NCERT Solutions for Class 7 Math Chapter 12 - Algebraic Expressions

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Mathematics NCERT Grade 7, Chapter 12: Algebraic Expressions- The chapters discuss concepts related to algebraic expressions.
After a short introduction, the first section discusses How are expressions formed?

Algebraic expressions are formed from variables and constants.

This is followed by the topics terms, factors, and coefficients.

Expressions are made up of terms.
A term is a product of factors.
Coefficient is the numerical factor in the term

Before moving to terms like monomials, binomials, and polynomials, like and unlike terms are discussed.

When terms have the same algebraic factors, they are like terms.
When terms have different algebraic factors, they are unlike terms.

Any expression with one or more terms is called a polynomial.

A one-term expression is called a monomial.
A two-term expression is called binomial.
An expression that contains three terms is called a trinomial.

Emphasis is given to important topics like the Addition and Subtraction of Algebraic Expressions.
2 sub parts are discussed in the following section:

Adding and subtracting like terms: The sum of two or more like terms is a like term with a numerical coefficient equal to the sum of the numerical coefficients of all the like terms.
The difference between two like terms is a like term with a numerical coefficient equal to the difference between the numerical coefficients of the two like terms.
Adding and subtracting general algebraic expressions
- The like terms are added together while the unlike terms are left as they are.

The next part of the section explains about Finding the Value of an Expression where value of a variable is given depending on which value of the expression is calculated.

This is followed by the topic- Using Algebraic Expressions- Formulas and Rules where one understands how algebraic expressions can be used to write formulas and patterns. These patterns are related to numbers and geometrical patterns.

Perimeter formulas
Area formulas
Rules for number patterns
Some more number patterns
Pattern in geometry

Unsolved, solved exercises, questions in different patterns will make the chapter more comprehensible to students.
All important points of the chapter are cited in the end under the title- What Have We Discussed?

Page No 234:

Question 1:

Get the algebraicexpressions in the following cases using variables, constants and arithmetic operations.

(i) Subtraction of z from y.

(ii) One-half of the sum of numbers x and y.

(iii) The number z multiplied by itself.

(iv) One-fourth of the product of numbers p and q.

(v) Numbers x and y both squared and added.

(vi) Number 5 added to three times the product of number m and n.

(vii) Product of numbers y and z subtracted from 10.

(viii)Sum of numbers a and b subtracted from their product.

Answer:

(i) y − z

(ii)

(iii) z²

(iv)

(v) x² + y²

(vi) 5 + 3 (mn)

(vii) 10 − yz

(viii) ab − (a + b)

Page No 234:

Question 2:

(i) Identify the terms and their factors in the following expressions

Show the terms and factors by tree diagrams.

(a) x − 3 (b) 1 + x + x² (c) y − y³

(d) (e) − ab + 2b² − 3a²

(ii) Identify terms and factors in the expressions given below:

(a) − 4x + 5 (b) − 4x + 5y (c) 5y + 3y²

(d) (e) pq + q

(f) 1.2 ab − 2.4 b + 3.6 a (g)

(h) 0.1p² + 0.2 q²

Answer:

(i)

(a)

(b)

(c)

(d)

(e)

(ii)

Row	Expression	Terms	Factors
(a)	− 4x + 5	− 4x 5	− 4, x 5
(b)	− 4x + 5y	− 4x 5y	− 4, x 5, y
(c)	5y + 3y²	5y 3y²	5, y 3, y, y
(d)	xy + 2x²y²	xy 2x²y²	x, y 2, x, x, y, y
(e)	pq + q	pq q	p, q q
(f)	1.2ab − 2.4b + 3.6a	1.2ab − 2.4b 3.6a	1.2, a, b − 2.4, b 3.6, a
(g)
(h)	0.1p² + 0.2q²	0.1p² 0.2q²	0.1, p, p 0.2, q, q

Page No 235:

Question 3:

Identify the numerical coefficients of terms (other than constants) in the following expressions:

(i) 5 − 3t² (ii) 1 + t + t² + t³ (iii) x + 2xy+ 3y

(iv) 100m + 1000n (v) − p²q² + 7pq (vi) 1.2a + 0.8b

(vii) 3.14 r² (viii) 2 (l + b) (ix) 0.1y + 0.01 y²

Answer:

Row	Expression	Terms	Coefficients
(i)	5 − 3t²	− 3t²	− 3
(ii)	1 + t + t² + t³	t t² t³	1 1 1
(iii)	x + 2xy + 3y	x 2xy 3y	1 2 3
(iv)	100m + 1000n	100m 1000n	100 1000
(v)	− p²q² + 7pq	− p²q² 7pq	− 1 7
(vi)	1.2a +0.8b	1.2a 0.8b	1.2 0.8
(vii)	3.14 r²	3.14 r²	3.14
(viii)	2(l + b)	2l 2b	2 2
(ix)	0.1y + 0.01y²	0.1y 0.01y²	0.1 0.01

Page No 235:

Question 4:

(a) Identify terms which contain x and give the coefficient of x.

(i) y²x + y (ii) 13y²− 8yx (iii) x + y + 2

(iv) 5 + z + zx (v) 1 + x+ xy (vi) 12xy² + 25

(vii) 7x + xy²

(b) Identify terms which contain y² and give the coefficient of y².

(i) 8 − xy² (ii) 5y² + 7x (iii) 2x²y −15xy² + 7y²

Answer:

(a)

Row	Expression	Terms with x	Coefficient of x
(i)	y²x + y	y²x	y²
(ii)	13y² − 8yx	− 8yx	−8y
(iii)	x + y + 2	x	1
(iv)	5 + z + zx	zx	z
(v)	1 + x + xy	x xy	1 y
(vi)	12xy² + 25	12xy²	12y²
(vii)	7x+ xy²	7x xy²	7 y²

(b)

Row	Expression	Terms with y²	Coefficient of y²
(i)	8 − xy²	−xy²	− x
(ii)	5y² + 7x	5y²	5
(iii)	2x²y + 7y² −15xy²	7y² −15xy²	7 −15x

Page No 235:

Question 5:

Classify into monomials, binomials and trinomials.

(i) 4y − 7z (ii) y² (iii) x + y − xy

(iv) 100 (v) ab − a − b (vi) 5 − 3t

(vii) 4p²q − 4pq² (viii) 7mn (ix) z² − 3z + 8

(x) a² + b² (xi) z² + z (xii) 1 + x + x²

Answer:

The monomials, binomials, and trinomials have 1, 2, and 3 unlike terms in it respectively.

(i) 4y − 7z

Binomial

(ii) y²

Monomial

(iii) x + y − xy

Trinomial

(iv) 100

Monomial

(v) ab − a − b

Trinomial

(vi) 5 − 3t

Binomial

(vii) 4p²q − 4pq²

Binomial

(viii) 7mn

Monomial

(ix) z² − 3z + 8

Trinomial

(x) a² + b²

Binomial

(xi) z² + z

Binomial

(xii) 1 + x + x²

Trinomial

Page No 235:

Question 6:

State whether a given pair of terms is of like or unlike terms.

(i) 1, 100 (ii) (iii) − 29x, − 29y

(iv) 14xy, 42yx (v) 4m²p, 4mp² (vi) 12xz, 12 x²z²

Answer:

The terms which have the same algebraic factors are called like terms. However, when the terms have different algebraic factors, these are called unlike terms.

(i) 1, 100

(ii) − 7x,

(iii) −29x, −29y

Unlike

(iv) 14xy, 42yx

(v) 4m²p, 4mp²

Unlike

(vi) 12xz, 12x²z²

Unlike

Page No 235:

Question 7:

Identify like terms in the following:

(a) −xy², − 4yx², 8x², 2xy², 7y, − 11x², − 100x, −11yx, 20x²y, −6x², y, 2xy,3x

(b) 10pq, 7p, 8q, − p²q², − 7qp, − 100q, − 23, 12q²p², − 5p², 41, 2405p, 78qp, 13p²q, qp², 701p²

Answer:

(a) −xy², 2xy²

−4yx², 20x²y

8x², −11x², −6x²

7y, y

−100x, 3x

−11xy, 2xy

(b) 10pq, −7qp, 78qp

7p, 2405p

8q, −100q

−p²q², 12p²q²

−23, 41

−5p², 701p²

13p²q, qp²

Page No 239:

Question 1:

Simplify combining like terms:

(i) 21b − 32 + 7b − 20b

(ii) − z² + 13z² − 5z + 7z³ − 15z

(iii) p − (p − q) − q − (q − p)

(iv) 3a − 2b − ab − (a − b + ab) + 3ab + b − a

(v) 5x²y − 5x² + 3y x² − 3y² + x²− y²+ 8xy² −3y²

(vi) (3 y²+ 5y − 4) − (8y − y² − 4)

Answer:

(i) 21b − 32 + 7b − 20b = 21b + 7b − 20b − 32

= b (21 + 7 − 20) −32

= 8b − 32

(ii) − z² + 13z² − 5z + 7z³ − 15z = 7z³ − z² + 13z² − 5z − 15z

= 7z³ + z² (−1 + 13) + z (−5 − 15)

= 7z³ + 12z² − 20z

(iii) p − (p − q) − q − (q − p) = p − p + q − q − q + p

= p − q

(iv) 3a − 2b − ab − (a − b + ab) + 3ba + b − a

= 3a − 2b − ab − a + b − ab + 3ab + b − a

= 3a − a − a − 2b + b + b − ab − ab + 3ab

= a (3 − 1 − 1) + b (− 2 + 1 + 1) + ab (−1 −1 + 3)

= a + ab

(v) 5x²y − 5x² + 3yx² − 3y² + x² − y² + 8xy² − 3y²

= 5x²y + 3yx²− 5x² + x² − 3y² − y² − 3y²+ 8xy²

= x²y (5 + 3) + x² (−5 + 1) + y²(−3 − 1 − 3) + 8xy²

= 8x²y − 4x² − 7y² + 8xy²

(vi) (3y²+ 5y − 4) − (8y − y² − 4)

= 3y² + 5y − 4 − 8y + y² + 4

= 3y² + y² + 5y − 8y − 4 + 4

= y² (3 + 1) + y (5 − 8) + 4 (1 − 1)

= 4y² − 3y

Page No 239:

Question 2:

Add:

(i) 3mn, − 5mn, 8mn, −4mn

(ii) t − 8tz, 3tz − z, z − t

(iii) − 7mn + 5, 12mn + 2, 9mn − 8, − 2mn − 3

(iv) a + b − 3, b − a + 3, a − b + 3

(v) 14x + 10y − 12xy − 13, 18 − 7x − 10y + 8xy, 4xy

(vi) 5m − 7n, 3n − 4m + 2, 2m − 3mn − 5

(vii) 4x²y, − 3xy², − 5xy², 5x²y

(viii) 3p²q² − 4pq + 5, − 10p²q², 15 + 9pq + 7p²q²

(ix) ab − 4a, 4b − ab, 4a − 4b

(x) x²− y² − 1 , y² − 1 − x², 1− x² − y²

Answer:

(i) 3mn + (−5mn) + 8mn + (−4mn) = mn (3 − 5 + 8 − 4)

= 2mn

(ii) (t − 8tz) + (3tz − z) + (z − t) = t − 8tz + 3tz − z + z − t

= t − t − 8tz + 3tz − z + z

= t (1 − 1) + tz (− 8 + 3) + z (− 1 + 1)

= −5tz

(iii) (− 7mn + 5) + (12mn + 2) + (9mn − 8) + (− 2mn − 3)

= − 7mn + 5 + 12mn + 2 + 9mn − 8 − 2mn − 3

= − 7mn + 12mn + 9mn − 2mn + 5 + 2 − 8 − 3

= mn (− 7 + 12 + 9 − 2) + (5 + 2 − 8 − 3)

= 12mn − 4

(iv) (a + b − 3) + (b − a + 3) + (a − b + 3)

= a + b − 3 + b − a + 3 + a − b + 3

= a − a + a + b + b − b − 3 + 3 + 3

= a (1 − 1 + 1) + b (1 + 1 − 1) + 3 (− 1 + 1 + 1)

= a + b + 3

(v) (14x + 10y − 12xy − 13) + (18 − 7x − 10y + 8yx) + 4xy

= 14x + 10y − 12xy − 13 + 18 − 7x − 10y + 8yx + 4xy

= 14x − 7x + 10y − 10y − 12xy + 8yx + 4xy − 13 + 18

= x (14 − 7) + y (10 − 10) + xy (− 12 + 8 + 4) − 13 + 18

= 7x + 5

(vi) (5m − 7n) + (3n − 4m + 2) + (2m − 3mn − 5)

= 5m − 7n + 3n − 4m + 2 + 2m − 3mn − 5

= 5m − 4m + 2m − 7n + 3n − 3mn + 2 − 5

= m (5 − 4 + 2) + n (− 7 + 3) −3mn + 2 − 5

= 3m − 4n − 3mn − 3

(vii) 4x² y − 3xy² − 5xy² + 5x²y = 4x² y + 5x²y − 3xy² − 5xy²

= x² y (4 + 5) + xy² (− 3 − 5)

= 9x²y − 8xy²

(viii) (3p²q² − 4pq + 5) + (−10 p²q²) + (15 + 9pq + 7p²q²)

= 3p²q² − 4pq + 5 − 10 p²q² + 15 + 9pq + 7p²q²

= 3p²q² − 10 p²q²+ 7p²q²− 4pq + 9pq + 5 + 15

= p²q² (3 − 10 + 7) + pq (− 4 + 9) + 5 + 15

= 5pq + 20

(ix) (ab − 4a) + (4b − ab) + (4a − 4b)

= ab − 4a + 4b − ab + 4a − 4b

= ab − ab − 4a + 4a + 4b − 4b

= ab (1 − 1) + a (− 4 + 4) + b(4 − 4)

= 0

(x) (x² − y² − 1) + (y² − 1 − x²) + (1 − x² − y²)

= x² − y² − 1 + y² − 1 − x² + 1 − x² − y²

= x² − x²− x²− y² + y²− y²− 1 − 1 + 1

= x²(1 − 1 − 1) + y² (−1 + 1 − 1) + (− 1 − 1 + 1)

= − x² − y² − 1

Page No 240:

Question 3:

Subtract:

(i) − 5y²from y²

(ii) 6xy from − 12xy

(iii) (a − b) from (a + b)

(iv) a (b − 5) from b (5 − a)

(v) − m² + 5mn from 4m² − 3mn + 8

(vi) − x² + 10x − 5 from 5x − 10

(vii) 5a² − 7ab + 5b² from 3ab − 2a² −2b²

(viii) 4pq − 5q² − 3p² from 5p² + 3q²− pq

Answer:

(i) y² − (−5y²) = y² + 5y² = 6y²

(ii) − 12xy − (6xy) = −18xy

(iii) (a + b) − (a − b) = a + b − a + b = 2b

(iv) b (5 − a) − a (b − 5) = 5b − ab − ab + 5a

= 5a + 5b − 2ab

(v) (4m² − 3mn + 8) − (− m² + 5mn) = 4m² − 3mn + 8 + m² − 5 mn

= 4m² + m² − 3mn − 5 mn + 8

= 5m² − 8mn + 8

(vi) (5x − 10) − (− x² + 10x − 5) = 5x − 10 + x² − 10x + 5

= x² + 5x − 10x − 10 + 5

= x² − 5x − 5

(vii) (3ab − 2a² − 2b²) − (5a²− 7ab + 5b²)

= 3ab − 2a² − 2b² − 5a² + 7ab − 5 b²

= 3ab + 7ab − 2a²− 5a² − 2b² − 5 b²

= 10ab − 7a² − 7b²

(viii) 4pq − 5q² − 3p² from 5p² + 3q² − pq

(5p² + 3q² − pq) − (4pq − 5q²− 3p²)

= 5p² + 3q² − pq − 4pq + 5q² + 3p²

= 5p² + 3p² + 3q² + 5q² − pq − 4pq

= 8p² + 8q² − 5pq

Page No 240:

Question 4:

(a) What should be added to x² + xy + y² to obtain 2x² + 3xy?

(b) What should be subtracted from 2a + 8b + 10 to get − 3a + 7b + 16?

Answer:

(a) Let a be the required term.

a + (x² + y² + xy) = 2x²+ 3xy
a = 2x² + 3xy − (x² + y² + xy)

a = 2x² + 3xy − x² − y² − xy

a = 2x² − x² − y² + 3xy − xy

= x² − y² + 2xy

(b) Let p be the required term.

(2a + 8b + 10) − p = − 3a + 7b + 16

p = 2a + 8b + 10 − (− 3a + 7b + 16)

= 2a + 8b + 10 + 3a − 7b − 16

= 2a + 3a + 8b − 7b + 10− 16

= 5a + b − 6

Video Solution for algebraic expressions (Page: 240 , Q.No.: 4)

NCERT Solution for Class 7 math - algebraic expressions 240 , Question 4

Page No 240:

Question 5:

What should be taken away from 3x² − 4y² + 5xy + 20 to obtain

− x² − y²+ 6xy + 20?

Answer:

Let p be the required term.

(3x² − 4y² + 5xy + 20) − p = − x² − y² + 6xy + 20

p = (3x² − 4y² + 5xy + 20) − (− x² − y² + 6xy + 20)

= 3x² − 4y² + 5xy + 20 + x² + y² − 6xy − 20

= 3x² + x²− 4y² + y² + 5xy − 6xy + 20 − 20

= 4x² − 3y² − xy

Video Solution for algebraic expressions (Page: 240 , Q.No.: 5)

NCERT Solution for Class 7 math - algebraic expressions 240 , Question 5

Page No 240:

Question 6:

(a) From the sum of 3x − y + 11 and − y − 11, subtract 3x − y − 11.

(b) From the sum of 4 + 3x and 5 − 4x + 2x², subtract the sum of 3x² − 5x and
− x² + 2x+ 5.

Answer:

(a) (3x − y + 11) + (− y − 11)

= 3x − y + 11 − y − 11

= 3x − y − y + 11 − 11

= 3x − 2y

(3x − 2y) − (3x − y − 11)

= 3x − 2y − 3x + y + 11

= 3x − 3x − 2y + y + 11

= − y + 11

(b) (4 + 3x) + (5 − 4x + 2x²) = 4 + 3x + 5 − 4x + 2x²

= 3x − 4x + 2x² + 4 + 5

= − x + 2x² + 9

(3x² − 5x) + (− x² + 2x + 5) = 3x² − 5x − x² + 2x + 5

= 3x² − x²− 5x + 2x + 5

= 2x² − 3x + 5

(− x + 2x² + 9) − (2x² − 3x + 5)

= − x + 2x² + 9 − 2x² + 3x − 5

= − x + 3x + 2x² − 2x²+ 9 − 5

= 2x + 4

Video Solution for algebraic expressions (Page: 240 , Q.No.: 6)

NCERT Solution for Class 7 math - algebraic expressions 240 , Question 6

Page No 242:

Question 1:

If m = 2, find the value of:

(i) m − 2 (ii) 3m − 5 (iii) 9 − 5m

(iv) 3m² − 2m − 7 (v)

Answer:

(i) m − 2 = 2 − 2 = 0

(ii) 3m − 5 = (3 × 2) − 5 = 6 − 5 = 1

(iii) 9 − 5m = 9 − (5 × 2) = 9 −10 = −1

(iv) 3m² − 2m − 7 = 3 × (2 × 2) − (2 × 2) − 7

= 12 − 4 − 7 = 1

(v)

Page No 242:

Question 2:

If p = −2, find the value of:

(i) 4p + 7

(ii) −3p² + 4p + 7

(iii) −2p³ − 3p² + 4p + 7

Answer:

(i) 4p + 7 = 4 × (−2) + 7 = − 8 + 7 = −1

(ii) − 3p² + 4p + 7 = −3 (−2) × (−2) + 4 × (−2) + 7

= − 12 − 8 + 7 = −13

(iii) −2p³ − 3p² + 4p + 7

= −2 (−2) × (−2) × (−2) − 3 (−2) × (−2) + 4 × (−2) + 7

= 16 − 12 − 8 + 7 = 3

Page No 242:

Question 3:

Find the value of the following expressions, when x = − 1:

(i) 2x − 7 (ii) − x + 2 (iii) x² + 2x + 1

(iv) 2x² − x − 2

Answer:

(i) 2x − 7

= 2 × (−1) − 7 = −9

(ii) − x + 2 = − (−1) + 2 = 1 + 2 = 3

(iii) x² + 2x + 1 = (−1) × (−1) + 2 × (−1) + 1

= 1 − 2 + 1 = 0

(iv) 2x² − x − 2 = 2 (−1) × (−1) − (−1) − 2

= 2 + 1 − 2 = 1

Page No 242:

Question 4:

If a = 2, b = − 2, find the value of:

(i) a²+ b² (ii) a² + ab + b² (iii) a² − b²

Answer:

(i) a² + b²

= (2)² + (−2)² = 4 + 4 = 8

(ii) a² + ab + b²

= (2 × 2) + 2 × (−2) + (−2) × (−2)

= 4 − 4 + 4 = 4

(iii) a² − b²

= (2)² − (−2)² = 4 − 4 = 0

Page No 242:

Question 5:

When a = 0, b = − 1, find the value of the given expressions:

(i) 2a + 2b (ii) 2a²+ b² + 1

(iii) 2a²b + 2ab² + ab (iv) a² + ab + 2

Answer:

(i) 2a + 2b = 2 × (0) + 2 × (−1) = 0 − 2 = −2

(ii) 2a² + b² + 1

= 2 × (0)² + (−1) × (−1) + 1

= 0 + 1 + 1 = 2

(iii) 2a²b + 2ab² + ab

= 2 × (0)² × (−1) + 2 × (0) × (−1) × (−1) + 0 × (−1)

= 0 + 0 + 0 = 0

(iv) a² + ab + 2

= (0)² + 0 × (−1) + 2

= 0 + 0 + 2 = 2

Video Solution for algebraic expressions (Page: 242 , Q.No.: 5)

NCERT Solution for Class 7 math - algebraic expressions 242 , Question 5

Page No 242:

Question 6:

Simplify the expressions and find the value if x is equal to 2

(i) x + 7 + 4 (x − 5) (ii) 3 (x + 2) + 5x − 7

(iii) 6x + 5 (x − 2) (iv) 4 (2x −1) + 3x + 11

Answer:

(i) x + 7 + 4 (x − 5) = x + 7 + 4x − 20

= x + 4x + 7 − 20

= 5x − 13

= (5 × 2) − 13

= 10 − 13 = −3

(ii) 3 (x + 2) + 5x − 7 = 3x + 6 + 5x − 7

= 3x + 5x + 6 − 7 = 8x − 1

= (8 × 2) − 1 = 16 − 1 =15

(iii) 6x + 5 (x − 2) = 6x + 5x − 10

= 11x − 10

= (11 × 2) − 10 = 22 − 10 = 12

(iv) 4 (2x − 1) + 3x + 11 = 8x − 4 + 3x + 11

= 11x + 7

= (11 × 2) + 7

= 22 + 7 = 29

Page No 242:

Question 7:

Simplify these expressions and find their values if x = 3, a = − 1, b = − 2.

(i) 3x − 5 − x + 9 (ii) 2 − 8x + 4x + 4

(iii) 3a + 5 − 8a + 1 (iv) 10 − 3b − 4 − 5b

(v) 2a − 2b − 4 − 5 + a

Answer:

(i) 3x − 5 − x + 9 = 3x − x − 5 + 9

= 2x + 4 = (2 × 3) + 4 = 10

(ii) 2 − 8x + 4x + 4 = 2 + 4 − 8x + 4x

= 6 − 4x = 6 − (4 × 3) = 6 − 12 = −6

(iii) 3a + 5 − 8a + 1 = 3a − 8a + 5 + 1

= − 5a + 6 = −5 × (−1) + 6

= 5 + 6 = 11

(iv) 10 − 3b − 4 − 5b = 10 − 4− 3b − 5b

= 6 − 8b = 6 − 8 × (−2)

= 6 + 16 = 22

(v) 2a − 2b − 4 − 5 + a = 2a + a − 2b − 4 − 5

= 3a − 2b − 9s

= 3 × (−1) − 2 (−2) − 9

= − 3 + 4 − 9 = −8

Page No 242:

Question 8:

(i) If z = 10, find the value of z³ − 3 (z − 10).

(ii) If p = − 10, find the value of p² − 2p − 100

Answer:

(i) z³ − 3 (z − 10) = z³ − 3z + 30

= (10 × 10 × 10) − (3 × 10) + 30

= 1000 − 30 + 30 = 1000

(ii) p² − 2p − 100

= (−10) × (−10) − 2 (−10) − 100

= 100 + 20 − 100 = 20

Video Solution for algebraic expressions (Page: 242 , Q.No.: 8)

NCERT Solution for Class 7 math - algebraic expressions 242 , Question 8

Page No 242:

Question 9:

What should be the value of a if the value of 2x² + x − a equals to 5, when x = 0?

Answer:

2x² + x − a = 5, when x = 0

(2 × 0) + 0 − a = 5

0 − a = 5

a = −5

Video Solution for algebraic expressions (Page: 242 , Q.No.: 9)

NCERT Solution for Class 7 math - algebraic expressions 242 , Question 9

Page No 242:

Question 10:

Simplify the expression and find its value when a = 5 and b = −3.

2 (a²+ ab) + 3 − ab

Answer:

2 (a² + ab) + 3 − ab = 2a² + 2ab + 3 − ab

= 2a² + 2ab − ab + 3

= 2a² + ab + 3

= 2 × (5 × 5) + 5 × (−3) + 3

= 50 − 15 + 3 = 38

Page No 246:

Question 1:

Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.

(a)

(b)

(c)

If the number of digits formed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern.

How many segments are required to form 5, 10, 100 digits of the kind −

, , .

Answer:

(a) It is given that the number of segments required to form n digits of the kind

is (5n + 1).

Number of segments required to form 5 digits = (5 × 5 + 1)

= 25 + 1 = 26

Number of segments required to form 10 digits = (5 × 10 + 1)

= 50 + 1 = 51

Number of segments required to form 100 digits = (5 × 100 + 1)

= 500 + 1 = 501

(b) It is given that the number of segments required to form n digits of the kind is (3n + 1).

Number of segments required to form 5 digits = (3 × 5 + 1)

= 15 + 1 = 16

Number of segments required to form 10 digits = (3 × 10 + 1)

= 30 + 1 = 31

Number of segments required to form 100 digits = (3 × 100 + 1)

= 300 + 1 = 301

(c)It is given that the number of segments required to form n digits of the kind is (5n + 2).

Number of segments required to form 5 digits = (5 × 5 + 2)

= 25 + 2 = 27

Number of segments required to form 10 digits = (5 × 10 + 2)

= 50 + 2 = 52

Number of segments required to form 100 digits = (5 × 100 + 2)

= 500 + 2 = 502

Page No 247:

Question 2:

Use the given algebraic expression to complete the table of number patterns.

S. No	Expression	Terms
S. No	Expression	1^st	2^nd	3^rd	4^th	5^th	…	10^th	…	100^th	…
(i)	2n − 1	1	3	5	7	9	-	19	-	-	-
(ii)	3n + 2	2	5	8	11	-	-	-	-	-	-
(iii)	4n + 1	5	9	13	17	-	-	-	-	-	-
(iv)	7n + 20	27	34	41	48	-	-	-	-	-	-
(v)	n² + 1	2	5	10	17	-	-	-	-	10, 001	-

Answer:

The given table can be completed as follows.

S.No.	Expression	Terms
S.No.	Expression	1^st	2^nd	3^rd	4^th	5^th	…	10^th	…	100^th	…
(i)	2n − 1	1	3	5	7	9	-	19	-	199	-
(ii)	3n + 2	2	5	8	11	17	-	32	-	302	-
(iii)	4n + 1	5	9	13	17	21	-	41	-	401	-
(iv)	7n + 20	27	34	41	48	55	-	90	-	720	-
(v)	n² + 1	2	5	10	17	26	-	101	-	10,001-	-

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