} x^{5} y^{4}$

$∴$ $(2 x y) \times (\frac{x^{2} y}{4}) \times (x^{2}) \times (y^{2}) = \frac{1}{2} x^{5} y^{4}$

Substituting x = 2 and y = $-$ 1 in the result, we get:

$\frac{1}{2} x^{5} y^{4} = \frac{1}{2} {(2)}^{5} {(- 1)}^{4} = \frac{1}{2} \times 32 \times 1 = 16$

Thus, the answer is $16$ .

Page No 6.14:

Question 33:

Evaluate each of the following when x = 2, y = −1.
$(\frac{3}{5} x^{2} y) \times (- \frac{15}{4} x y^{2}) \times (\frac{7}{9} x^{2} y^{2})$

Answer:

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

$(\frac{3}{5} x^{2} y) \times (- \frac{15}{4} x y^{2}) \times (\frac{7}{9} x^{2} y^{2}) = \{\frac{3}{5} \times (- \frac{15}{4}) \times \frac{7}{9}\} \times (x^{2} \times x \times x^{2}) \times (y \times y^{2} \times y^{2}) = \{\frac{3}{5} \times (- \frac{15}{4}) \times \frac{7}{9}\} \times (x^{2 + 1 + 2}) \times (y^{1 + 2 + 2}) = - \frac{7}{4} x^{5} y^{5}$

$∴$ $(\frac{3}{5} x^{2} y) \times (- \frac{15}{4} x y^{2}) \times (\frac{7}{9} x^{2} y^{2}) = - \frac{7}{4} x^{5} y^{5}$ .

Substituting x = 2 and y = $-$ 1 in the result, we get:

$- \frac{7}{4} x^{5} y^{5} = - \frac{7}{4} {(2)}^{5} {(- 1)}^{5} = (- \frac{7}{4}) \times 32 \times (- 1) = 56$

Thus, the answer is 56.

Page No 6.2:

Question 1:

Identify the terms, their coefficients for each of the following expressions:
(i) 7x²yz − 5xy
(ii) x² + x + 1
(iii) 3x²y² − 5x²y²z² + z²
(iv) 9 − ab + bc − ca
(v) $\frac{a}{2} + \frac{b}{2} - a b$
(vi) 0.2x − 0.3xy + 0.5y

Answer:

Definitions:

A term in an algebraic expression can be a constant, a variable or a product of constants and variables separated by the signs of addition (+) or subtraction ( $-$ ) . Examples: 27, x, xyz, $\frac{1}{2} x^{2} y z$ etc.
The number factor of the term is called its coefficient.

(i) The expression $7 x^{2} y z - 5 x y$ consists of two terms, i.e., $7 x^{2} y z$ and $- 5 x y$ .
The coefficient of $7 x^{2} y z$ is 7 and the coefficient of $- 5 x y$ is $- 5$ .

(ii) The expression $x^{2} + x + 1$ consists of three terms , i.e., $x^{2}$ , x and 1.
The coefficient of each term is 1.

(iii) The expression $3 x^{2} y^{2} - 5 x^{2} y^{2} z^{2} + z^{2}$ consists of three terms , i.e., $3 x^{2} y^{2}$ , $- 5 x^{2} y^{2} z^{2}$ and $z^{2}$ . The coefficient of $3 x^{2} y^{2}$ is 3. The coefficient of $- 5 x^{2} y^{2} z^{2}$ is $- 5$ and the coefficient of $z^{2}$ is 1.
(iv) The expression $9 - a b + b c - c a$ consists of four terms — $9, - a b, b c and - c a$ . The coefficient of the term 9 is 9. The coefficient of $- a b$ is $-$ 1. The coefficient of $b c$ is 1, and the coefficient of $- c a$ is $-$ 1.

(v) The expression $\frac{a}{2} + \frac{b}{2} - a b$ consists of three terms , i.e., $\frac{a}{2}, \frac{b}{2} and - a b$ . The coefficient of $\frac{a}{2}$ is $\frac{1}{2}$ . The coefficient of $\frac{b}{2}$ is $\frac{1}{2}$ , and the coefficient of $- a b$ is $-$ 1.

(vi) The expression $0.2 x - 0.3 x y + 0.5 y$ consists of three terms , i.e., $0.2 x, - 0.3 x y and 0.5 y .$ The coefficient of $0.2 x$ is $0.2$ . The coefficient of $- 0.3 x y$ is $- 0.3$ , and the coefficient of $0.5 y$ is $0.5$ .

Page No 6.2:

Question 2:

Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any category?
(i) x + y
(ii) 1000
(iii) x + x² + x³ + 4y⁴
(iv) 7 + a + 5b
(v) 2b − 3b²
(vi) 2y − 3y² + 4y³
(vii) 5x − 4y + 3x
(viii) 4a − 15a²
(ix) xy + yz + zt + tx
(x) pqr
(xi) p²q + pq²
(xii) 2p + 2q

Answer:

Definitions:

A polynomial is monomial if it has exactly one term. It is called binomial if it has exactly two non-zero terms. A polynomial is a trinomial if it has exactly three non-zero terms.

(i) The polynomial $x + y$ has exactly two non zero terms , i.e., x and y. Therefore, it is a binomial.
(ii) The polynomial 1000 has exactly one term, i.e., 1000. Therefore, it is a monomial.
(iii) The polynomial $x + x^{2} + x^{3} + x^{4}$ has exactly four terms, i.e., $x, x^{2}, x^{3} and x^{4}$ . Therefore, it doesn't belong to any of the categories.
(iv) The polynomial $7 + a + 5 b$ has exactly three terms, i.e., 7, a and 5b. Therefore, it is a trinomial.
(v) The polynomial $2 b - 3 b^{2}$ has exactly two terms, i.e., 2b and $- 3 b^{2}$ . Therefore, it is a binomial.
(vi) The polynomial $2 y - 3 y^{2} + 4 y^{3}$ has exactly three terms, i.e., 2y, $- 3 y^{2}$ and $4 y^{3}$ . Therefore, it is a trinomial.
(vii) The polynomial $5 x - 4 y + 3 x$ has exactly three terms, i.e., 5x, $-$ 4y and 3x. Therefore, it is a trinomial.
(viii) The polynomial $4 a - 15 a^{2}$ has exactly two terms, i.e., 4a and $- 15 a^{2}$ . Therefore, it is a binomial.
(ix) The polynomial $x y + y z + z t + t x$ has exactly four terms xy, yz, zt and tx. Therefore, it doesn't belong to any of the categories.
(x) The polynomial pqr has exactly one term, i.e., pqr. Therefore, it is a monomial.
(xi) The polynomial $p^{2} q + p q^{2}$ has exactly two terms, i.e., $p^{2} q$ and $p q^{2}$ . Therefore, it is a binomial.
(xi) The polynomial $2 p + 2 q$ has two terms, i.e., 2p and 2q. Therefore, it is a binomial.

Page No 6.21:

Question 1:

Find the following product:
2a³(3a + 5b)

Answer:

To find the product, we will use distributive law as follows:

$2 a^{3} (3 a + 5 b) = 2 a^{3} \times 3 a + 2 a^{3} \times 5 b = (2 \times 3) (a^{3} \times a) + (2 \times 5) a^{3} b = (2 \times 3) a^{3 + 1} + (2 \times 5) a^{3} b = 6 a^{4} + 10 a^{3} b$

Thus, the answer is $6 a^{4} + 10 a^{3} b$ .

Page No 6.21:

Question 2:

Find the following product:
−11a(3a + 2b)

Answer:

To find the product, we will use distributive law as follows:

$- 11 a (3 a + 2 b) = (- 11 a) \times 3 a + (- 11 a) \times 2 b = (- 11 \times 3) \times (a \times a) + (- 11 \times 2) \times (a \times b) = (- 33) \times (a^{1 + 1}) + (- 22) \times (a \times b) = - 33 a^{2} - 22 a b$

Thus, the answer is $- 33 a^{2} - 22 a b$ .

Page No 6.21:

Question 3:

Find the following product:
−5a(7a − 2b)

Answer:

To find the product, we will use distributive law as follows:

$- 5 a (7 a - 2 b) = (- 5 a) \times 7 a + (- 5 a) \times (- 2 b) = (- 5 \times 7) \times (a \times a) + (- 5 \times (- 2)) \times (a \times b) = (- 35) \times (a^{1 + 1}) + (10) \times (a \times b) = - 35 a^{2} + 10 a b$

Thus, the answer is $- 35 a^{2} + 10 a b$ .

Page No 6.21:

Question 4:

Find the following product:
−11y²(3y + 7)

Answer:

To find the product, we will use distributive law as follows:

$- 11 y^{2} (3 y + 7) = (- 11 y^{2}) \times 3 y + (- 11 y^{2}) \times 7 = (- 11 \times 3) (y^{2} \times y) + (- 11 \times 7) \times (y^{2}) = (- 33) (y^{2 + 1}) + (- 77) \times (y^{2}) = - 33 y^{3} - 77 y^{2}$

Thus, the answer is $- 33 y^{3} - 77 y^{2}$ .

Page No 6.21:

Question 5:

Find the following product:
$\frac{6 x}{5} (x^{3} + y^{3})$

Answer:

To find the product, we will use distributive law as follows:

$\frac{6 x}{5} (x^{3} + y^{3}) = \frac{6 x}{5} \times x^{3} + \frac{6 x}{5} \times y^{3} = \frac{6}{5} \times (x \times x^{3}) + \frac{6}{5} \times (x \times y^{3}) = \frac{6}{5} \times (x^{1 + 3}) + \frac{6}{5} \times (x \times y^{3}) = \frac{6 x^{4}}{5} + \frac{6 x y^{3}}{5}$

Thus, the answer is $\frac{6 x^{4}}{5} + \frac{6 x y^{3}}{5}$ .

Page No 6.21:

Question 6:

xy(x³ − y³)

Answer:

To find the product, we will use the distributive law in the following way:

$x y (x^{3} - y^{3}) = x y \times x^{3} - x y \times y^{3} = (x \times x^{3}) \times y - x \times (y \times y^{3}) = x^{1 + 3} y - x y^{1 + 3} = x^{4} y - x y^{4}$

Thus, the answer is $x^{4} y - x y^{4}$ .

Page No 6.21:

Question 7:

Find the following product:
0.1y(0.1x⁵ + 0.1y)

Answer:

To find the product, we will use distributive law as follows:

$0.1 y (0.1 x^{5} + 0.1 y) = (0.1 y) (0.1 x^{5}) + (0.1 y) (0.1 y) = (0.1 \times 0.1) (y \times x^{5}) + (0.1 \times 0.1) (y \times y) = (0.1 \times 0.1) (x^{5} \times y) + (0.1 \times 0.1) (y^{1 + 1}) = 0.01 x^{5} y + 0.01 y^{2}$

Thus, the answer is $0.01 x^{5} y + 0.01 y^{2}$ .

Page No 6.21:

Question 8:

Find the following product:
$(- \frac{7}{4} a b^{2} c - \frac{6}{25} a^{2} c^{2}) (- 50 a^{2} b^{2} c^{2})$

Answer:

To find the product, we will use distributive law as follows:

$(- \frac{7}{4} a b^{2} c - \frac{6}{25} a^{2} c^{2}) (- 50 a^{2} b^{2} c^{2}) = \{(- \frac{7}{4} a b^{2} c) (- 50 a^{2} b^{2} c^{2})\} - \{(\frac{6}{25} a^{2} c^{2}) (- 50 a^{2} b^{2} c^{2})\} = \{\{- \frac{7}{4} \times (- 50)\} (a \times a^{2}) \times (b^{2} \times b^{2}) \times (c \times c^{2})\} - \{(\frac{6}{25}) (- 50) (a^{2} \times a^{2}) \times (b^{2}) \times (c^{2} \times c^{2})\} = \{- \frac{7}{4} \times (- 50)\} (a^{1 + 2} b^{2 + 2} c^{1 + 2}) - \{(\frac{6}{25}) (- 50) (a^{2 + 2} b^{2} c^{2 + 2})\} = \frac{175}{2} a^{3} b^{4} c^{3} - (- 12 a^{4} b^{2} c^{4}) = \frac{175}{2} a^{3} b^{4} c^{3} + 12 a^{4} b^{2} c^{4}$

Thus, the answer is $\frac{175}{2} a^{3} b^{4} c^{3} + 12 a^{4} b^{2} c^{4}$ .

Page No 6.21:

Question 9:

Find the following product:
$- \frac{8}{27} x y z (\frac{3}{2} x y z^{2} - \frac{9}{4} x y^{2} z^{3})$

Answer:

To find the product, we will use the distributive law in the following way:

$- \frac{8}{27} x y z (\frac{3}{2} x y z^{2} - \frac{9}{4} x y^{2} z^{3}) = \{(- \frac{8}{27} x y z) (\frac{3}{2} x y z^{2})\} - \{(- \frac{8}{27} x y z) (\frac{9}{4} x y^{2} z^{3})\} = \{(- \frac{8}{27} \times \frac{3}{2}) (x \times x) \times (y \times y) \times (z \times z^{2})\} - \{(- \frac{8}{27} \times \frac{9}{4}) (x \times x) \times (y \times y^{2}) \times (z \times z^{3})\} = \{(- \frac{8}{27} \times \frac{3}{2}) (x^{1 + 1} y^{1 + 1} z^{1 + 2})\} - \{(- \frac{8}{27} \times \frac{9}{4}) (x^{1 + 1} y^{1 + 2} z^{1 + 3})\} = \{(- \frac{8^{4}}{27_{9}} \times \frac{3}{2}) (x^{1 + 1} y^{1 + 1} z^{1 + 2})\} - \{(- \frac{8^{2}}{27_{3}} \times \frac{9}{4}) (x^{1 + 1} y^{1 + 2} z^{1 + 3})\} = - \frac{4}{9} x^{2} y^{2} z^{3} + \frac{2}{3} x^{2} y^{3} z^{4}$

Thus, the answer is $- \frac{4}{9} x^{2} y^{2} z^{3} + \frac{2}{3} x^{2} y^{3} z^{4}$ .

Page No 6.21:

Question 10:

Find the following product:
$- \frac{4}{27} x y z (\frac{9}{2} x^{2} y z - \frac{3}{4} x y z^{2})$

Answer:

To find the product, we will use distributive law as follows:

$- \frac{4}{27} x y z (\frac{9}{2} x^{2} y z - \frac{3}{4} x y z^{2}) = \{(- \frac{4}{27} x y z) (\frac{9}{2} x^{2} y z)\} - \{(- \frac{4}{27} x y z) (\frac{3}{4} x y z^{2})\} = \{(- \frac{4}{27} \times \frac{9}{2}) (x^{1 + 2} y^{1 + 1} z^{1 + 1})\} - \{(- \frac{4}{27} \times \frac{3}{4}) (x^{1 + 1} y^{1 + 1} z^{1 + 2})\} = \{(- \frac{4^{2}}{27_{3}} \times \frac{9}{2}) (x^{1 + 2} y^{1 + 1} z^{1 + 1})\} - \{(- \frac{4^{1}}{27_{9}} \times \frac{3}{4}) (x^{1 + 1} y^{1 + 1} z^{1 + 2})\} = - \frac{2}{3} x^{3} y^{2} z^{2} + \frac{1}{9} x^{2} y^{2} z^{3}$

Thus, the answer is $- \frac{2}{3} x^{3} y^{2} z^{2} + \frac{1}{9} x^{2} y^{2} z^{3}$ .

Page No 6.21:

Question 11:

Find the following product:
1.5x(10x²y − 100xy²)

Answer:

To find the product, we will use distributive law as follows:

$1.5 x (10 x^{2} y - 100 x y^{2}) = (1.5 x \times 10 x^{2} y) - (1.5 x \times 100 x y^{2}) = (15 x^{1 + 2} y) - (150 x^{1 + 1} y^{2}) = 15 x^{3} y - 150 x^{2} y^{2}$

Thus, the answer is $15 x^{3} y - 150 x^{2} y^{2}$ .

Page No 6.21:

Question 12:

Find the following product:
4.1xy(1.1x − y)

Answer:

To find the product, we will use distributive law as follows:

$4.1 x y (1.1 x - y) = (4.1 x y \times 1.1 x) - (4.1 x y \times y) = \{(4.1 \times 1.1) \times x y \times x\} - (4.1 x y \times y) = (4.51 x^{1 + 1} y) - (4.1 x y^{1 + 1}) = 4.51 x^{2} y - 4.1 x y^{2}$

Thus, the answer is $4.51 x^{2} y - 4.1 x y^{2}$ .

Page No 6.21:

Question 13:

Find the following product:
250.5xy $(x z + \frac{y}{10})$

Answer:

To find the product, we will use distributive law as follows:

$250.5 x y (x z + \frac{y}{10}) = 250.5 x y \times x z + 250.5 x y \times \frac{y}{10} = 250.5 x^{1 + 1} y z + 25.05 x y^{1 + 1} = 250.5 x^{2} y z + 25.05 x y^{2}$

Thus, the answer is $250.5 x^{2} y z + 25.05 x y^{2}$ .

Page No 6.21:

Question 14:

Find the following product:
$\frac{7}{5} x^{2} y (\frac{3}{5} x y^{2} + \frac{2}{5} x)$

Answer:

To find the product, we will use distributive law as follows:

$\frac{7}{5} x^{2} y (\frac{3}{5} x y^{2} + \frac{2}{5} x) = \frac{7}{5} x^{2} y \times \frac{3}{5} x y^{2} + \frac{7}{5} x^{2} y \times \frac{2}{5} x = \frac{21}{25} x^{2 + 1} y^{1 + 2} + \frac{14}{25} x^{2 + 1} y = \frac{21}{25} x^{3} y^{3} + \frac{14}{25} x^{3} y$

Thus, the answer is $\frac{21}{25} x^{3} y^{3} + \frac{14}{25} x^{3} y$ .

Page No 6.21:

Question 15:

Find the following product:
$\frac{4}{3} a (a^{2} + b^{2} - 3 c^{2})$

Answer:

To find the product, we will use distributive law as follows:

$\frac{4}{3} a (a^{2} + b^{2} - 3 c^{2}) = \frac{4}{3} a \times a^{2} + \frac{4}{3} a \times b^{2} - \frac{4}{3} a \times 3 c^{2} = \frac{4}{3} a^{1 + 2} + \frac{4}{3} a b^{2} - 4 a c^{2} = \frac{4}{3} a^{3} + \frac{4}{3} a b^{2} - 4 a c^{2}$

Thus, the answer is $\frac{4}{3} a^{3} + \frac{4}{3} a b^{2} - 4 a c^{2}$ .

Page No 6.21:

Question 16:

Find the product 24x² (1 − 2x) and evaluate its value for x = 3.

Answer:

To find the product, we will use distributive law as follows:

$24 x^{2} (1 - 2 x) = 24 x^{2} \times 1 - 24 x^{2} \times 2 x = 24 x^{2} - 48 x^{1 + 2} = 24 x^{2} - 48 x^{3}$

Substituting x = 3 in the result, we get:

$24 x^{2} - 48 x^{3} = 24 {(3)}^{2} - 48 {(3)}^{3} = 24 \times 9 - 48 \times 27 = 216 - 1296 = - 1080$

Thus, the product is $(24 x^{2} - 48 x^{3}) and its value fo r x = 3 is (- 1080)$ .

Page No 6.21:

Question 17:

Find the product −3y(xy + y²) and find its value for x = 4 and y = 5.

Answer:

To find the product, we will use distributive law as follows:

$- 3 y (x y + y^{2}) = - 3 y \times x y + (- 3 y) \times y^{2} = - 3 x y^{1 + 1} - 3 y^{1 + 2} = - 3 x y^{2} - 3 y^{3}$

Substituting x = 4 and y = 5 in the result, we get:

$- 3 x y^{2} - 3 y^{3} = - 3 (4) {(5)}^{2} - 3 {(5)}^{3} = - 3 (4) (25) - 3 (125) = - 300 - 375 = - 675$

Thus, the product is ( $- 3 x y^{2} - 3 y^{3}$ ), and its value for x = 4 and y = 5 is ( $-$ 675).

Page No 6.21:

Question 18:

Multiply $- \frac{3}{2} x^{2} y^{3} by (2 x - y)$ and verify the answer for x = 1 and y = 2.

Answer:

To find the product, we will use distributive law as follows:

$- \frac{3}{2} x^{2} y^{3} \times (2 x - y) = (- \frac{3}{2} x^{2} y^{3} \times 2 x) - (- \frac{3}{2} x^{2} y^{3} \times y) = (- 3 x^{2 + 1} y^{3}) - (- \frac{3}{2} x^{2} y^{3 + 1}) = - 3 x^{3} y^{3} + \frac{3}{2} x^{2} y^{4}$

Substituting x = 1 and y = 2 in the result, we get:

$- 3 x^{3} y^{3} + \frac{3}{2} x^{2} y^{4} = - 3 {(1)}^{3} {(2)}^{3} + \frac{3}{2} {(1)}^{2} {(2)}^{4} = - 3 \times 1 \times 8 + \frac{3}{2} \times 1 \times 16 = - 24 + 24 = 0$

Thus, the product is $- 3 x^{3} y^{3} + \frac{3}{2} x^{2} y^{4}$ , and its value for x = 1 and y = 2 is 0.

Page No 6.21:

Question 19:

Multiply the monomial by the binomial and find the value of each for x = −1, y = 0.25 and z = 0.05:
(i) 15y²(2 − 3x)
(ii) −3x(y² + z²)
(iii) z²(x − y)
(iv) xz(x² + y²)

Answer:

(i) To find the product, we will use distributive law as follows:

$15 y^{2} (2 - 3 x) = 15 y^{2} \times 2 - 15 y^{2} \times 3 x = 30 y^{2} - 45 x y^{2}$

Substituting x = $-$ 1 and y = 0.25 in the result, we get:

$30 y^{2} - 45 x y^{2} = 30 {(0.25)}^{2} - 45 (- 1) {(0.25)}^{2} = 30 \times 0.0625 - \{45 \times (- 1) \times 0.0625\} = 30 \times 0.0625 - \{45 \times (- 1) \times 0.0625\} = 1.875 - (- 2.8125) = 1.875 + 2.8125 = 4.6875$

(ii) To find the product, we will use distributive law as follows:

$- 3 x (y^{2} + z^{2}) = - 3 x \times y^{2} + (- 3 x) \times z^{2} = - 3 x y^{2} - 3 x z^{2}$

Substituting x = $-$ 1, y = 0.25 and z = 0.05 in the result, we get:

$- 3 x y^{2} - 3 x z^{2} = - 3 (- 1) {(0.25)}^{2} - 3 (- 1) {(0.05)}^{2} = - 3 (- 1) (0.0625) - 3 (- 1) (0.0025) = 01875 + 0.0075 = 0.195$

(iii) To find the product, we will use distributive law as follows:

$z^{2} (x - y) = z^{2} \times x - z^{2} \times y = x z^{2} - y z^{2}$

Substituting x = $-$ 1, y = 0.25 and z = 0.05 in the result, we get:

$x z^{2} - y z^{2} = (- 1) {(0.05)}^{2} - (0.25) {(0.05)}^{2} = (- 1) (0.0025) - (0.25) (0.0025) = - 0.0025 - 0.000625 = - 0.003125$

(iv) To find the product, we will use distributive law as follows:

$x z (x^{2} + y^{2}) = x z \times x^{2} + x z \times y^{2} = x^{3} z + x y^{2} z$

Substituting x = $-$ 1, y = 0.25 and z = 0.05 in the result, we get:

$x^{3} z + x y^{2} z = {(- 1)}^{3} (0.05) + (- 1) {(0.25)}^{2} (0.05) = (- 1) (0.05) + (- 1) (0.0625) (0.05) = - 0.05 - 0.003125 = - 0.053125$

Page No 6.21:

Question 20:

Simplify:
(i) 2x²(x³ − x) − 3x(x⁴ + 2x) − 2(x⁴ − 3x²)
(ii) x³y(x² − 2x) + 2xy(x³ − x⁴)
(iii) 3a² + 2(a + 2) − 3a(2a + 1)
(iv) x(x + 4) + 3x(2x² − 1) + 4x² + 4
(v) a(b − c) − b(c − a) − c(a − b)
(vi) a(b − c) + b(c − a) + c(a − b)
(vii) 4ab(a − b) − 6a²(b − b²) − 3b²(2a² − a) + 2ab(b − a)
(viii) x²(x² + 1) − x³(x + 1) − x(x³ − x)
(ix) 2a² + 3a(1 − 2a³) + a(a + 1)
(x) a²(2a − 1) + 3a + a³ − 8
(xi) $\frac{3}{2} x^{2} (x^{2} - 1) + \frac{1}{4} x^{2} (x^{2} + x) - \frac{3}{4} x (x^{3} - 1)$
(xii) a²b(a − b²⁾ + ab²(4ab − 2a²) − a³b(1 − 2b)
(xiii) a²b(a³ − a + 1) − ab(a⁴ − 2a² + 2a) − b (a³ − a² − 1)

Answer:

(i) To simplify, we will use distributive law as follows:

$2 x^{2} (x^{3} - x) - 3 x (x^{4} + 2 x) - 2 (x^{4} - 3 x^{2}) = 2 x^{5} - 2 x^{3} - 3 x^{5} - 6 x^{2} - 2 x^{4} + 6 x^{2} = 2 x^{5} - 3 x^{5} - 2 x^{4} - 2 x^{3} - 6 x^{2} + 6 x^{2} = - x^{5} - 2 x^{4} - 2 x^{3}$

(ii) To simplify, we will use distributive law as follows:

$x^{3} y (x^{2} - 2 x) + 2 x y (x^{3} - x^{4}) = x^{5} y - 2 x^{4} y + 2 x^{4} y - 2 x^{5} y = x^{5} y - 2 x^{5} y - 2 x^{4} y + 2 x^{4} y = - x^{5} y$

(iii) To simplify, we will use distributive law as follows:

$3 a^{2} + 2 (a + 2) - 3 a (2 a + 1) = 3 a^{2} + 2 a + 4 - 6 a^{2} - 3 a = 3 a^{2} - 6 a^{2} + 2 a - 3 a + 4 = - 3 a^{2} - a + 4$

(iv) To simplify, we will use distributive law as follows:

$x (x + 4) + 3 x (2 x^{2} - 1) + 4 x^{2} + 4 = x^{2} + 4 x + 6 x^{3} - 3 x + 4 x^{2} + 4 = x^{2} + 4 x^{2} + 4 x - 3 x + 6 x^{3} + 4 = 5 x^{2} + x + 6 x^{3} + 4$

(v) To simplify, we will use distributive law as follows:

$a (b - c) - b (c - a) - c (a - b) = a b - a c - b c + b a - c a + c b = a b + b a - a c - c a - b c + c b = 2 a b - 2 a c$

(vi) To simplify, we will use distributive law as follows:

$a (b - c) + b (c - a) + c (a - b) = a b - a c + b c - b a + c a - c b = a b - b a - a c + c a + b c - c b = 0$

(vii) To simplify, we will use distributive law as follows:

$4 a b (a - b) - 6 a^{2} (b - b^{2}) - 3 b^{2} (2 a^{2} - a) + 2 a b (b - a) = 4 a^{2} b - 4 a b^{2} - 6 a^{2} b + 6 a^{2} b^{2} - 6 b^{2} a^{2} + 3 b^{2} a + 2 a b^{2} - 2 a^{2} b = 4 a^{2} b - 6 a^{2} b - 2 a^{2} b - 4 a b^{2} + 3 b^{2} a + 2 a b^{2} + 6 a^{2} b^{2} - 6 b^{2} a^{2} = - 4 a^{2} b + a b^{2}$

(viii) To simplify, we will use distributive law as follows:

$x^{2} (x^{2} + 1) - x^{3} (x + 1) - x (x^{3} - x) = x^{4} + x^{2} - x^{4} - x^{3} - x^{4} + x^{2} = x^{4} - x^{4} - x^{4} - x^{3} + x^{2} + x^{2} = - x^{4} - x^{3} + 2 x^{2}$

(ix) To simplify, we will use distributive law as follows:

$2 a^{2} + 3 a (1 - 2 a^{3}) + a (a + 1) = 2 a^{2} + 3 a - 6 a^{4} + a^{2} + a = 2 a^{2} + a^{2} + 3 a + a - 6 a^{4} = 3 a^{2} + 4 a - 6 a^{4}$

(x) To simplify, we will use distributive law as follows:

$a^{2} (2 a - 1) + 3 a + a^{3} - 8 = 2 a^{3} - a^{2} + 3 a + a^{3} - 8 = 2 a^{3} + a^{3} - a^{2} + 3 a - 8 = 3 a^{3} - a^{2} + 3 a - 8$

(xi) To simplify, we will use distributive law as follows:

$\frac{3}{2} x^{2} (x^{2} - 1) + \frac{1}{4} x^{2} (x^{2} + x) - \frac{3}{4} x (x^{3} - 1) = \frac{3}{2} x^{4} - \frac{3}{2} x^{2} + \frac{1}{4} x^{4} + \frac{1}{4} x^{3} - \frac{3}{4} x^{4} + \frac{3}{4} x = \frac{3}{2} x^{4} + \frac{1}{4} x^{4} - \frac{3}{4} x^{4} + \frac{1}{4} x^{3} - \frac{3}{2} x^{2} + \frac{3}{4} x = (\frac{6 + 1 - 3}{4}) x^{4} + \frac{1}{4} x^{3} - \frac{3}{2} x^{2} + \frac{3}{4} x = x^{4} + \frac{1}{4} x^{3} - \frac{3}{2} x^{2} + \frac{3}{4} x$

(xii) To simplify, we will use distributive law as follows:

$a^{2} b (a - b^{2}) + a b^{2} (4 a b - 2 a^{2}) - a^{3} b (1 - 2 b) = a^{3} b - a^{2} b^{3} + 4 a^{2} b^{3} - 2 a^{3} b^{2} - a^{3} b + 2 a^{3} b^{2} = a^{3} b - a^{3} b - a^{2} b^{3} + 4 a^{2} b^{3} - 2 a^{3} b^{2} + 2 a^{3} b^{2} = 3 a^{2} b^{3}$

(xiii) To simplify, we will use distributive law as follows:

$a^{2} b (a^{3} - a + 1) - a b (a^{4} - 2 a^{2} + 2 a) - b (a^{3} - a^{2} - 1) = a^{5} b - a^{3} b + a^{2} b - a^{5} b + 2 a^{3} b - 2 a^{2} b - a^{3} b + a^{2} b + b = a^{5} b - a^{5} b - a^{3} b + 2 a^{3} b - a^{3} b + a^{2} b - 2 a^{2} b + a^{2} b + b = b$

Page No 6.30:

Question 1:

Multiply:
(5x + 3) by (7x + 2)

Answer:

To multiply, we will use distributive law as follows:

$(5 x + 3) (7 x + 2) = 5 x (7 x + 2) + 3 (7 x + 2) = (5 x \times 7 x + 5 x \times 2) + (3 \times 7 x + 3 \times 2) = (35 x^{2} + 10 x) + (21 x + 6) = 35 x^{2} + 10 x + 21 x + 6 = 35 x^{2} + 31 x + 6$

Thus, the answer is $35 x^{2} + 31 x + 6$ .

Page No 6.30:

Question 2:

Multiply:
(2x + 8) by (x − 3)

Answer:

To multiply the expressions, we will use the distributive law in the following way:

$(2 x + 8) (x - 3) = 2 x (x - 3) + 8 (x - 3) = (2 x \times x - 2 x \times 3) + (8 x - 8 \times 3) = (2 x^{2} - 6 x) + (8 x - 24) = 2 x^{2} - 6 x + 8 x - 24 = 2 x^{2} + 2 x - 24$

Thus, the answer is $2 x^{2} + 2 x - 24$ .

Page No 6.30:

Question 3:

Multiply:
(7x + y) by (x + 5y)

Answer:

To multiply, we will use distributive law as follows:

$(7 x + y) (x + 5 y) = 7 x (x + 5 y) + y (x + 5 y) = 7 x^{2} + 35 x y + x y + 5 y^{2} = 7 x^{2} + 36 x y + 5 y^{2}$

Thus, the answer is $7 x^{2} + 36 x y + 5 y^{2}$ .

Page No 6.30:

Question 4:

Multiply:
(a − 1) by (0.1a² + 3)

Answer:

To multiply, we will use distributive law as follows:

$(a - 1) (0.1 a^{2} + 3) = 0.1 a^{2} (a - 1) + 3 (a - 1) = 0.1 a^{3} - 0.1 a^{2} + 3 a - 3$

Thus, the answer is $0.1 a^{3} - 0.1 a^{2} + 3 a - 3$ .

Page No 6.30:

Question 5:

Multiply:
(3x² + y²) by (2x² + 3y²)

Answer:

To multiply, we will use distributive law as follows:

$(3 x^{2} + y^{2}) (2 x^{2} + 3 y^{2}) = 3 x^{2} (2 x^{2} + 3 y^{2}) + y^{2} (2 x^{2} + 3 y^{2}) = 6 x^{4} + 9 x^{2} y^{2} + 2 x^{2} y^{2} + 3 y^{4} = 6 x^{4} + 11 x^{2} y^{2} + 3 y^{4}$

Thus, the answer is $6 x^{4} + 11 x^{2} y^{2} + 3 y^{4}$ .

Page No 6.30:

Question 6:

Multiply:
$(\frac{3}{5} x + \frac{1}{2} y) by (\frac{5}{6} x + 4 y)$

Answer:

To multiply, we will use distributive law as follows:

$(\frac{3}{5} x + \frac{1}{2} y) (\frac{5}{6} x + 4 y) = \frac{3}{5} x (\frac{5}{6} x + 4 y) + \frac{1}{2} y (\frac{5}{6} x + 4 y) = \frac{1}{2} x^{2} + \frac{12}{5} x y + \frac{5}{12} x y + 2 y^{2} = \frac{1}{2} x^{2} + (\frac{144 + 25}{60}) x y + 2 y^{2} = \frac{1}{2} x^{2} + \frac{169}{60} x y + 2 y^{2}$

Thus, the answer is $\frac{1}{2} x^{2} + \frac{169}{60} x y + 2 y^{2}$ .

Page No 6.31:

Question 7:

Multiply:
(x⁶ − y⁶) by (x²+ y²)

Answer:

To multiply, we will use distributive law as follows:

$(x^{6} - y^{6}) (x^{2} + y^{2}) = x^{6} (x^{2} + y^{2}) - y^{6} (x^{2} + y^{2}) = (x^{8} + x^{6} y^{2}) - (y^{6} x^{2} + y^{8}) = x^{8} + x^{6} y^{2} - y^{6} x^{2} - y^{8}$

Thus, the answer is $x^{8} + x^{6} y^{2} - y^{6} x^{2} - y^{8}$ .

Page No 6.31:

Question 8:

Multiply:
(x² + y²) by (3a + 2b)

Answer:

To multiply, we will use distributive law as follows:

$(x^{2} + y^{2}) (3 a + 2 b) = x^{2} (3 a + 2 b) + y^{2} (3 a + 2 b) = 3 a x^{2} + 2 b x^{2} + 3 a y^{2} + 2 b y^{2}$

Thus, the answer is $3 a x^{2} + 2 b x^{2} + 3 a y^{2} + 2 b y^{2}$ .

Page No 6.31:

Question 9:

Multiply:
[−3d + (−7f)] by (5d + f)

Answer:

To multiply, we will use distributive law as follows:

$[- 3 d + (- 7 f)] (5 d + f) = (- 3 d) (5 d + f) + (- 7 f) (5 d + f) = (- 15 d^{2} - 3 d f) + (- 35 d f - 7 f^{2}) = - 15 d^{2} - 3 d f - 35 d f - 7 f^{2} = - 15 d^{2} - 38 d f - 7 f^{2}$

Thus, the answer is $- 15 d^{2} - 38 d f - 7 f^{2}$ .

Page No 6.31:

Question 10:

Multiply:
(0.8a − 0.5b) by (1.5a − 3b)

Answer:

To multiply, we will use distributive law as follows:

$(0.8 a - 0.5 b) (1.5 a - 3 b) = 0.8 a (1.5 a - 3 b) - 0.5 b (1.5 a - 3 b) = 1.2 a^{2} - 2.4 a b - 0.75 a b + 1.5 b^{2} = 1.2 a^{2} - 3.15 a b + 1.5 b^{2}$

Thus, the answer is $1.2 a^{2} - 3.15 a b + 1.5 b^{2}$ .

Page No 6.31:

Question 11:

Multiply:
(2x²y² − 5xy²) by (x² − y²)

Answer:

To multiply, we will use distributive law as follows:

$(2 x^{2} y^{2} - 5 x y^{2}) (x^{2} - y^{2}) = 2 x^{2} y^{2} (x^{2} - y^{2}) - 5 x y^{2} (x^{2} - y^{2}) = 2 x^{4} y^{2} - 2 x^{2} y^{4} - 5 x^{3} y^{2} + 5 x y^{4}$

Thus, the answer is $2 x^{4} y^{2} - 2 x^{2} y^{4} - 5 x^{3} y^{2} + 5 x y^{4}$ .

Page No 6.31:

Question 12:

Multiply:
$(\frac{x}{7} + \frac{x^{2}}{2}) b y (\frac{2}{5} + \frac{9 x}{4})$

Answer:

To multiply the expressions, we will use the distributive law in the following way:

$(\frac{x}{7} + \frac{x^{2}}{2}) (\frac{2}{5} + \frac{9 x}{4}) = \frac{x}{7} (\frac{2}{5} + \frac{9 x}{4}) + \frac{x^{2}}{2} (\frac{2}{5} + \frac{9 x}{4}) = \frac{2 x}{35} + \frac{9 x^{2}}{28} + \frac{x^{2}}{5} + \frac{9 x^{3}}{8} = \frac{2 x}{35} + (\frac{45 + 28}{140}) x^{2} + \frac{9 x^{3}}{8} = \frac{2 x}{35} + \frac{73 x^{2}}{140} + \frac{9 x^{2}}{8}$
Thus, the answer is $\frac{2 x}{35} + \frac{73 x^{2}}{140} + \frac{9 x^{3}}{8}$

Page No 6.31:

Question 13:

Multiply:
$(- \frac{a}{7} + \frac{a^{2}}{9}) b y (\frac{b}{2} - \frac{b^{2}}{3})$

Answer:

To multiply, we will use distributive law as follows:

$(- \frac{a}{7} + \frac{a^{2}}{9}) (\frac{b}{2} - \frac{b^{2}}{3}) = (- \frac{a}{7}) (\frac{b}{2} - \frac{b^{2}}{3}) + (\frac{a^{2}}{9}) (\frac{b}{2} - \frac{b^{2}}{3}) = (- \frac{a b}{14} + \frac{a b^{2}}{21}) + (\frac{a^{2} b}{18} - \frac{a^{2} b^{2}}{27}) = - \frac{a b}{14} + \frac{a b^{2}}{21} + \frac{a^{2} b}{18} - \frac{a^{2} b^{2}}{27}$

Thus, the answer is $- \frac{a b}{14} + \frac{a b^{2}}{21} + \frac{a^{2} b}{18} - \frac{a^{2} b^{2}}{27}$ .

Page No 6.31:

Question 14:

Multiply:
(3x²y − 5xy²) by $(\frac{1}{5} x^{2} + \frac{1}{3} y^{2})$

Answer:

To multiply, we will use distributive law as follows:

$(3 x^{2} y - 5 x y^{2}) (\frac{1}{5} x^{2} + \frac{1}{3} y^{2}) = \frac{1}{5} x^{2} (3 x^{2} y - 5 x y^{2}) + \frac{1}{3} y^{2} (3 x^{2} y - 5 x y^{2}) = \frac{3}{5} x^{4} y - x^{3} y^{2} + x^{2} y^{3} - \frac{5}{3} x y^{4}$

Thus, the answer is $\frac{3}{5} x^{4} y - x^{3} y^{2} + x^{2} y^{3} - \frac{5}{3} x y^{4}$ .

Page No 6.31:

Question 15:

Multiply:
(2x² − 1) by (4x³ + 5x²)

Answer:

To multiply, we will use distributive law as follows:

$(2 x^{2} - 1) (4 x^{3} + 5 x^{2}) = 2 x^{2} (4 x^{3} + 5 x^{2}) - 1 (4 x^{3} + 5 x^{2}) = 8 x^{5} + 10 x^{4} - 4 x^{3} - 5 x^{2}$

Thus, the answer is $8 x^{5} + 10 x^{4} - 4 x^{3} - 5 x^{2}$ .

Page No 6.31:

Question 16:

(2xy + 3y²) (3y² − 2)

Answer:

To multiply, we will use distributive law as follows:

$(2 x y + 3 y^{2}) (3 y^{2} - 2) = 2 x y (3 y^{2} - 2) + 3 y^{2} (3 y^{2} - 2) = 6 x y^{3} - 4 x y + 9 y^{4} - 6 y^{2} = 9 y^{4} + 6 x y^{3} - 6 y^{2} - 4 x y$

Thus, the answer is $9 y^{4} + 6 x y^{3} - 6 y^{2} - 4 x y$ .

Page No 6.31:

Question 17:

Find the following product and verify the result for x = − 1, y = − 2:
(3x − 5y) (x + y)

Answer:

To multiply, we will use distributive law as follows:

$(3 x - 5 y) (x + y) = 3 x (x + y) - 5 y (x + y) = 3 x^{2} + 3 x y - 5 x y - 5 y^{2} = 3 x^{2} - 2 x y - 5 y^{2}$

$∴$ $(3 x - 5 y) (x + y) = 3 x^{2} - 2 x y - 5 y^{2}$ .
Now, we put x = $-$ 1 and y = $-$ 2 on both sides to verify the result.

$LHS = (3 x - 5 y) (x + y) = \{3 (- 1) - 5 (- 2)\} \{- 1 + (- 2)\} = (- 3 + 10) (- 3) = (7) (- 3) = - 21$

$RHS = 3 x^{2} - 2 x y - 5 y^{2} = 3 {(- 1)}^{2} - 2 (- 1) (- 2) - 5 {(- 2)}^{2} = 3 \times 1 - 4 - 5 \times 4 = 3 - 4 - 20 = - 21$

Because LHS is equal to RHS, the result is verified.

Thus, the answer is $3 x^{2} - 2 x y - 5 y^{2}$ .

Page No 6.31:

Question 18:

Find the following product and verify the result for x = − 1, y = − 2:
(x²y − 1) (3 − 2x²y)

Answer:

To multiply, we will use distributive law as follows:

$(x^{2} y - 1) (3 - 2 x^{2} y) = x^{2} y (3 - 2 x^{2} y) - 1 \times (3 - 2 x^{2} y) = 3 x^{2} y - 2 x^{4} y^{2} - 3 + 2 x^{2} y = 5 x^{2} y - 2 x^{4} y^{2} - 3$

$∴$ $(x^{2} y - 1) (3 - 2 x^{2} y) = 5 x^{2} y - 2 x^{4} y^{2} - 3$

Now, we put x = $-$ 1 and y = $-$ 2 on both sides to verify the result.

$LHS = (x^{2} y - 1) (3 - 2 x^{2} y) = [{(- 1)}^{2} (- 2) - 1] [3 - 2 {(- 1)}^{2} (- 2)] = [1 \times (- 2) - 1] [3 - 2 \times 1 \times (- 2)] = (- 2 - 1) (3 + 4) = - 3 \times 7 = - 21$

$RHS = 5 x^{2} y - 2 x^{4} y^{2} - 3 = 5 {(- 1)}^{2} (- 2) - 2 {(- 1)}^{4} {(- 2)}^{2} - 3 = [5 \times 1 \times (- 2)] - [2 \times 1 \times 4] - 3 = - 10 - 8 - 3 = - 21$

Because LHS is equal to RHS, the result is verified.

Thus, the answer is $5 x^{2} y - 2 x^{4} y^{2} - 3$ .

Page No 6.31:

Question 19:

Find the following product and verify the result for x = − 1, y = − 2:
$(\frac{1}{3} x - \frac{y^{2}}{5}) (\frac{1}{3} x + \frac{y^{2}}{5})$

Answer:

To multiply, we will use distributive law as follows:

$(\frac{1}{3} x - \frac{y^{2}}{5}) (\frac{1}{3} x + \frac{y^{2}}{5}) = [\frac{1}{3} x (\frac{1}{3} x + \frac{y^{2}}{5})] - [\frac{y^{2}}{5} (\frac{1}{3} x + \frac{y^{2}}{5})] = [\frac{1}{9} x^{2} + \frac{x y^{2}}{15}] - [\frac{x y^{2}}{15} + \frac{y^{4}}{25}] = \frac{1}{9} x^{2} + \frac{x y^{2}}{15} - \frac{x y^{2}}{15} - \frac{y^{4}}{25} = \frac{1}{9} x^{2} - \frac{y^{4}}{25}$

$∴$ $(\frac{1}{3} x - \frac{y^{2}}{5}) (\frac{1}{3} x + \frac{y^{2}}{5}) = \frac{1}{9} x^{2} - \frac{y^{4}}{25}$

Now, we will put x = $-$ 1 and y = $-$ 2 on both the sides to verify the result.

$LHS = (\frac{1}{3} x - \frac{y^{2}}{5}) (\frac{1}{3} x + \frac{y^{2}}{5}) = [\frac{1}{3} (- 1) - \frac{{(- 2)}^{2}}{5}] [\frac{1}{3} (- 1) + \frac{{(- 2)}^{2}}{5}] = (- \frac{1}{3} - \frac{4}{5}) (- \frac{1}{3} + \frac{4}{5}) = (\frac{- 17}{15}) (\frac{7}{15}) = \frac{- 119}{225}$

$RHS = \frac{1}{9} x^{2} - \frac{y^{4}}{25} = \frac{1}{9} {(- 1)}^{2} - \frac{{(- 2)}^{4}}{25} = \frac{1}{9} \times 1 - \frac{16}{25} = \frac{1}{9} - \frac{16}{25} = - \frac{119}{225}$

Because LHS is equal to RHS, the result is verified.

Thus, the answer is $\frac{1}{9} x^{2} - \frac{y^{4}}{25}$ .

Page No 6.31:

Question 20:

Simplify:
x²(x + 2y) (x − 3y)

Answer:

To simplify, we will proceed as follows:

$x^{2} (x + 2 y) (x - 3 y) = [x^{2} (x + 2 y)] (x - 3 y) = (x^{3} + 2 x^{2} y) (x - 3 y) = x^{3} (x - 3 y) + 2 x^{2} y (x - 3 y) = x^{4} - 3 x^{3} y + 2 x^{3} y - 6 x^{2} y^{2} = x^{4} - x^{3} y - 6 x^{2} y^{2}$

Thus, the answer is $x^{4} - x^{3} y - 6 x^{2} y^{2}$ .

Page No 6.31:

Question 21:

Simplify:
(x² − 2y²) (x + 4y) x²y²

Answer:

To simplify, we will proceed as follows:

$(x^{2} - 2 y^{2}) (x + 4 y) x^{2} y^{2} = [x^{2} (x + 4 y) - 2 y^{2} (x + 4 y)] x^{2} y^{2} = (x^{3} + 4 x^{2} y - 2 x y^{2} - 8 y^{3}) x^{2} y^{2} = x^{5} y^{2} + 4 x^{4} y^{3} - 2 x^{3} y^{4} - 8 x^{2} y^{5}$

Thus, the answer is $x^{5} y^{2} + 4 x^{4} y^{3} - 2 x^{3} y^{4} - 8 x^{2} y^{5}$ .

Page No 6.31:

Question 22:

Simplify:
a²b²(a + 2b)(3a + b)

Answer:

To simplify, we will proceed as follows:

$a^{2} b^{2} (a + 2 b) (3 a + b) = [a^{2} b^{2} (a + 2 b)] (3 a + b) = (a^{3} b^{2} + 2 a^{2} b^{3}) (3 a + b) = 3 a (a^{3} b^{2} + 2 a^{2} b^{3}) + b (a^{3} b^{2} + 2 a^{2} b^{3}) = 3 a^{4} b^{2} + 6 a^{3} b^{3} + a^{3} b^{3} + 2 a^{2} b^{4} = 3 a^{4} b^{2} + 7 a^{3} b^{3} + 2 a^{2} b^{4}$

Thus, the answer is $3 a^{4} b^{2} + 7 a^{3} b^{3} + 2 a^{2} b^{4}$ .

Page No 6.31:

Question 23:

Simplify:
x²(x − y) y²(x + 2y)

Answer:

To simplify, we will proceed as follows:

$x^{2} (x - y) y^{2} (x + 2 y) = [x^{2} (x - y)] [y^{2} (x + 2 y)] = (x^{3} - x^{2} y) (x y^{2} + 2 y^{3}) = x^{3} (x y^{2} + 2 y^{3}) - x^{2} y (x y^{2} + 2 y^{3}) = x^{4} y^{2} + 2 x^{3} y^{3} - [x^{3} y^{3} + 2 x^{2} y^{4}] = x^{4} y^{2} + 2 x^{3} y^{3} - x^{3} y^{3} - 2 x^{2} y^{4} = x^{4} y^{2} + x^{3} y^{3} - 2 x^{2} y^{4}$

Thus, the answer is $x^{4} y^{2} + x^{3} y^{3} - 2 x^{2} y^{4}$ .

Page No 6.31:

Question 24:

Simplify:
(x³ − 2x² + 5x − 7)(2x − 3)

Answer:

To simplify, we will proceed as follows:

$(x^{3} - 2 x^{2} + 5 x - 7) (2 x - 3) = 2 x (x^{3} - 2 x^{2} + 5 x - 7) - 3 (x^{3} - 2 x^{2} + 5 x - 7) = 2 x^{4} - 4 x^{3} + 10 x^{2} - 14 x - 3 x^{3} + 6 x^{2} - 15 x + 21$
$= 2 x^{4} - 4 x^{3} - 3 x^{3} + 10 x^{2} + 6 x^{2} - 14 x - 15 x + 21$ (Rearranging)
$= 2 x^{4} - 7 x^{3} + 16 x^{2} - 29 x + 21$ (Combining like terms)

Thus, the answer is $2 x^{4} - 7 x^{3} + 16 x^{2} - 29 x + 21$ .

Page No 6.31:

Question 25:

Simplify:
(5x + 3)(x − 1)(3x − 2)

Answer:

To simplify, we will proceed as follows:

$(5 x + 3) (x - 1) (3 x - 2) = [(5 x + 3) (x - 1)] (3 x - 2)$
$= [5 x (x - 1) + 3 (x - 1)] (3 x - 2)$ (Distributive law)
$= [5 x^{2} - 5 x + 3 x - 3] (3 x - 2) = [5 x^{2} - 2 x - 3] (3 x - 2) = 3 x (5 x^{2} - 2 x - 3) - 2 (5 x^{2} - 2 x - 3) = 15 x^{3} - 6 x^{2} - 9 x - [10 x^{2} - 4 x - 6] = 15 x^{3} - 6 x^{2} - 9 x - 10 x^{2} + 4 x + 6$
$= 15 x^{3} - 6 x^{2} - 10 x^{2} - 9 x + 4 x + 6$ (Rearranging)
$= 15 x^{3} - 16 x^{2} - 5 x + 6$ (Combining like terms)

Thus, the answer is $15 x^{3} - 16 x^{2} - 5 x + 6$ .

Page No 6.31:

Question 26:

Simplify:
(5 − x)(6 − 5x)( 2 − x)

Answer:

To simplify, we will proceed as follows:

$(5 - x) (6 - 5 x) (2 - x) = [(5 - x) (6 - 5 x)] (2 - x)$
$= [5 (6 - 5 x) - x (6 - 5 x)] (2 - x)$ (Distributive law)
$= (30 - 25 x - 6 x + 5 x^{2}) (2 - x) = (30 - 31 x + 5 x^{2}) (2 - x) = 2 (30 - 31 x + 5 x^{2}) - x (30 - 31 x + 5 x^{2}) = 60 - 62 x + 10 x^{2} - 30 x + 31 x^{2} - 5 x^{3}$
$= 60 - 62 x - 30 x + 10 x^{2} + 31 x^{2} - 5 x^{3}$ (Rearranging)
$= 60 - 92 x + 41 x^{2} - 5 x^{3}$ (Combining like terms)

Thus, the answer is $60 - 92 x + 41 x^{2} - 5 x^{3}$ .

Page No 6.31:

Question 27:

Simplify:
(2x² + 3x − 5)(3x² − 5x + 4)

Answer:

To simplify, we will proceed as follows:

$(2 x^{2} + 3 x - 5) (3 x^{2} - 5 x + 4)$
$= 2 x^{2} (3 x^{2} - 5 x + 4) + 3 x (3 x^{2} - 5 x + 4) - 5 (3 x^{2} - 5 x + 4)$ (Distributive law)
$= 6 x^{4} - 10 x^{3} + 8 x^{2} + 9 x^{3} - 15 x^{2} + 12 x - 15 x^{2} + 25 x - 20$
$= 6 x^{4} - 10 x^{3} + 9 x^{3} + 8 x^{2} - 15 x^{2} - 15 x^{2} + 12 x + 25 x - 20$ (Rearranging)
$= 6 x^{4} - x^{3} - 22 x^{2} + 36 x - 20$ (Combining like terms)

Thus, the answer is $6 x^{4} - x^{3} - 22 x^{2} + 36 x - 20$ .

Page No 6.31:

Question 28:

Simplify:
(3x − 2)(2x − 3) + (5x − 3)(x + 1)

Answer:

To simplify, we will proceed as follows:

$(3 x - 2) (2 x - 3) + (5 x - 3) (x + 1) = [(3 x - 2) (2 x - 3)] + [(5 x - 3) (x + 1)]$
$= [3 x (2 x - 3) - 2 (2 x - 3)] + [5 x (x + 1) - 3 (x + 1)]$ (Distributive law)
$= 6 x^{2} - 9 x - 4 x + 6 + 5 x^{2} + 5 x - 3 x - 3$
$= 6 x^{2} + 5 x^{2} - 9 x - 4 x + 5 x - 3 x - 3 + 6$ (Rearranging)
$= 11 x^{2} - 11 x + 3$ (Combining like terms)

Thus, the answer is $11 x^{2} - 11 x + 3$ .

Page No 6.31:

Question 29:

Simplify:
(5x − 3)(x + 2) − (2x + 5)(4x − 3)

Answer:

To simplify, we will proceed as follows:

$(5 x - 3) (x + 2) - (2 x + 5) (4 x - 3) = [(5 x - 3) (x + 2)] - [(2 x + 5) (4 x - 3)]$
$= [5 x (x + 2) - 3 (x + 2)] - [2 x (4 x - 3) + 5 (4 x - 3)]$ (Distributive law)
$= 5 x^{2} + 10 x - 3 x - 6 - 8 x^{2} + 6 x - 20 x + 15$
$= 5 x^{2} - 8 x^{2} + 10 x - 3 x + 6 x - 20 x - 6 + 15$ (Rearranging)
$= 5 x^{2} - 8 x^{2} + 10 x - 3 x + 6 x - 20 x - 6 + 15 = - 3 x^{2} - 7 x + 9$ (Combining like terms)

Hence, the answer is $- 3 x^{2} - 7 x + 9$ .

Page No 6.31:

Question 30:

Simplify:
(3x + 2y)(4x + 3y) − (2x − y)(7x − 3y)

Answer:

To simplify, we will proceed as follows:

$(3 x + 2 y) (4 x + 3 y) - (2 x - y) (7 x - 3 y) = [(3 x + 2 y) (4 x + 3 y)] - [(2 x - y) (7 x - 3 y)]$
$= [3 x (4 x + 3 y) + 2 y (4 x + 3 y)] - [2 x (7 x - 3 y) - y (7 x - 3 y)]$ (Distributive law)
$= 12 x^{2} + 9 x y + 8 x y + 6 y^{2} - [14 x^{2} - 6 x y - 7 x y + 3 y^{2}] = 12 x^{2} + 9 x y + 8 x y + 6 y^{2} - 14 x^{2} + 6 x y + 7 x y - 3 y^{2}$
$= 12 x^{2} - 14 x^{2} + 9 x y + 8 x y + 6 x y + 7 x y + 6 y^{2} - 3 y^{2}$ (Rearranging)
$= - 2 x^{2} + 30 x y + 3 y^{2}$ (Combining like terms)

Thus, the answer is $- 2 x^{2} + 30 x y + 3 y^{2}$ .

Page No 6.31:

Question 31:

Simplify:
(x² − 3x + 2)(5x − 2) − (3x² + 4x − 5)(2x − 1)

Answer:

To simplify, we will proceed as follows:

$(x^{2} - 3 x + 2) (5 x - 2) - (3 x^{2} + 4 x - 5) (2 x - 1) = [(x^{2} - 3 x + 2) (5 x - 2)] - [(3 x^{2} + 4 x - 5) (2 x - 1)]$
$= [5 x (x^{2} - 3 x + 2) - 2 (x^{2} - 3 x + 2)] - [2 x (3 x^{2} + 4 x - 5) - 1 \times (3 x^{2} + 4 x - 5)]$ (Distributive law)
$= [5 x^{3} - 15 x^{2} + 10 x - (2 x^{2} - 6 x + 4)] - [6 x^{3} + 8 x^{2} - 10 x - 3 x^{2} - 4 x + 5] = [5 x^{3} - 15 x^{2} + 10 x - 2 x^{2} + 6 x - 4] - [6 x^{3} + 8 x^{2} - 10 x - 3 x^{2} - 4 x + 5] = 5 x^{3} - 15 x^{2} + 10 x - 2 x^{2} + 6 x - 4 - 6 x^{3} - 8 x^{2} + 10 x + 3 x^{2} + 4 x - 5$
$= 5 x^{3} - 6 x^{3} - 15 x^{2} - 2 x^{2} - 8 x^{2} + 3 x^{2} + 10 x + 6 x + 10 x + 4 x - 5 - 4$ (Rearranging)
$= - x^{3} - 22 x^{2} + 30 x - 9$ (Combining like terms)

Thus, the answer is $- x^{3} - 22 x^{2} + 30 x - 9$ .

Page No 6.31:

Question 32:

Simplify:
(x³ − 2x² + 3x − 4) (x −1) − (2x − 3)(x² − x + 1)

Answer:

To simplify,we will proceed as follows:

$(x^{3} - 2 x^{2} + 3 x - 4) (x - 1) - (2 x - 3) (x^{2} - x + 1) = [(x^{3} - 2 x^{2} + 3 x - 4) (x - 1)] - [(2 x - 3) (x^{2} - x + 1)]$
$= [x (x^{3} - 2 x^{2} + 3 x - 4) - 1 (x^{3} - 2 x^{2} + 3 x - 4)] - [2 x (x^{2} - x + 1) - 3 (x^{2} - x + 1)]$           (Distributive law)
$= [x (x^{3} - 2 x^{2} + 3 x - 4) - 1 (x^{3} - 2 x^{2} + 3 x - 4)] - [2 x (x^{2} - x + 1) - 3 (x^{2} - x + 1)] = x^{4} - 2 x^{3} + 3 x^{2} - 4 x - x^{3} + 2 x^{2} - 3 x + 4 - [2 x^{3} - 2 x^{2} + 2 x - 3 x^{2} + 3 x - 3] = x^{4} - 2 x^{3} + 3 x^{2} - 4 x - x^{3} + 2 x^{2} - 3 x + 4 - 2 x^{3} + 2 x^{2} - 2 x + 3 x^{2} - 3 x + 3$
$= x^{4} - 2 x^{3} - 2 x^{3} - x^{3} + 3 x^{2} + 2 x^{2} + 2 x^{2} + 3 x^{2} - 4 x - 3 x - 2 x - 3 x + 4 + 3$    (Rearranging)
$= x^{4} - 5 x^{3} + 10 x^{2} - 12 x + 7$                                                                                             (Combining like terms)

Thus, the answer is $x^{4} - 5 x^{3} + 10 x^{2} - 12 x + 7$ .

Page No 6.43:

Question 1:

Write the following squares of binomials as trinomials:
(i) (x + 2)²
(ii) (8a + 3b)²
(iii) (2m + 1)²
(iv) ${(9 a + \frac{1}{6})}^{2}$
(v) ${(x + \frac{x^{2}}{2})}^{2}$
(vi) $(\frac{x}{4} - \frac{y}{3})$
(vii) ${(3 x - \frac{1}{3 x})}^{2}$
(viii) ${(\frac{x}{y} - \frac{y}{x})}^{2}$
(ix) ${(\frac{3 a}{2} - \frac{5 b}{4})}^{2}$
(x) (a²b − bc²)²
(xi) ${(\frac{2 a}{3 b} + \frac{2 b}{3 a})}^{2}$
(xii) (x² − ay)²

Answer:

We will use the identities ${(a + b)}^{2} = a^{2} + 2 a b + b^{2} and {(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ to convert the squares of binomials as trinomials.

$(i) {(x + 2)}^{2} = x^{2} + 2 \times x \times 2 + b^{2} = x^{2} + 4 x + b^{2}$

$(ii) {(8 a + 3 b)}^{2} = {(8 a)}^{2} + 2 (8 a) (3 b) + {(6 b)}^{2} = 64 a^{2} + 48 a b + 36 b^{2}$

^{$(iii) {(2 m + 1)}^{2} = {(2 m)}^{2} + 2 (2 m) (1) + 1^{2} = 4 m^{2} + 4 m + 1$

$(iv) {(9 a + \frac{1}{6})}^{2} = {(9 a)}^{2} + 2 (9 a) (\frac{1}{6}) + {(\frac{1}{6})}^{2} = 81 a^{2} + 3 a + \frac{1}{36}$

$(v) {(x + \frac{x^{2}}{2})}^{2} = x^{2} + 2 x (\frac{x^{2}}{2}) + {(\frac{x^{2}}{2})}^{2} = x^{2} + x^{3} + \frac{x^{4}}{4}$

$(vi) {(\frac{x}{4} - \frac{y}{3})}^{2} = {(\frac{x}{4})}^{2} - 2 (\frac{x}{4}) (\frac{y}{3}) + {(\frac{y}{3})}^{2} = \frac{x^{2}}{16} - \frac{1}{6} x y + \frac{y^{2}}{9}$

$(vii) {(3 x - \frac{1}{3 x})}^{2} = {(3 x)}^{2} - 2 (3 x) (\frac{1}{3 x}) + {(\frac{1}{3 x})}^{2} = 9 x^{2} - 2 + \frac{1}{9 x^{2}}$

$(viii) {(\frac{x}{y} - \frac{y}{x})}^{2} = {(\frac{x}{y})}^{2} - 2 (\frac{x}{y}) (\frac{y}{x}) + {(\frac{y}{x})}^{2} = \frac{x^{2}}{y^{2}} - 2 + \frac{y^{2}}{x^{2}}$

$(ix) {(\frac{3 a}{2} - \frac{5 b}{4})}^{2} = {(\frac{3 a}{2})}^{2} - 2 (\frac{3 a}{2}) (\frac{5 b}{4}) + {(\frac{5 b}{4})}^{2} = \frac{9 a^{2}}{4} - \frac{15 a b}{4} + \frac{25 b^{2}}{16}$

$(x) {(a^{2} b - b c^{2})}^{2} = {(a^{2} b)}^{2} - 2 (a^{2} b) (b c^{2}) + {(b c^{2})}^{2} = a^{4} b^{2} - 2 a^{2} b^{2} c^{2} + b^{2} c^{4}$}

$(xi) {(\frac{2 a}{3 b} + \frac{2 b}{3 a})}^{2} = {(\frac{2 a}{3 b})}^{2} + 2 (\frac{2 a}{3 b}) (\frac{2 b}{3 a}) + {(\frac{2 b}{3 a})}^{2} = \frac{4 a^{2}}{9 b^{2}} + \frac{8}{9} + \frac{4 b^{2}}{9 a^{2}}$

$(xii) {(x^{2} - a y)}^{2} = {(x^{2})}^{2} - 2 x^{2} (a y) + {(a y)}^{2} = x^{4} - 2 x^{2} a y + a^{2} y^{2}$

Page No 6.43:

Question 2:

Find the product of the following binomials:
(i) (2x + y)(2x + y)
(ii) (a + 2b)(a − 2b)
(iii) (a² + bc)(a²− bc)
(iv) $(\frac{4 x}{5} - \frac{3 y}{4}) (\frac{4 x}{5} + \frac{3 y}{4})$
(v) $(2 x + \frac{3}{y}) (2 x - \frac{3}{y})$
(vi) (2a³ + b³)(2a³ − b³)
(vii) $(x^{4} + \frac{2}{x^{2}}) (x^{4} - \frac{2}{x^{2}})$
(viii) $(x^{3} + \frac{1}{x^{3}}) (x^{3} - \frac{1}{x^{3}})$

Answer:

(i) We will use the identity ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ in the given expression to find the product.
$(2 x + y) (2 x + y) = {(2 x + y)}^{2} = {(2 x)}^{2} + 2 (2 x) (y) + y^{2} = 4 x^{2} + 4 x y + y^{2}$

(ii) We will use the identity $(a + b) (a - b) = a^{2} - b^{2}$ in the given expression to find the product.
$(a + 2 b) (a - 2 b) = a^{2} - {(2 b)}^{2} = a^{2} - 4 b^{2}$

(iii) We will use the identity $(a + b) (a - b) = a^{2} - b^{2}$ in the given expression to find the product.
$(a^{2} + b c) (a^{2} - b c) = {(a^{2})}^{2} - {(b c)}^{2} = a^{4} - b^{2} c^{2}$

(iv)We will use the identity $(a + b) (a - b) = a^{2} - b^{2}$ in the given expression to find the product.
$(\frac{4 x}{5} - \frac{3 y}{4}) (\frac{4 x}{5} + \frac{3 y}{4}) = {(\frac{4 x}{5})}^{2} - {(\frac{3 y}{4})}^{2} = \frac{16 x^{2}}{25} - \frac{9 y^{2}}{16}$

(v) We will use the identity $(a + b) (a - b) = a^{2} - b^{2}$ in the given expression to find the product.
$(2 x + \frac{3}{y}) (2 x - \frac{3}{y}) = {(2 x)}^{2} - {(\frac{3}{y})}^{2} = 4 x^{2} - \frac{9}{y^{2}}$

(vi) We will use the identity $(a + b) (a - b) = a^{2} - b^{2}$ in the given expression to find the product.
$(2 a^{3} + b^{3}) (2 a^{3} - b^{3}) = {(2 a^{3})}^{2} - {(b^{3})}^{2} = 4 a^{6} - b^{6}$

(vii) We will use the identity $(a + b) (a - b) = a^{2} - b^{2}$ in the given expression to find the product.
$(x^{4} + \frac{2}{x^{2}}) (x^{4} - \frac{2}{x^{2}}) = {(x^{4})}^{2} - {(\frac{2}{x^{2}})}^{2} = x^{8} - \frac{4}{x^{4}}$

(viii) We will use the identity $(a + b) (a - b) = a^{2} - b^{2}$ in the given expression to find the product.

$(x^{3} + \frac{1}{x^{3}}) (x^{3} - \frac{1}{x^{3}}) = {(x^{3})}^{2} - {(\frac{1}{x^{3}})}^{2} = x^{6} - \frac{1}{x^{6}}$

Page No 6.43:

Question 3:

Using the formula for squaring a binomial, evaluate the following:
(i) (102)²
(ii) (99)²
(iii) (1001)²
(iv) (999)²
(v) (703)²

Answer:

(i) Here, we will use the identity ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$
${(102)}^{2} = {(100 + 2)}^{2} = {(100)}^{2} + 2 \times 100 \times 2 + 2^{2} = 10000 + 400 + 4 = 10404$

(ii) Here, we will use the identity ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$
${(99)}^{2} = {(100 - 1)}^{2} = {(100)}^{2} - 2 \times 100 \times 1 + 1^{2} = 10000 - 200 + 1 = 9801$

(iii) Here, we will use the identity ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$
${(1001)}^{2} = {(1000 + 1)}^{2} = {(1000)}^{2} + 2 \times 1000 \times 1 + 1^{2} = 1000000 + 2000 + 1 = 1002001$

(iv) Here, we will use the identity ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$
${(999)}^{2} = {(1000 - 1)}^{2} = {(1000)}^{2} - 2 \times 1000 \times 1 + 1^{2} = 1000000 - 2000 + 1 = 998001$

(v) Here, we will use the identity ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$
${(703)}^{2} = {(700 + 3)}^{2} = {(700)}^{2} + 2 \times 700 \times 3 + 3^{2} = 490000 + 4200 + 9 = 494209$

Page No 6.43:

Question 4:

Simplify the following using the formula: (a − b)(a + b) = a² − b²:
(i) (82)² − (18)²
(ii) (467)² − (33)²
(iii) (79)² − (69)²
(iv) 197 × 203
(v) 113 × 87
(vi) 95 × 105
(vii) 1.8 × 2.2
(viii) 9.8 × 10.2

Answer:

Here, we will use the identity $(a - b) (a + b) = a^{2} - b^{2}$

(i) Let us consider the following expression:

${(82)}^{2} - {(18)}^{2} = (82 + 18) (82 - 18) = 100 \times 64 = 6400$

(ii) Let us consider the following expression:

${(467)}^{2} - {(33)}^{2} = (467 + 33) (467 - 33) = 500 \times 434 = 217000$

(iii) Let us consider the following expression:

${(79)}^{2} - {(69)}^{2} = (79 + 69) (79 - 69) = 148 \times 10 = 1480$

(iv) Let us consider the following product:

$197 \times 203$

$∵ \frac{197 + 203}{2} = \frac{400}{2} = 200$ ; therefore, we will write the above product as:
$197 \times 203 = (200 - 3) (200 + 3) = {(200)}^{2} - {(3)}^{2} = 40000 - 9 = 39991$

Thus, the answer is $39991$ .

(v) Let us consider the following product:

$113 \times 87$

$∵ \frac{113 + 87}{2} = \frac{200}{2} = 100$ ; therefore, we will write the above product as:
$113 \times 87 = (100 + 13) (100 - 13) = {(100)}^{2} - {(13)}^{2} = 10000 - 169 = 9831$

Thus, the answer is 9831.

(vi) Let us consider the following product:

$95 \times 105$

$∵ \frac{95 + 105}{2} = \frac{200}{2} = 100$ ; therefore, we will write the above product as:
$95 \times 105 = (100 + 5) (100 - 5) = {(100)}^{2} - {(5)}^{2} = 10000 - 25 = 9975$

Thus, the answer is 9975.

(vii) Let us consider the following product:

$1.8 \times 2.2$

$∵ \frac{1.8 + 2.2}{2} = \frac{4}{2} = 2$ ; therefore, we will write the above product as:
$1.8 \times 2.2 = (2 - 0.2) (2 + 0.2) = {(2)}^{2} - {(0.2)}^{2} = 4 - 0.04 = 3.96$

Thus, the answer is 3.96.

(viii) Let us consider the following product:

$9.8 \times 10.2$

$∵ \frac{9.8 + 10.2}{2} = \frac{20}{2} = 10$ ; therefore, we will write the above product as:
$9.8 \times 10.2 = (10 - 0.2) (10 + 0.2) = {(10)}^{2} - {(0.2)}^{2} = 100 - 0.04 = 99.96$

Thus, the answer is 99.96.

Page No 6.43:

Question 5:

Simplify the following using the identities:
(i) $\frac{58^{2} - 42^{2}}{16}$
(ii) 178 × 178 − 22 × 22
(iii) $\frac{198 \times 198 - 102 \times 102}{96}$
(iv) 1.73 × 1.73 − 0.27 × 0.27
(v) $\frac{8.63 \times 8.63 - 1.37 \times 1.37}{0.726}$

Answer:

(i) Let us consider the following expression:

$\frac{58^{2} - 42^{2}}{16}$
Using the identity $(a + b) (a - b) = a^{2} - b^{2}$ , we get:

$\frac{58^{2} - 42^{2}}{16} = \frac{(58 + 42) (58 - 42)}{16}$
$\Rightarrow \frac{58^{2} - 42^{2}}{16} = \frac{100 \times 16}{16} \Rightarrow \frac{58^{2} - 42^{2}}{16} = 100$

Thus, the answer is 100.

(ii) Let us consider the following expression:

$178 \times 178 - 22 \times 22$
Using the identity $(a + b) (a - b) = a^{2} - b^{2}$ , we get:

$178 \times 178 - 22 \times 22 = 178^{2} - 22^{2} = (178 + 22) (178 - 22) = 200 \times 156 = 31200$
Thus, the answer is 31200.

(iii) Let us consider the following expression:

$\frac{198 \times 198 - 102 \times 102}{96} = \frac{198^{2} - 102^{2}}{96}$
Using the identity $(a + b) (a - b) = a^{2} - b^{2}$ , we get:

$\frac{198 \times 198 - 102 \times 102}{96} = \frac{198^{2} - 102^{2}}{96} = \frac{(198 + 102) (198 - 102)}{96}$
$\Rightarrow \frac{198 \times 198 - 102 \times 102}{96} = \frac{(198 + 102) (198 - 102)}{96} \Rightarrow \frac{198 \times 198 - 102 \times 102}{96} = \frac{300 \times 96}{96} \Rightarrow \frac{198 \times 198 - 102 \times 102}{96} = 300$

Thus, the answer is 300.

(iv) Let us consider the following expression:

$1.73 \times 1.73 - 0.27 \times 0.27$
Using the identity $(a + b) (a - b) = a^{2} - b^{2}$ , we get:

$1.73 \times 1.73 - 0.27 \times 0.27 = 1 . 73^{2} - 0 . 27^{2} = (1.73 + 0.27) (1.73 - 0.27) = 2 \times 1.46 = 2.92$
Thus, the answer is 2.92.

(v) Let us consider the following expression:

$\frac{8.63 \times 8.63 - 1.37 \times 1.37}{0.726} = \frac{8 . 63^{2} - 1 . 37^{2}}{0.726}$
Using the identity $(a + b) (a - b) = a^{2} - b^{2}$ , we get:

$\frac{8.63 \times 8.63 - 1.37 \times 1.37}{0.726} = \frac{8 . 63^{2} - 1 . 37^{2}}{0.726} = \frac{(8.63 + 1.37) (8.63 - 1.37)}{0.726}$
$\Rightarrow \frac{8.63 \times 8.63 - 1.37 \times 1.37}{0.726} = \frac{(8.63 + 1.37) (8.63 - 1.37)}{0.726} \Rightarrow \frac{8.63 \times 8.63 - 1.37 \times 1.37}{0.726} = \frac{(8.63 + 1.37) (8.63 - 1.37)}{0.726} \Rightarrow \frac{8.63 \times 8.63 - 1.37 \times 1.37}{0.726} = \frac{10 \times 7.26}{0.726} \Rightarrow \frac{8.63 \times 8.63 - 1.37 \times 1.37}{0.726} = \frac{10 \times {7.26}^{10}}{0.726} \Rightarrow \frac{8.63 \times 8.63 - 1.37 \times 1.37}{0.726} = 100$

Thus, the answer is 100.

Page No 6.43:

Question 6:

Find the value of x, if:
(i) 4x = (52)² − (48)²
(ii) 14x = (47)² − (33)²
(iii) 5x = (50)² − (40)²

Answer:

(i) Let us consider the following equation:

$4 x = {(52)}^{2} - {(48)}^{2}$
Using the identity $(a + b) (a - b) = a^{2} - b^{2}$ , we get:
$4 x = {(52)}^{2} - {(48)}^{2} 4 x = (52 + 48) (52 - 48) 4 x = 100 \times 4 = 400$
$\Rightarrow 4 x = 400$
$\Rightarrow x = 100$         (Dividing both sides by 4)

(ii) Let us consider the following equation:

$14 x = {(47)}^{2} - {(33)}^{2}$
Using the identity $(a + b) (a - b) = a^{2} - b^{2}$ , we get:
$14 x = {(47)}^{2} - {(33)}^{2} 14 x = (47 + 33) (47 - 33) 14 x = 80 \times 14 = 1120$
$\Rightarrow 14 x = 1120$
$\Rightarrow x = 80$         (Dividing both sides by 14)

(iii) Let us consider the following equation:

$5 x = {(50)}^{2} - {(40)}^{2}$
Using the identity $(a + b) (a - b) = a^{2} - b^{2}$ , we get:
$5 x = {(50)}^{2} - {(40)}^{2} 5 x = (50 + 40) (50 - 40) 5 x = 90 \times 10 = 900$
$\Rightarrow 5 x = 900$
$\Rightarrow x = 180$         (Dividing both sides by 5)

Page No 6.43:

Question 7:

If $x + \frac{1}{x} = 20,$ find the value of $x^{2} + \frac{1}{x^{2}} .$

Answer:

Let us consider the following equation:

$x + \frac{1}{x} = 20$
Squaring both sides, we get:

${(x + \frac{1}{x})}^{2} = {(20)}^{2} = 400 \Rightarrow {(x + \frac{1}{x})}^{2} = 400 \Rightarrow x^{2} + 2 \times x \times \frac{1}{x} + {(\frac{1}{x})}^{2} = 400 [(a + b)^{2} = a^{2} + b^{2} + 2 a b] \Rightarrow x^{2} + 2 + \frac{1}{x^{2}} = 400$
$\Rightarrow x^{2} + \frac{1}{x^{2}} = 398$ (Subtracting 2 from both sides)

Thus, the answer is 398.

Page No 6.43:

Question 8:

If $x - \frac{1}{x} = 3,$ find the values of $x^{2} + \frac{1}{x^{2}}$ and $x^{4} + \frac{1}{x^{4}} .$

Answer:

Let us consider the following equation:

$x - \frac{1}{x} = 3$

Squaring both sides, we get:

${(x - \frac{1}{x})}^{2} = {(3)}^{2} = 9 \Rightarrow {(x - \frac{1}{x})}^{2} = 9 \Rightarrow x^{2} - 2 \times x \times \frac{1}{x} + {(\frac{1}{x})}^{2} = 9 \Rightarrow x^{2} - 2 + \frac{1}{x^{2}} = 9$
$\Rightarrow x^{2} + \frac{1}{x^{2}} = 11$ (Adding 2 to both sides)

Squaring both sides again, we get:

${(x^{2} + \frac{1}{x^{2}})}^{2} = {(11)}^{2} = 121 \Rightarrow {(x^{2} + \frac{1}{x^{2}})}^{2} = 121 \Rightarrow {(x^{2})}^{2} + 2 (x^{2}) (\frac{1}{x^{2}}) + {(\frac{1}{x^{2}})}^{2} = 121 \Rightarrow x^{4} + 2 + \frac{1}{x^{4}} = 121$
$\Rightarrow x^{4} + \frac{1}{x^{4}} = 119$

Page No 6.43:

Question 9:

If $x^{2} + \frac{1}{x^{2}} = 18,$ find the values of $x + \frac{1}{x} and x - \frac{1}{x} .$

Answer:

Let us consider the following expression:

$x + \frac{1}{x}$
Squaring the above expression, we get:

${(x + \frac{1}{x})}^{2} = x^{2} + 2 \times x \times \frac{1}{x} + {(\frac{1}{x})}^{2} = x^{2} - 2 + \frac{1}{x^{2}} [(a + b)^{2} = a^{2} + b^{2} + 2 a b] \Rightarrow {(x + \frac{1}{x})}^{2} = x^{2} + 2 + \frac{1}{x^{2}}$
$\Rightarrow {(x + \frac{1}{x})}^{2} = 20$                                        ( $∵$ $x^{2} + \frac{1}{x^{2}} = 18$ )
$\Rightarrow x + \frac{1}{x} = \pm \sqrt{20}$                                                     (Taking square root of both sides)

Now, let us consider the following expression:

$x - \frac{1}{x}$
Squaring the above expression, we get:

${(x - \frac{1}{x})}^{2} = x^{2} - 2 \times x \times \frac{1}{x} + {(\frac{1}{x})}^{2} = x^{2} - 2 + \frac{1}{x^{2}} [(a - b)^{2} = a^{2} + b^{2} - 2 a b] \Rightarrow {(x - \frac{1}{x})}^{2} = x^{2} - 2 + \frac{1}{x^{2}}$
$\Rightarrow {(x - \frac{1}{x})}^{2} = 16$                                  ( $∵$ $x^{2} + \frac{1}{x^{2}} = 18$ )
$\Rightarrow x - \frac{1}{x} = \pm 4$                                      (Taking square root of both sides)

Page No 6.43:

Question 10:

If x + y = 4 and xy = 2, find the value of x² + y²

Answer:

We have:

${(x + y)}^{2} = x^{2} + 2 x y + y^{2} \Rightarrow x^{2} + y^{2} = {(x + y)}^{2} - 2 x y$
$\Rightarrow x^{2} + y^{2} = 4^{2} - 2 \times 2$ ( $∵$ $x + y = 4 and x y = 2$ )
$\Rightarrow x^{2} + y^{2} = 16 - 4 \Rightarrow x^{2} + y^{2} = 12$

Page No 6.43:

Question 11:

If x − y = 7 and xy = 9, find the value of x² + y²

Answer:

We have:

${(x - y)}^{2} = x^{2} - 2 x y + y^{2} \Rightarrow x^{2} + y^{2} = {(x - y)}^{2} + 2 x y$
$\Rightarrow x^{2} + y^{2} = 7^{2} + 2 \times 9$ ( $∵$ $x - y = 7 and x y = 9$ )
$\Rightarrow x^{2} + y^{2} = 7^{2} + 2 \times 9 \Rightarrow x^{2} + y^{2} = 49 + 18 \Rightarrow x^{2} + y^{2} = 67$

Page No 6.44:

Question 12:

If 3x + 5y = 11 and xy = 2, find the value of 9x²+ 25y²

Answer:

We have:

${(3 x + 5 y)}^{2} = {(3 x)}^{2} + 2 (3 x) (5 y) + {(5 y)}^{2} \Rightarrow {(3 x + 5 y)}^{2} = 9 x^{2} + 30 x y + 25 y^{2} \Rightarrow 9 x^{2} + 25 y^{2} = {(3 x + 5 y)}^{2} - 30 x y$
$\Rightarrow 9 x^{2} + 25 y^{2} = 11^{2} - 30 \times 2$ ( $∵$ $3 x + 5 y = 11 and x y = 2$ )
$\Rightarrow 9 x^{2} + 25 y^{2} = 121 - 60 \Rightarrow 9 x^{2} + 25 y^{2} = 61$

Page No 6.44:

Question 13:

Find the values of the following expressions:
(i) 16x² + 24x + 9, when $x = \frac{7}{4}$
(ii) 64x² + 81y² + 144xy, when x = 11 and $y = \frac{4}{3}$
(iii) 81x² + 16y² − 72xy, when $x = \frac{2}{3}$ and $y = \frac{3}{4}$

Answer:

(i) Let us consider the following expression:

$16 x^{2} + 24 x + 9$

Now

$16 x^{2} + 24 x + 9 = {(4 x + 3)}^{2}$                              (Using identity ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ )
$\Rightarrow 16 x^{2} + 24 x + 9 = {(4 \times \frac{7}{4} + 3)}^{2} (Substituting x = \frac{7}{4}) \Rightarrow 16 x^{2} + 24 x + 9 = {(7 + 3)}^{2} \Rightarrow 16 x^{2} + 24 x + 9 = 10^{2} \Rightarrow 16 x^{2} + 24 x + 9 = 100$

(ii) Let us consider the following expression:

$64 x^{2} + 81 y^{2} + 144 x y$

Now

$64 x^{2} + 81 y^{2} + 144 x y = {(8 x + 9 y)}^{2}$                           (Using identity ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ )
$\Rightarrow 64 x^{2} + 81 y^{2} + 144 x y = {[8 (11) + 9 (\frac{4}{3})]}^{2} (Substituting x = 11 and y = \frac{4}{3}) \Rightarrow 64 x^{2} + 81 y^{2} + 144 x y = {[88 + 12]}^{2} \Rightarrow 64 x^{2} + 81 y^{2} + 144 x y = 100^{2} \Rightarrow 64 x^{2} + 81 y^{2} + 144 x y = 10000$

(iii) Let us consider the following expression:

$81 x^{2} + 16 y^{2} - 72 x y$

Now

$81 x^{2} + 16 y^{2} - 72 x y = {(9 x - 4 y)}^{2}$                                   (Using identity ${(a + b)}^{2} = a^{2} - 2 a b + b^{2}$ )
$\Rightarrow 81 x^{2} + 16 y^{2} - 72 x y = {[9 (\frac{2}{3}) - 4 (\frac{3}{4})]}^{2} (Substituting x = \frac{2}{3} and y = \frac{3}{4}) \Rightarrow 81 x^{2} + 16 y^{2} - 72 x y = {[6 - 3]}^{2} \Rightarrow 81 x^{2} + 16 y^{2} - 72 x y = 3^{2} \Rightarrow 81 x^{2} + 16 y^{2} - 72 x y = 9$

Page No 6.44:

Question 14:

If $x + \frac{1}{x} = 9,$ find the value of $x^{4} + \frac{1}{x^{4}} .$

Answer:

Let us consider the following equation:

$x + \frac{1}{x} = 9$

Squaring both sides, we get:

${(x + \frac{1}{x})}^{2} = {(9)}^{2} = 81 \Rightarrow {(x + \frac{1}{x})}^{2} = 81 \Rightarrow x^{2} + 2 \times x \times \frac{1}{x} + {(\frac{1}{x})}^{2} = 81 \Rightarrow x^{2} + 2 + \frac{1}{x^{2}} = 81$
$\Rightarrow x^{2} + \frac{1}{x^{2}} = 79$ (Subtracting 2 from both sides)

Now, squaring both sides again, we get:

${(x^{2} + \frac{1}{x^{2}})}^{2} = {(79)}^{2} = 6241 \Rightarrow {(x^{2} + \frac{1}{x^{2}})}^{2} = 6241 \Rightarrow {(x^{2})}^{2} + 2 (x^{2}) (\frac{1}{x^{2}}) + {(\frac{1}{x^{2}})}^{2} = 6241 \Rightarrow x^{4} + 2 + \frac{1}{x^{4}} = 6241$
$\Rightarrow x^{4} + \frac{1}{x^{4}} = 6239$

Page No 6.44:

Question 15:

If $x + \frac{1}{x} = 12,$ find the value of $x - \frac{1}{x} .$

Answer:

Let us consider the following equation:

$x + \frac{1}{x} = 12$

Squaring both sides, we get:

${(x + \frac{1}{x})}^{2} = {(12)}^{2} = 144 \Rightarrow {(x + \frac{1}{x})}^{2} = 144 \Rightarrow x^{2} + 2 \times x \times \frac{1}{x} + {(\frac{1}{x})}^{2} = 144 [(a + b)^{2} = a^{2} + b^{2} + 2 a b] \Rightarrow x^{2} + 2 + \frac{1}{x^{2}} = 144$
$\Rightarrow x^{2} + \frac{1}{x^{2}} = 142$ (Subtracting 2 from both sides)

Now

${(x - \frac{1}{x})}^{2} = x^{2} - 2 \times x \times \frac{1}{x} + {(\frac{1}{x})}^{2} = x^{2} - 2 + \frac{1}{x^{2}} [(a - b)^{2} = a^{2} + b^{2} - 2 a b] \Rightarrow {(x - \frac{1}{x})}^{2} = x^{2} - 2 + \frac{1}{x^{2}} \Rightarrow {(x - \frac{1}{x})}^{2} = 142 - 2 (∵ x^{2} + \frac{1}{x^{2}} = 142) \Rightarrow {(x - \frac{1}{x})}^{2} = 140 \Rightarrow x - \frac{1}{x} = \pm \sqrt{140} (Taking square root)$

Page No 6.44:

Question 16:

If 2x + 3y = 14 and 2x − 3y = 2, find the value of xy.
[Hint: Use (2x + 3y)² − (2x − 3y)² = 24xy]

Answer:

We will use the identity $(a + b) (a - b) = a^{2} - b^{2}$ to obtain the value of xy.
$Squaring (2 x + 3 y) and (2 x - 3 y) both and then subtracting them, we get :$
${(2 x + 3 y)}^{2} - {(2 x - 3 y)}^{2} = \{(2 x + 3 y) + (2 x - 3 y)\} \{(2 x + 3 y) - (2 x - 3 y)\} = 4 x \times 6 y = 24 x y \Rightarrow {(2 x + 3 y)}^{2} - {(2 x - 3 y)}^{2} = 24 x y$
$\Rightarrow 24 x y = {(2 x + 3 y)}^{2} - {(2 x - 3 y)}^{2} \Rightarrow 24 x y = {(14)}^{2} - {(2)}^{2} \Rightarrow 24 x y = (14 + 2) (14 - 2) (∵ (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow 24 x y = 16 \times 12 \Rightarrow x y = \frac{16 \times 12}{24} (Dividing both sides by 24) \Rightarrow x y = 8$

Page No 6.44:

Question 17:

If x² + y² = 29 and xy = 2, find the value of
(i) x + y
(ii) x − y
(iii) x⁴ + y⁴

Answer:

(i) We have:

${(x + y)}^{2} = x^{2} + 2 x y + y^{2} \Rightarrow (x + y) = \pm \sqrt{x^{2} + 2 x y + y^{2}} \Rightarrow (x + y) = \pm \sqrt{29 + 2 \times 2} (∵ x^{2} + y^{2} = 29 and x y = 2) \Rightarrow (x + y) = \pm \sqrt{29 + 4} \Rightarrow (x + y) = \pm \sqrt{33}$

(ii) We have:

${(x - y)}^{2} = x^{2} - 2 x y + y^{2} \Rightarrow (x - y) = \pm \sqrt{x^{2} - 2 x y + y^{2}} \Rightarrow (x + y) = \pm \sqrt{29 - 2 \times 2} (∵ x^{2} + y^{2} = 29 and x y = 2) \Rightarrow (x + y) = \pm \sqrt{29 - 4} \Rightarrow (x + y) = \pm \sqrt{25} \Rightarrow (x + y) = \pm 5$

(iii) We have:

${(x^{2} + y^{2})}^{2} = x^{4} + 2 x^{2} y^{2} + y^{4} \Rightarrow x^{4} + y^{4} = {(x^{2} + y^{2})}^{2} - 2 x^{2} y^{2} \Rightarrow x^{4} + y^{4} = {(x^{2} + y^{2})}^{2} - 2 {(x y)}^{2} \Rightarrow x^{4} + y^{4} = 29^{2} - 2 {(2)}^{2} (∵ x^{2} + y^{2} = 29 and x y = 2) \Rightarrow x^{4} + y^{4} = 841 - 8 \Rightarrow x^{4} + y^{4} = 833$

Page No 6.44:

Question 18:

What must be added to each of the following expressions to make it a whole square?
(i) 4x² − 12x + 7
(ii) 4x²− 20x + 20

Answer:

(i) Let us consider the following expression:

$4 x^{2} - 12 x + 7$
The above expression can be written as:

$4 x^{2} - 12 x + 7 = {(2 x)}^{2} - 2 \times 2 x \times 3 + 7$
It is evident that if 2x is considered as the first term and 3 is considered as the second term, 2 is required to be added to the above expression to make it a perfect square. Therefore, 7 must become 9.

Therefore, adding and subtracting 2 in the above expression, we get:

$(4 x^{2} - 12 x + 7) + 2 - 2 = \{{(2 x)}^{2} - 2 \times 2 x \times 3 + 7\} + 2 - 2 = \{{(2 x)}^{2} - 2 \times 2 x \times 3 + 9\} - 2 = {(2 x + 3)}^{2} - 2$
Thus, the answer is 2.

(ii) Let's consider the following expression:

$4 x^{2} - 20 x + 20$
The above expression can be written as:

$4 x^{2} - 20 x + 20 = {(2 x)}^{2} - 2 \times 2 x \times 5 + 20$
It is evident that if 2x is considered as the first term and 5 is considered as the second term, 5 is required to be added to the above expression to make it a perfect square. Therefore, number 20 must become 25.

Therefore, adding and subtracting 5 in the above expression, we get:

$(4 x^{2} - 20 x + 20 + 5) - 5 = \{{(2 x)}^{2} - 2 \times 2 x \times 5 + 20\} + 5 - 5 = \{{(2 x)}^{2} - 2 \times 2 x \times 5 + 25\} - 5 = {(2 x + 5)}^{2} - 5$
Thus, the answer is 5.

Page No 6.44:

Question 19:

Simplify:
(i) (x − y)(x + y) (x² + y²)(x⁴ + y²)
(ii) (2x − 1)(2x + 1)(4x² + 1)(16x⁴ + 1)
(iii) (4m − 8n)² + (7m + 8n)²
(iv) (2.5p − 1.5q)² − (1.5p − 2.5q)²
(v) (m² − n²m)² + 2m³n²

Answer:

To simplify, we will proceed as follows:
(i)
$(i) (x - y) (x + y) (x^{2} + y^{2}) (x^{4} + y^{4}) = (x^{2} - y^{2}) (x^{2} + y^{2}) (x^{4} + y^{4}) [∵ (a + b) (a - b) = a^{2} - b^{2}] = (x^{4} - y^{4}) (x^{4} + y^{4}) [∵ (a + b) (a - b) = a^{2} - b^{2}] = x^{8} - x^{8} [∵ (a + b) (a - b) = a^{2} - b^{2}]$

$(ii) (2 x - 1) (2 x + 1) (4 x^{2} + 1) (16 x^{4} + 1) = ({(2 x)}^{2} - 1^{2}) (4 x^{2} + 1) (16 x^{4} + 1) [∵ (a + b) (a - b) = a^{2} - b^{2}] = (4 x^{2} - 1) (4 x^{2} + 1) (16 x^{4} + 1) = \{{(4 x^{2})}^{2} - {(1^{2})}^{2}\} (16 x^{4} + 1) [∵ (a + b) (a - b) = a^{2} - b^{2}] = (16 x^{4} - 1) (16 x^{4} + 1) = {(16 x^{4})}^{2} - 1^{2} [∵ (a + b) (a - b) = a^{2} - b^{2}] = 256 x^{8} - 1$

$(iii) {(7 m - 8 n)}^{2} + {(7 m + 8 n)}^{2} = 2 {(7 m)}^{2} + 2 {(8 n)}^{2} [∵ {(a - b)}^{2} + {(a + b)}^{2} = 2 a^{2} + 2 b^{2}] = 98 m^{2} + 128 n^{2}$

$(iv) {(2.5 p - 1.5 q)}^{2} - {(1.5 p - 2.5 q)}^{2} = {(2.5 p)}^{2} + {(1.5 q)}^{2} - 2 (2.5 p) (1.5 q) - [{(1.5 p)}^{2} + {(2.5 q)}^{2} - 2 (1.5 p) (2.5 q)] = {(2.5 p)}^{2} + {(1.5 q)}^{2} - 2 (2.5 p) (1.5 q) - {(1.5 p)}^{2} - {(2.5 q)}^{2} + 2 (1.5 p) (2.5 q) = {(2.5 p)}^{2} - {(1.5 p)}^{2} + {(1.5 q)}^{2} - {(2.5 q)}^{2} = [(2.5 p + 1.5 p) (2.5 p - 1.5 p)] + [(1.5 q + 2.5) (1.5 q - 2.5 q)] [∵ (a + b) (a - b) = a^{2} - b^{2}] = 4 p \times p + 4 q \times (- q) = 4 p^{2} - 4 q^{2} = 4 (p^{2} - 4 q^{2})$

$v) {(m^{2} - n^{2} m)}^{2} + 2 m^{3} n^{2} = {(m^{2})}^{2} + {(n^{2} m)}^{2} [∵ {(a - b)}^{2} + 2 a b = a^{2} + b^{2}] = m^{4} + n^{4} m^{2}$

Page No 6.44:

Question 20:

Show that:
(i) (3x + 7)² − 84x = (3x − 7)²
(ii) (9a − 5b)² + 180ab = (9a + 5b)²
(iii) ${(\frac{4 m}{3} - \frac{3 n}{4})}^{2} + 2 m n = \frac{16 m^{2}}{9} + \frac{9 n^{2}}{16}$
(iv) (4pq + 3q)² − (4pq − 3q)² = 48pq²
(v) (a − b)(a + b) + (b − c)(b + c) + (c − a)( c + a) = 0

Answer:

$(i) LHS = {(3 x + 7)}^{2} - 84 x = {(3 x + 7)}^{2} - 4 \times 3 x \times 7 = {(3 x - 7)}^{2} [∵ {(a + b)}^{2} - 4 a b = {(a - b)}^{2}] = RHS Because LHS is equal to RHS, the given equation is verified.$

$(ii) LHS = {(9 a - 5 b)}^{2} + 180 a b = {(9 a - 5 b)}^{2} + 4 \times 9 a \times 5 b = {(9 a + 5 b)}^{2} [∵ {(a - b)}^{2} + 4 a b = {(a + b)}^{2}] = RHS Because LHS is equal to RHS, the given equation is verified.$

$(iii) LHS = {(\frac{4 m}{3} - \frac{3 n}{4})}^{2} + 2 m n = {(\frac{4 m}{3} - \frac{3 n}{4})}^{2} + 2 \times \frac{4 m}{3} \times \frac{3 n}{4} = {(\frac{4 m}{3})}^{2} + {(\frac{3 n}{4})}^{2} [∵ {(a - b)}^{2} + 2 a b = a^{2} + b^{2}] = \frac{16 m^{2}}{9} + \frac{9 n^{2}}{16} = RHS Because LHS is equal to RHS, the given equation is verified.$

$(iv) LHS = {(4 p q + 3 q)}^{2} - {(4 p q - 3 q)}^{2} = 4 (4 p q) (3 q) [∵ {(a + b)}^{2} - {(a + b)}^{2} = 4 a b] = 48 p q^{2} = RHS Because LHS is equal to RHS, the given equation is verified.$

$(v) LHS = (a - b) (a + b) + (b - c) (b + c) + (c + a) (c - a) = a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - a^{2} [∵ (a + b) (a - b) = a^{2} - b^{2}] = a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - a^{2} = 0 = RHS Because LHS is equal to RHS, the given equation is verified.$

Page No 6.47:

Question 1:

Find the following products:
(i) (x + 4) (x + 7)
(ii) (x − 11) (x + 4)
(iii) (x + 7) (x − 5)
(iv) (x − 3) ( x − 2)
(v) (y² − 4) (y² − 3)
(vi) $(x + \frac{4}{3}) (x + \frac{3}{4})$
(vii) (3x + 5) (3x + 11)
(viii) (2x² − 3) (2x² + 5)
(ix) (z² + 2) (z² − 3)
(x) (3x − 4y) (2x − 4y)
(xi) (3x² − 4xy) (3x² − 3xy)
(xii) $(x + \frac{1}{5}) (x + 5)$
(xiii) $(z + \frac{3}{4}) (z + \frac{4}{3})$
(xiv) (x² + 4) (x² + 9)
(xv) (y²+ 12) (y²+ 6)
(xvi) $(y^{2} + \frac{5}{7}) (y^{2} - \frac{14}{5})$
(xvii) (p² + 16) $(p^{2} - \frac{1}{4})$

Answer:

(i) Here, we will use the identity $(x + a) (x + b) = x^{2} + (a + b) x + a b .$
$(x + 4) (x + 7) = x^{2} + (4 + 7) x + 4 \times 7 = x^{2} + 11 x + 28$

(ii) Here, we will use the identity $(x - a) (x + b) = x^{2} + (b - a) x - a b$ .
$(x - 11) (x + 4) = x^{2} + (4 - 11) x - 11 \times 4 = x^{2} - 7 x - 44$

(iii) Here, we will use the identity $(x + a) (x - b) = x^{2} + (a - b) x - a b$ .
$(x + 7) (x - 5) = x^{2} + (7 - 5) x - 7 \times 5 = x^{2} + 2 x - 35$

(iv) Here, we will use the identity $(x - a) (x - b) = x^{2} - (a + b) x + a b$ .
$(x - 3) (x - 2) = x^{2} - (3 + 2) x + 3 \times 2 = x^{2} - 5 x + 6$

(v) Here, we will use the identity $(x - a) (x - b) = x^{2} - (a + b) x + a b$ .
$(y^{2} - 4) (y^{2} - 3) = {(y^{2})}^{2} - (4 + 3) (y^{2}) + 4 \times 3 = y^{4} - 7 y^{2} + 12$

(vi) Here, we will use the identity $(x + a) (x + b) = x^{2} + (a + b) x + a b$ .
$(x + \frac{4}{3}) (x + \frac{3}{4}) = x^{2} + (\frac{4}{3} + \frac{3}{4}) x + \frac{4}{3} \times \frac{3}{4} = x^{2} + \frac{25}{12} x + 1$

(vii) Here, we will use the identity $(x + a) (x + b) = x^{2} + (a + b) x + a b$ .
$(3 x + 5) (3 x + 11) = {(3 x)}^{2} + (5 + 11) (3 x) + 5 \times 11 = 9 x^{2} + 48 x + 55$

(viii) Here, we will use the identity $(x - a) (x + b) = x^{2} + (b - a) x - a b$ .
$(2 x^{2} - 3) (2 x^{2} + 5) = {(2 x^{2})}^{2} + (5 - 3) (2 x^{2}) - 3 \times 5 = 4 x^{4} + 4 x^{2} - 15$

(ix) Here, we will use the identity $(x + a) (x - b) = x^{2} + (a - b) x - a b$ .
$(z^{2} + 2) (z^{2} - 3) = {(z^{2})}^{2} + (2 - 3) (z^{2}) - 2 \times 3 = z^{4} - z^{2} - 6$

(x) Here, we will use the identity $(x - a) (x - b) = x^{2} - (a + b) x + a b$ .
$(3 x - 4 y) (2 x - 4 y) = (4 y - 3 x) (4 y - 2 x) (Taking common - 1 from both parentheses) = {(4 y)}^{2} - (3 x + 2 x) (4 y) + 3 x \times 2 x = 16 y^{2} - (12 x y + 8 x y) + 6 x^{2} = 16 y^{2} - 20 x y + 6 x^{2}$

(xi) Here, we will use the identity $(x - a) (x - b) = x^{2} - (a + b) x + a b$ .

$(3 x^{2} - 4 x y) (3 x^{2} - 3 x y) = {(3 x^{2})}^{2} - (4 x y + 3 x y) (3 x^{2}) + 4 x y \times 3 x y = 9 x^{4} - (12 x^{3} y + 9 x^{3} y) + 12 x^{2} y^{2} = 9 x^{4} - 21 x^{3} y + 12 x^{2} y^{2}$

(xii) Here, we will use the identity $(x + a) (x + b) = x^{2} + (a + b) x + a b$ .
$(x + \frac{1}{5}) (x + 5) = x^{2} + (\frac{1}{5} + 5) x + \frac{1}{5} \times 5 = x^{2} + \frac{26}{5} x + 1$

(xiii) Here, we will use the identity $(x + a) (x + b) = x^{2} + (a + b) x + a b$ .
$(z + \frac{3}{4}) (z + \frac{4}{3}) = z^{2} + (\frac{3}{4} + \frac{4}{3}) x + \frac{3}{4} \times \frac{4}{3} = z^{2} + \frac{25}{12} z + 1$

(xiv) Here, we will use the identity $(x + a) (x + b) = x^{2} + (a + b) x + a b$ .
$(x^{2} + 4) (x^{2} + 9) = {(x^{2})}^{2} + (4 + 9) (x^{2}) + 4 \times 9 = x^{4} + 13 x^{2} + 36$

(xv) Here, we will use the identity $(x + a) (x + b) = x^{2} + (a + b) x + a b$ .
$(y^{2} + 12) (y^{2} + 6) = {(y^{2})}^{2} + (12 + 6) (y^{2}) + 12 \times 6 = y^{4} + 18 y^{2} + 72$

(xvi) Here, we will use the identity $(x + a) (x - b) = x^{2} + (a - b) x - a b$ .
$(y^{2} + \frac{5}{7}) (y^{2} - \frac{14}{5}) = {(y^{2})}^{2} + (\frac{5}{7} - \frac{14}{5}) (y^{2}) - \frac{5}{7} \times \frac{14}{5} = y^{4} - \frac{73}{35} y^{2} - 2$

(xvii) Here, we will use the identity $(x + a) (x - b) = x^{2} + (a - b) x - a b$ .
$(p^{2} + 16) (p^{2} - \frac{1}{4}) = {(p^{2})}^{2} + (16 - \frac{1}{4}) (p^{2}) - 16 \times \frac{1}{4} = p^{4} + \frac{63}{4} p^{2} - 4$

Page No 6.47:

Question 2:

Evaluate the following:
(i) 102 × 106
(ii) 109 × 107
(iii) 35 × 37
(iv) 53 × 55
(v) 103 × 96
(vi) 34 × 36
(vii) 994 × 1006

Answer:

(i) Here, we will use the identity $(x + a) (x + b) = x^{2} + (a + b) x + a b$
$102 \times 106 = (100 + 2) (100 + 6) = 100^{2} + (2 + 6) 100 + 2 \times 6 = 10000 + 800 + 12 = 10812$

(ii) Here, we will use the identity $(x + a) (x + b) = x^{2} + (a + b) x + a b$
$109 × 107 = (100 + 9) (100 + 7) = 100^{2} + (9 + 7) 100 + 9 \times 7 = 10000 + 1600 + 63 = 11663$

(iii) Here, we will use the identity $(x + a) (x + b) = x^{2} + (a + b) x + a b$
$35 × 37 = (30 + 5) (30 + 7) = 30^{2} + (5 + 7) 30 + 5 \times 7 = 900 + 360 + 35 = 1295$

(iv) Here, we will use the identity $(x + a) (x + b) = x^{2} + (a + b) x + a b$
$53 × 55 = (50 + 3) (50 + 5) = 50^{2} + (3 + 5) 50 + 3 \times 5 = 2500 + 400 + 15 = 2915$

(v) Here, we will use the identity $(x + a) (x - b) = x^{2} + (a - b) x - a b$
$103 × 96 = (100 + 3) (100 - 4) = 100^{2} + (3 - 4) 100 - 3 \times 4 = 10000 - 100 - 12 = 9888$

(vi) Here, we will use the identity $(x + a) (x + b) = x^{2} + (a + b) x + a b$
$34 × 36 = (30 + 4) (30 + 6) = 30^{2} + (4 + 6) 30 + 4 \times 6 = 900 + 300 + 24 = 1224$

(vii) Here, we will use the identity $(x - a) (x + b) = x^{2} + (b - a) x - a b$
$994 × 1006 = (1000 - 6) \times (1000 + 6) = 1000^{2} + (6 - 6) \times 1000 - 6 \times 6 = 1000000 - 36 = 999964$

Page No 6.5:

Question 1:

Add the following algebraic expressions:
(i) 3a²b, − 4a²b, 9a²b
(ii) $\frac{2}{3} a, \frac{3}{5} a, - \frac{6}{5} a$
(iii) 4xy² − 7x²y, 12x²y − 6xy², − 3x²y +5xy²
(iv) $\frac{3}{2} a - \frac{5}{4} b + \frac{2}{5} c, \frac{2}{3} a - \frac{7}{2} b + \frac{7}{2} c, \frac{5}{3} a + \frac{5}{2} b - \frac{5}{4} c$
(v) $\frac{11}{2} x y + \frac{12}{5} y + \frac{13}{7} x, - \frac{11}{2} y - \frac{12}{5} x - \frac{13}{7} x y$
(vi) $\frac{7}{2} x^{3} - \frac{1}{2} x^{2} + \frac{5}{3}, \frac{3}{2} x^{3} + \frac{7}{4} x^{2} - x + \frac{1}{3}, \frac{3}{2} x^{2} - \frac{5}{2} x - 2$

Answer:

(i) To add the like terms, we proceed as follows:

$3 a^{2} b + (- 4 a^{2} b) + 9 a^{2} b = 3 a^{2} b - 4 a^{2} b + 9 a^{2} b = (3 - 4 + 9) a^{2} b (Distributive law) = 8 a^{2} b$

(ii) To add the like terms, we proceed as follows:

$\frac{2}{3} a + \frac{3}{5} a + (- \frac{6}{5} a) = \frac{2}{3} a + \frac{3}{5} a - \frac{6}{5} a = (\frac{2}{3} + \frac{3}{5} - \frac{6}{5}) a (Distributive law) = \frac{1}{15} a$

(iii) To add, we proceed as follows:

$(4 {xy}^{2} - 7 x^{2} y) + (12 x^{2} y) + (- 6 {xy}^{2}) + (- 3 x^{2} y + 5 {xy}^{2}) = 4 {xy}^{2} - 7 x^{2} y + 12 x^{2} y - 6 {xy}^{2} - 3 x^{2} y + 5 {xy}^{2} = 4 {xy}^{2} - 6 {xy}^{2} + 5 {xy}^{2} - 7 x^{2} y + 12 x^{2} y - 3 x^{2} y (Collecting like terms) = 3 {xy}^{2} + 2 x^{2} y (Combining like terms)$

(iv) To add, we proceed as follows:

$(\frac{3}{2} a - \frac{5}{4} b + \frac{2}{5} c) + (\frac{2}{3} a - \frac{7}{2} b + \frac{7}{2} c) + (\frac{5}{3} a + \frac{5}{2} b - \frac{5}{4} c) = \frac{3}{2} a - \frac{5}{4} b + \frac{2}{5} c + \frac{2}{3} a - \frac{7}{2} b + \frac{7}{2} c + \frac{5}{3} a + \frac{5}{2} b - \frac{5}{4} c = \frac{3}{2} a + \frac{2}{3} a + \frac{5}{3} a - \frac{5}{4} b - \frac{7}{2} b + \frac{5}{2} b + \frac{2}{5} c + \frac{7}{2} c - \frac{5}{4} c (Collecting like terms) = \frac{23}{6} a - \frac{9}{4} b + \frac{53}{20} c (Combining like terms)$

(v) To add, we proceed as follows:

$(\frac{11}{2} x y + \frac{12}{5} y + \frac{13}{7} x) + (- \frac{11}{2} y - \frac{12}{5} x - \frac{13}{7} x y) = \frac{11}{2} x y + \frac{12}{5} y + \frac{13}{7} x - \frac{11}{2} y - \frac{12}{5} x - \frac{13}{7} x y = \frac{11}{2} x y - \frac{13}{7} x y + \frac{12}{5} y - \frac{11}{2} y + \frac{13}{7} x - \frac{12}{5} x (Collecting like terms) = \frac{51}{14} x y - \frac{31}{10} y - \frac{19}{35} x (Combining like terms)$

(vi) To add, we proceed as follows:

$(\frac{7}{2} x^{3} - \frac{1}{2} x^{2} + \frac{5}{3}) + (\frac{3}{2} x^{3} + \frac{7}{4} x^{2} - x + \frac{1}{3}) + (\frac{3}{2} x^{2} - \frac{5}{2} x - 2) = \frac{7}{2} x^{3} - \frac{1}{2} x^{2} + \frac{5}{3} + \frac{3}{2} x^{3} + \frac{7}{4} x^{2} - x + \frac{1}{3} + \frac{3}{2} x^{2} - \frac{5}{2} x - 2 = \frac{7}{2} x^{3} + \frac{3}{2} x^{3} - \frac{1}{2} x^{2} + \frac{7}{4} x^{2} + \frac{3}{2} x^{2} - x - \frac{5}{2} x + \frac{5}{3} + \frac{1}{3} - 2 (Collecting like terms) = 5 x^{3} + \frac{11}{4} x^{2} - \frac{7}{2} x (Combining like terms)$

Page No 6.5:

Question 2:

Subtract:
(i) − 5xy from 12xy
(ii) 2a² from − 7a²
(iii) 2a − b from 3a − 5b
(iv) 2x³ − 4x² + 3x + 5 from 4x³ + x² + x + 6
(v) $\frac{2}{3} y^{3} - \frac{2}{7} y^{2} - 5 from \frac{1}{3} y^{3} + \frac{5}{7} y^{2} + y - 2$
(vi) $\frac{3}{2} x - \frac{5}{4} y - \frac{7}{2} z from \frac{2}{3} x + \frac{3}{2} y - \frac{4}{3} z$
(vii) $x^{2} y - \frac{4}{5} x y^{2} + \frac{4}{3} x y from \frac{2}{3} x^{2} y + \frac{3}{2} x y^{2} - \frac{1}{3} x y$
(viii) $\frac{a b}{7} - \frac{35}{3} b c + \frac{6}{5} a c from \frac{3}{5} b c - \frac{4}{5} a c$

Answer:

$(i) 12 x y - (- 5 x y) = 12 x y + 5 x y = 17 x y$

$(ii) - 7 a^{2} - (2 a^{2}) = - 7 a^{2} - 2 a^{2} = - 9 a^{2}$

$(iii) (3 a - 5 b) - (2 a - b) = (3 a - 5 b) - 2 a + b = 3 a - 5 b - 2 a + b = 3 a - 2 a - 5 b + b = a - 4 b$

$(iv) (4 x^{3} + x^{2} + x + 6) - (2 x^{3} - 4 x^{2} + 3 x + 5) = 4 x^{3} + x^{2} + x + 6 - 2 x^{3} + 4 x^{2} - 3 x - 5 = 4 x^{3} - 2 x^{3} + x^{2} + 4 x^{2} + x - 3 x + 6 - 5 (Collecting like terms) = 2 x^{3} + 5 x^{2} - 2 x + 1 (Combining like terms)$

$(v) (\frac{1}{3} y^{3} + \frac{5}{7} y^{2} + y - 2) - (\frac{2}{3} y^{3} - \frac{2}{7} y^{2} - 5) = \frac{1}{3} y^{3} + \frac{5}{7} y^{2} + y - 2 - \frac{2}{3} y^{3} + \frac{2}{7} y^{2} + 5 = \frac{1}{3} y^{3} - \frac{2}{3} y^{3} + \frac{5}{7} y^{2} + \frac{2}{7} y^{2} + y - 2 + 5 (Collecting like terms) = - \frac{1}{3} y^{3} + y^{2} + y + 3 (Combining like terms)$

$(vi) (\frac{2}{3} x + \frac{3}{2} y - \frac{4}{3} z) - (\frac{3}{2} x - \frac{5}{4} y - \frac{7}{2} z) = \frac{2}{3} x + \frac{3}{2} y - \frac{4}{3} z - \frac{3}{2} x + \frac{5}{4} y + \frac{7}{2} z = \frac{2}{3} x - \frac{3}{2} x + \frac{3}{2} y + \frac{5}{4} y - \frac{4}{3} z + \frac{7}{2} z (Collecting like terms) = - \frac{5}{6} x + \frac{11}{4} y + \frac{13}{6} z (Combining like terms)$

$(vii) (\frac{2}{3} x^{2} y + \frac{3}{2} x y^{2} - \frac{1}{3} x y) - (x^{2} y - \frac{4}{5} x y^{2} + \frac{4}{3} x y) = \frac{2}{3} x^{2} y + \frac{3}{2} x y^{2} - \frac{1}{3} x y - x^{2} y + \frac{4}{5} x y^{2} - \frac{4}{3} x y = \frac{2}{3} x^{2} y - x^{2} y + \frac{3}{2} x y^{2} + \frac{4}{5} x y^{2} - \frac{1}{3} x y - \frac{4}{3} x y (Collecting like terms) = - \frac{1}{3} x^{2} y + \frac{23}{10} x y^{2} - \frac{5}{3} x y (Combining like terms)$

$(viii) (\frac{3}{5} b c - \frac{4}{5} a c) - (\frac{a b}{7} - \frac{35}{3} b c + \frac{6}{5} a c) = \frac{3}{5} b c - \frac{4}{5} a c - \frac{a b}{7} + \frac{35}{3} b c - \frac{6}{5} a c = \frac{3}{5} b c + \frac{35}{3} b c - \frac{4}{5} a c - \frac{6}{5} a c - \frac{a b}{7} (Collecting like terms) = \frac{184}{15} b c - 2 a c - \frac{a b}{7} (Combining like terms)$

Page No 6.5:

Question 3:

Take away:
(i) $\frac{6}{5} x^{2} - \frac{4}{5} x^{3} + \frac{5}{6} + \frac{3}{2} x from \frac{x^{3}}{3} - \frac{5}{2} x^{2} + \frac{3}{5} x + \frac{1}{4}$
(ii) $\frac{5 a^{2}}{2} + \frac{3 a^{3}}{2} + \frac{a}{3} - \frac{6}{5} from \frac{1}{3} a^{3} - \frac{3}{4} a^{2} - \frac{5}{2}$
(iii) $\frac{7}{4} x^{3} + \frac{3}{5} x^{2} + \frac{1}{2} x + \frac{9}{2} from \frac{7}{2} - \frac{x}{3} - \frac{x^{2}}{5}$
(iv) $\frac{y^{3}}{3} + \frac{7}{3} y^{2} + \frac{1}{2} y + \frac{1}{2} from \frac{1}{3} - \frac{5}{3} y^{2}$
(v) $\frac{2}{3} a c - \frac{5}{7} a b + \frac{2}{3} b c from \frac{3}{2} a b - \frac{7}{4} a c - \frac{5}{6} b c$

Answer:

(i) The difference is given by:

$(\frac{x^{3}}{3} - \frac{5}{2} x^{2} + \frac{3}{5} x + \frac{1}{4}) - (\frac{6}{5} x^{2} - \frac{4}{5} x^{3} + \frac{5}{6} + \frac{3}{2} x) = \frac{x^{3}}{3} - \frac{5}{2} x^{2} + \frac{3}{5} x + \frac{1}{4} - \frac{6}{5} x^{2} + \frac{4}{5} x^{3} - \frac{5}{6} - \frac{3}{2} x$
$= \frac{x^{3}}{3} + \frac{4}{5} x^{3} - \frac{5}{2} x^{2} - \frac{6}{5} x^{2} + \frac{3}{5} x - \frac{3}{2} x + \frac{1}{4} - \frac{5}{6}$              (Collecting like terms)
= $(\frac{5 + 12}{15}) x^{3} + (\frac{- 25 - 12}{10}) x^{2} + (\frac{6 - 15}{10}) x + (\frac{6 - 20}{24})$
$= \frac{17}{15} x^{3} - \frac{37}{10} x^{2} - \frac{9}{10} x - \frac{7}{12}$                                             (Combining like terms)

(ii) The difference is given by:

$(\frac{1}{3} a^{3} - \frac{3 a^{2}}{4} - \frac{5}{2}) - (\frac{5 a^{2}}{2} + \frac{3 a^{3}}{2} + \frac{a}{3} - \frac{6}{5}) = \frac{1}{3} a^{3} - \frac{3 a^{2}}{4} - \frac{5}{2} - \frac{5 a^{2}}{2} - \frac{3 a^{3}}{2} - \frac{a}{3} + \frac{6}{5}$
$= \frac{1}{3} a^{3} - \frac{3 a^{3}}{2} - \frac{3 a^{2}}{4} - \frac{5 a^{2}}{2} - \frac{a}{3} - \frac{5}{2} + \frac{6}{5}$                       (Collecting like terms)
= $(\frac{2 - 9}{6}) a^{3} + (\frac{- 3 - 10}{4}) a^{2} - \frac{a}{3} + (\frac{- 25 + 12}{10})$
$= - \frac{7}{6} a^{3} - \frac{13}{4} a^{2} - \frac{a}{3} - \frac{13}{10}$                                              (Combining like terms)

(iii) The difference is given by:

$(\frac{7}{2} - \frac{x}{3} - \frac{x^{2}}{5}) - (\frac{7 x^{3}}{4} + \frac{3 x^{2}}{5} + \frac{x}{2} + \frac{9}{2}) = \frac{7}{2} - \frac{x}{3} - \frac{x^{2}}{5} - \frac{7 x^{3}}{4} - \frac{3 x^{2}}{5} - \frac{x}{2} - \frac{9}{2}$
$= \frac{7}{2} - \frac{9}{2} - \frac{x}{3} - \frac{x}{2} - \frac{x^{2}}{5} - \frac{3 x^{2}}{5} - \frac{7 x^{3}}{4}$                      (Collecting like terms)
= $(\frac{7 - 9}{2}) + (\frac{- 2 - 3}{6}) x + (\frac{- 1 - 3}{5}) x^{2} - \frac{7 x^{3}}{4}$
$= - 1 - \frac{5 x}{6} - \frac{4 x^{2}}{5} - \frac{7 x^{3}}{4}$                                            (Combining like terms )

(iv) The difference is given by:

$(\frac{1}{3} - \frac{5}{3} y^{2}) - (\frac{y^{3}}{3} + \frac{7 y^{2}}{3} + \frac{y}{2} + \frac{1}{2}) = \frac{1}{3} - \frac{5}{3} y^{2} - \frac{y^{3}}{3} - \frac{7 y^{2}}{3} - \frac{y}{2} - \frac{1}{2}$
$= \frac{1}{3} - \frac{1}{2} - \frac{y}{2} - \frac{5}{3} y^{2} - \frac{7 y^{2}}{3} - \frac{y^{3}}{3}$              ( Collecting like terms)
= $(\frac{2 - 3}{6}) - \frac{y}{2} + (\frac{- 5 - 7}{3}) y^{2} - \frac{y^{3}}{3}$
$= - \frac{1}{6} - \frac{y}{2} - 4 y^{2} - \frac{y^{3}}{3}$                                (Combining like terms. )

(v) The difference is given by:

$(\frac{3}{2} a b - \frac{7}{4} a c - \frac{5}{6} b c) - (\frac{2}{3} a c - \frac{5}{7} a b + \frac{2}{3} b c) = \frac{3}{2} a b - \frac{7}{4} a c - \frac{5}{6} b c - \frac{2}{3} a c + \frac{5}{7} a b - \frac{2}{3} b c$

$= \frac{3}{2} a b + \frac{5}{7} a b - \frac{7}{4} a c - \frac{2}{3} a c - \frac{5}{6} b c - \frac{2}{3} b c$              (Collecting like terms )
= $(\frac{21 + 10}{14}) a b + (\frac{- 21 - 8}{12}) a c + (\frac{- 5 - 4}{6}) b c$
$= \frac{31}{14} a b - \frac{29}{12} a c - \frac{3}{2} b c$                                  (Combining like terms )

Page No 6.6:

Question 4:

Subtract 3x − 4y − 7z from the sum of x − 3y + 2z and − 4x + 9y − 11z.

Answer:

Let first add the expressions $x - 3 y + 2 z and - 4 x + 9 y - 11 z$ . We get:

$(x - 3 y + 2 z) + (- 4 x + 9 y - 11 z)$
$= x - 3 y + 2 z - 4 x + 9 y - 11 z$
$= x - 4 x - 3 y + 9 y + 2 z - 11 z$            (Collecting like terms)
$= - 3 x + 6 y - 9 z$                                        (Combining like terms)

Now, subtracting the expression $3 x - 4 y - 7 z$ from the above sum; we get:

$(- 3 x + 6 y - 9 z) - (3 x - 4 y - 7 z) = - 3 x + 6 y - 9 z - 3 x + 4 y + 7 z$
$= - 3 x - 3 x + 6 y + 4 y - 9 z + 7 z$           (Collecting like terms)
$= - 6 x + 10 y - 2 z$                             (Combining like terms)

Thus, the answer is $- 6 x + 10 y - 2 z$ .

Page No 6.6:

Question 5:

Subtract the sum of 3l − 4m − 7n² and 2l + 3m − 4n² from the sum of 9l + 2m − 3n² and − 3l + m + 4n² .....

Answer:

We have to subtract the sum of (3l $-$ 4m $-$ 7n²) and (2l + 3m $-$ 4n²) from the sum of (9l + 2m $-$ 3n²) and ( $-$ 3l + m + 4n²)

$\{(9 l + 2 m - 3 n^{2}) + (- 3 l + m + 4 n^{2})\} - \{(3 l - 4 m - 7 n^{2}) + (2 l + 3 m - 4 n^{2})\}$
$= (9 l - 3 l + 2 m + m - 3 n^{2} + 4 n^{2}) - (3 l + 2 l - 4 m + 3 m - 7 n^{2} - 4 n^{2})$
$= (6 l + 3 m + n^{2}) - (5 l - m - 11 n^{2})$                                           (Combining like terms inside the parentheses)
$= 6 l + 3 m + n^{2} - 5 l + m + 11 n^{2}$
$= 6 l - 5 l + 3 m + m + n^{2} + 11 n^{2}$                                                         (Collecting like terms)
$= l + 4 m + 12 n^{2}$                                                                               (Combining like terms)

Thus, the required solution is $l + 4 m + 12 n^{2}$ .

Page No 6.6:

Question 6:

Subtract the sum of 2x − x² + 5 and − 4x − 3 + 7x² from 5.

Answer:

We have to subtract the sum of (2x − x² + 5) and (− 4x − 3 + 7x) from 5.

$5 - \{(2 x - x^{2} + 5) + (- 4 x - 3 + 7 x^{2})\} = 5 - (2 x - 4 x - x^{2} + 7 x^{2} + 5 - 3) = 5 - 2 x + 4 x + x^{2} - 7 x^{2} - 5 + 3$
$= 5 - 5 + 3 - 2 x + 4 x + x^{2} - 7 x^{2}$ (Collecting like terms)
$= 3 + 2 x - 6 x^{2}$ (Combining like terms)

Thus, the answer is $3 + 2 x - 6 x^{2}$ .

Page No 6.6:

Question 7:

Simplify each of the following:
(i) x² − 3x + 5 − $\frac{1}{2}$ (3x² − 5x + 7)
(ii) [5 − 3x + 2y − (2x − y)] − (3x − 7y + 9)
(iii) $\frac{11}{2} x^{2} y - \frac{9}{4} x y^{2} + \frac{1}{4} x y - \frac{1}{14} y^{2} x + \frac{1}{15} y x^{2} + \frac{1}{2} x y$
(iv) $(\frac{1}{3} y^{2} - \frac{4}{7} y + 11) - (\frac{1}{7} y - 3 + 2 y^{2}) - (\frac{2}{7} y - \frac{2}{3} y^{2} + 2)$
(v) $- \frac{1}{2} a^{2} b^{2} c + \frac{1}{3} a b^{2} c - \frac{1}{4} a b c^{2} - \frac{1}{5} c b^{2} a^{2} + \frac{1}{6} c b^{2} a - \frac{1}{7} c^{2} a b + \frac{1}{8} c a^{2} b .$

Answer:

$(i) x^{2} - 3 x + 5 - \frac{1}{2} (3 x^{2} - 5 x + 7) = x^{2} - 3 x + 5 - \frac{3 x^{2}}{2} + \frac{5 x}{2} - \frac{7}{2} = x^{2} - \frac{3 x^{2}}{2} - 3 x + \frac{5 x}{2} + 5 - \frac{7}{2} (Collecting like terms) = (\frac{1 - 3}{2}) x^{2} + (\frac{- 3 + 5}{2}) x + (\frac{10 - 7}{2}) = - \frac{x^{2}}{2} - \frac{x}{2} + \frac{3}{2}$

Thus, the answer is $- \frac{x^{2}}{2} - \frac{x}{2} + \frac{3}{2}$ .

$(ii) [5 - 3 x + 2 y - (2 x - y)] - (3 x - 7 y + 9) = [5 - 3 x + 2 y - 2 x + y] - (3 x - 7 y + 9) = [5 - 5 x + 3 y] - (3 x - 7 y + 9) = 5 - 5 x + 3 y - 3 x + 7 y - 9 = 5 - 9 - 5 x - 3 x + 3 y + 7 y = - 4 - 8 x + 10 y$

$(iii) \frac{11}{2} x^{2} y - \frac{9}{4} x y^{2} + \frac{1}{4} x y - \frac{1}{14} y^{2} x + \frac{1}{15} y x^{2} + \frac{1}{2} x y$
$= \frac{11}{2} x^{2} y + \frac{1}{15} y x^{2} - \frac{9}{4} x y^{2} - \frac{1}{14} y^{2} x + \frac{1}{4} x y + \frac{1}{2} x y$            (Collecting like terms)
= $(\frac{165 + 2}{30}) x^{2} y + (\frac{- 63 - 2}{28}) x y^{2} + (\frac{1 + 2}{4}) x y$
$= \frac{167}{30} x^{2} y - \frac{65}{28} y^{2} x + \frac{3}{2} x y$                                                 (Combining like terms)

$(iv) (\frac{1}{3} y^{2} - \frac{4}{7} y + 11) - (\frac{1}{7} y - 3 + 2 y^{2}) - (\frac{2}{7} y - \frac{2}{3} y^{2} + 2) = \frac{1}{3} y^{2} - \frac{4}{7} y + 11 - \frac{1}{7} y + 3 - 2 y^{2} - \frac{2}{7} y + \frac{2}{3} y^{2} - 2$

$= \frac{1}{3} y^{2} - 2 y^{2} + \frac{2}{3} y^{2} - \frac{4}{7} y - \frac{1}{7} y - \frac{2}{7} y + 11 + 3 - 2$         (Collecting like terms)
= $(\frac{1 - 6 + 2}{3}) y^{2} + (\frac{- 4 - 1 - 2}{7}) y + 12$
$= - y^{2} - 7 y + 12$                                            (Combining like terms)

$(v) - \frac{1}{2} a^{2} b^{2} c + \frac{1}{3} a b^{2} c - \frac{1}{4} a b c^{2} - \frac{1}{5} c b^{2} a^{2} + \frac{1}{6} c b^{2} a - \frac{1}{7} c^{2} a b + \frac{1}{8} c a^{2} b$

$= - \frac{1}{2} a^{2} b^{2} c - \frac{1}{5} c b^{2} a^{2} + \frac{1}{3} a b^{2} c + \frac{1}{6} c b^{2} a - \frac{1}{4} a b c^{2} - \frac{1}{7} c^{2} a b + \frac{1}{8} c a^{2} b$            (Collecting like terms)
= $(\frac{- 5 - 2}{10}) a^{2} b^{2} c + (\frac{2 + 1}{6}) c b^{2} a^{2} + (\frac{- 7 - 4}{28}) c^{2} a b + \frac{1}{8} c a^{2} b$
$= - \frac{7}{10} a^{2} b^{2} c + \frac{1}{2} a b^{2} c - \frac{11}{28} a b c^{2} + \frac{1}{8} a^{2} b c$                                               (Combining like terms)

View NCERT Solutions for all chapters of Class 8