Rd Sharma 2020 2021 for Class 7 Maths Chapter 8 - Linear Equations In One Variable

Answer:

x − 3 = 5
Adding 3 to both sides, we get
⇒ x − 3 + 3 = 5 + 3  
⇒ x = 8
Verification:
Substituting x= 8 in LHS, we get
LHS = x − 3 and RHS = 5
LHS = 8 − 3 = 5 and RHS = 5
LHS = RHS
Hence, verified.

Answer:

x + 9 = 13
Subtracting 9 from both sides, we get
=> x + 9 $-$ 9 = 13 $-$ 9
=> x = 4
Verification:
Substituting x = 4 on LHS, we get
LHS = 4 + 9 = 13 = RHS
LHS = RHS
Hence, verified.

Answer:

x − $\frac{3}{5}$ = $\frac{7}{5}$
Adding $\frac{3}{5}$ to both sides, we get
=> x − $\frac{3}{5}$ + $\frac{3}{5}$ = $\frac{7}{5}$ + $\frac{3}{5}$
=> x = $\frac{7+3}{5}$
=> x = $\frac{10}{5}$
⇒ x = 2
Verification:
Substituting x = 2 in LHS, we get
LHS = 2 − $\frac{3}{5}$=$\frac{10-3}{5}=\frac{7}{5}$, and RHS = $\frac{7}{5}$
LHS = RHS
Hence, verified.

Answer:

3x = 0
Dividing both sides by 3, we get
⇒ $\frac{3x}{3}$  = $\frac{0}{3}$
⇒ x = 0
Verification:
Substituting x = 0 in LHS = 3x, we get
LHS = 3 $\times$ 0 = 0 and RHS = 0
LHS = RHS
Hence, verified.

Answer:

 $\frac{x}{2}$ = 0
Multiplying both sides by 2, we get
⇒ $\frac{x}{2}\times 2 = 0 \times 2$
⇒ x = 0

Verification:
Substituting x= 0 in LHS, we get
LHS =$\frac{0}{2}$= 0 and RHS = 0
LHS = 0 and RHS = 0
LHS = RHS
Hence, verified.

Answer:

x − $\frac{1}{3}$ = $\frac{2}{3}$
⇒ Adding $\frac{1}{\mathrm{3 }}$ to both sides, we get
⇒ x − $\frac{1}{3}$ +$\frac{1}{\mathrm{3 }}$ = $\frac{2}{3}$ + $\frac{1}{\mathrm{3 }}$
=> x = $\frac{2+1}{3}$
=> x = $\frac{3}{3}$
⇒ x = 1
Verification:
Substituting x= 1 in LHS, we get
LHS = 1 −$$$\frac{1}{\mathrm{3 }}$ =$\frac{3-1}{3}$$=\frac{2}{\mathrm{3 }}$,  and RHS = $\frac{2}{\mathrm{3 }}$
LHS = RHS
Hence, verified.

Answer:

x + $\frac{1}{2}$ = $\frac{7}{2}$
⇒ Subtracting $\frac{1}{2}$ from both sides, we get
⇒ x + $\frac{1}{2}$ − $\frac{1}{2}$ = $\frac{7}{2}$ − $\frac{1}{2}$
$\Rightarrow \text{x=}\frac{7-1}{2}=\frac{6}{2}$
⇒ x = 3
Verification:
Substituting x = 3 in LHS, we get
LHS = 3 + $\frac{1}{2}$=$\frac{6+1}{2}=\frac{7}{2}$, and RHS = $\frac{7}{2}$
LHS = RHS
Hence, verified.

Answer:

10 − y = 6
Subtracting 10 from both sides, we get
⇒ 10 − y − 10 = 6 − 10
⇒ −y = −4.
⇒ Multiplying both sides by −1, we get
⇒ −y $\times$−1 = −4 $\times$−1
⇒ y = 4
Verification:
Substituting y = 4 in LHS, we get
LHS = 10 − y = 10$-$4 = 6 and RHS = 6
LHS = RHS
Hence, verified.

Answer:

7 + 4y = −5
Subtracting 7 from both sides, we get
⇒ 7 + 4y − 7 = −5 − 7
⇒  4y  = −12
Dividing both sides by 4, we get
⇒ $\text{y}$= $\frac{-12}{ 4}$
⇒  y  = −3

Verification :
Substituting y = −3 in LHS, we get
LHS = 7 + 4y = 7  + 4(−3) = 7 − 12 = −5, and RHS = −5
LHS = RHS
Hence, verified.

Answer:

$\frac{4}{5}$ − x = $\frac{3}{5}$
Subtracting$$$\frac{4}{5}$ from both sides, we get
⇒ $\frac{4}{5}$ − x − $\frac{4}{5}$= $\frac{3}{5}$  +    6+ − $\frac{4}{5}$
⇒ −x = $\frac{3-4}{5}$
$\Rightarrow -\text{x}=\frac{-1}{5}$
Multiplying both sides by -1, we get
⇒ −x $\times$ (−1) = −$\frac{1}{5}$ $\times$ (−1)
⇒ x = $\frac{1}{5}$
Verification:
Substituting x= $\frac{1}{5}$ in LHS, we get
LHS = $\frac{4}{5}$ − $\frac{1}{5}$=$\frac{4-1}{5}=\frac{3}{5}$, and RHS = $\frac{3}{5}$
LHS = RHS
Hence, verified.

Answer:

2y  − $\frac{1}{2}$$$$=-\frac{1}{3}$
Adding $\frac{1}{2}$$$to both sides, we get
⇒ 2y  − $\frac{1}{2}$$+$$\frac{1}{2}$$$$=-\frac{1}{3}$ + $\frac{1}{2}$$$
⇒ 2y = $\frac{-2 + 3}{6}$
⇒ 2y = $\frac{1}{6}$$$
Dividing both sides by 2, we get
⇒ $\frac{2y}{2} = \frac{1}{6\times 2}$
⇒ y = $\frac{1}{12}$
Verification:
Substituting y = $\frac{1}{12}$ in LHS, we get
LHS = $2\times \left(\frac{1}{12}\right) - \frac{1}{2} = \frac{1}{6} - \frac{1}{2}$ =$\frac{1-3}{6}=\frac{-2}{6}$ =$-\frac{1}{3}$,  and RHS = $-\frac{1}{3}$
LHS = RHS
Hence, verified.

Answer:

 $14=$$\frac{7\text{x}}{10}$ - 8
Adding 8 to both sides, we get
⇒ 14 + 8 = $\frac{7x}{10}$ − 8 + 8
⇒ 22 = $\frac{7x}{10}$
Multiplying both sides by 10, we get
⇒ 22 $\times 10$ = $\frac{7x}{10}$$\times 10$
⇒ 220 = 7x
Dividing both sides by 7, we get
⇒ $\frac{220}{7} = \frac{7x}{7}$
⇒ x = $\frac{220}{7}$
Verification:

Substituting x = $\frac{220}{7}$ in RHS, we get
LHS =$$ 14, and RHS = $\frac{7\left({\displaystyle \frac{220}{7}}\right)}{10} - 8$ = $\frac{220}{10} - 8 = 22 - 8 = 14$
LHS = RHS
Hence, verified.

Answer:

3 (x + 2) = 15
Dividing both sides by 3, we get
⇒ $$$\frac{\mathrm{3\; (x\; +\; 2)}}{3}= \frac{15}{3}$
⇒ (x + 2) = 5
Subtracting 2 from both sides, we get
⇒ x + 2 $-$ 2 = 5 $-$ 2
⇒ x = 3
Verification:
Substituting x = 3 in LHS, we get
LHS = 3 (x + 2)= 3 (3+2) = 3$\times$5 = 15, and RHS = 15
LHS = RHS
Hence, verified.

Answer:

 $\frac{x}{4} = \frac{7}{8}$
Multiplying both sides by 4, we get
⇒  $\frac{x}{4}\times 4 = \frac{7}{8} \times 4$
⇒ x = $\frac{7}{2}$
Verification:
Substituting x = $\frac{7}{2}$ in LHS, we get
LHS = $\frac{7}{2\times 4}$ = $\frac{7}{8}$, and RHS = $\frac{7}{8}$
LHS = RHS
Hence, verified.

Answer:

 $\frac{1}{3} - 2x = 0$
Subtracting $\frac{1}{3}$ from both sides, we get
⇒  $\frac{1}{3} - 2x - \frac{1}{3}$ = 0 $-$$\frac{1}{3}$
⇒ $-2x = -\frac{1}{3}$
Multiplying both sides by $-$1, we get
⇒ $-2x \times (-1) = -\frac{1}{3} \times (-1)$
⇒ 2x = $\frac{1}{3}$
Dividing both sides by 2, we get
⇒ $\frac{2x}{2} = \frac{1}{3\times 2}$
⇒ x = $\frac{1}{6}$
Verification:
Substituting x = $\frac{1}{6}$ in LHS, we get
LHS = $\frac{1}{3} - \left(2 \times \frac{1}{6}\right) = \frac{1}{3}- \frac{1}{3} = 0$, and RHS = 0
LHS = RHS
Hence, verified.

Answer:

3(x + 6) = 24

Dividing both sides by 3, we get
⇒ $$$\frac{\mathrm{3\; (x\; +\; 6)}}{3}=\frac{24}{3}$
⇒ (x + 6) = 8
Subtracting 6 from both sides, we get
⇒ x + 6 $-$ 6 = 8 $-$ 6
⇒ x = 2
Verification:
Substituting x = 2 in LHS, we get
LHS = 3 (x + 6) = 3 (2 + 6) = 3$\times$8 = 24, and RHS = 24
LHS = RHS
Hence, verified.

Answer:

3(x + 2) − 2(x − 1) = 7
On expanding the brackets, we get 
⇒ $\left(3\times x\right) +\left( 3 \times 2\right) - \left(2 \times x\right) + \left(2 \times 1\right) = 7$
⇒ 3x + 6 $-$ 2x + 2 = 7
⇒ 3x $-$ 2x + 6 + 2 = 7
⇒ x + 8 = 7
Subtracting 8 from both sides, we get
⇒ x + 8 $-$8 = 7 $-$ 8
⇒ x = $-$1
Verification:
Substituting x = $-$1 in LHS, we get
LHS = 3 (x + 2) $-$2(x $-$1), and RHS = 7
LHS = 3 ($-$1 + 2) $-$2($-$1$-$1) = (3$\times$1) $-$ (2$\times$$-$2) = 3 + 4  = 7, and RHS = 7
LHS = RHS
Hence, verified.

Answer:

8(2x − 5) − 6(3x − 7) = 1
On expanding the brackets, we get
⇒ (8$\times$2x) $-$ (8 $\times$ 5) $-$ (6 $\times$ 3x) + ($-$6)$\times$($-$7)  =  1
⇒ 16x $-$ 40 $-$ 18x + 42 = 1
⇒ 16x $-$ 18x + 42 $-$ 40 = 1
⇒ $-$2x + 2 = 1
Subtracting 2 from both sides, we get
⇒ $-$2x + 2 $-$ 2 = 1 $-$ 2
⇒ $-$2x  = $-$1
Multiplying both sides by $-$1, we get
⇒ $-$2x $\times$($-$1) = $-$1$\times$($-$1)
⇒ 2x = 1
Dividing both sides by 2, we get
⇒ $\frac{2x}{2} = \frac{1}{2}$
⇒ x = $\frac{1}{2}$
Verification:
Substituting x = $\frac{1}{2}$ in LHS, we get
= 8(2$\times$$\frac{1}{2}$ $-$ 5) $-$ 6(3$\times$$\frac{1}{2}$ − 7)

= 8(1 − 5) − 6($\frac{3}{2}$ − 7)

= 8$\times$(−4) − (6 $\times$$\frac{3}{2}$)  + (6 $\times$7)
= −32 − 9 + 42 = − 41 + 42 = 1 = RHS
LHS = RHS
Hence, verified.

Answer:

6(1 − 4x) + 7(2 + 5x) = 53
On expanding the brackets, we get
⇒ (6$\times$1) $-$ (6 $\times$ 4x) + (7 $\times$2) + (7$\times$5x)  =  53
⇒ 6 $-$ 24x + 14 + 35x = 53
⇒ 6 + 14 + 35x $-$ 24x = 53
⇒ 20 + 11x = 53
Subtracting 20 from both sides, we get
⇒ 20 + 11x $-$ 20  = 53 $-$ 20
⇒ 11x  = 33
⇒ Dividing both sides by 11, we get
⇒ $\frac{11x}{11} = \frac{33}{11}$
⇒ x = 3

Verification:
Substituting x = 3 in LHS, we get
= 6(1 − 4$\times$3) + 7(2 + 5$\times$3)
= 6(1 − 12) + 7(2 + 15)
= 6(−11) + 7(17)
= −66 + 119 = 53 = RHS
LHS = RHS
Hence, verified.

Answer:

5(2 − 3x) − 17(2x − 5) = 16
On expanding the brackets, we get
⇒ (5$\times$2) $-$ (5 $\times$ 3x) − (17 $\times$2x ) + (17$\times$5)  =  16
⇒ 10 $-$ 15x − 34x + 85 = 16
⇒ 10 + 85 − 34x $-$ 15x = 16
⇒ 95 - 49x = 16
Subtracting 95 from both sides, we get
⇒ - 49x + 95 $-$ 95  = 16 $-$ 95
⇒ - 49x  = -79
Dividing both sides by -49, we get
⇒ $\frac{-49x}{-49} = \frac{-79}{-49}$

⇒ x = $\frac{79}{49}$
Verification:
Substituting x = $\frac{79}{49}$  in LHS, we get
= 5(2 − 3$\times$$\frac{79}{49}$) − 17(2$\times$$\frac{79}{49}$ − 5)
= (5$\times$2) $-$ (5 $\times$ 3$\times$$\frac{79}{49}$ ) − (17 $\times$2$\times$$\frac{79}{49}$ ) + (17$\times$5) 
= 10 $-$ $\frac{1185}{49}$ − $\frac{2686}{49}$ + 85
= $\frac{490 - 1185 - 2686 + 4165}{49}$
= $\frac{784}{49}$
= 16
= RHS
So, LHS = RHS
Hence, verified.

Answer:

 $\frac{x - 3}{5}- 2 = -1$
Adding 2 to both sides, we get
⇒ $\frac{x - 3}{5 }- 2 + 2 = -1 + 2$
⇒ $\frac{x - 3}{5} = 1$
Multiplying both sides by 5, we get
⇒ $\left(\frac{x - 3 }{5}\right) \times 5 = 1 \times 5$
⇒ x $-$ 3 = 5
Adding 3 to both sides, we get
⇒ x $-$ 3 + 3  = 5 + 3
⇒ x = 8
Verification:
Substituting x = 8  in LHS, we get
 = $\frac{8 -3 }{5}-2$
 = $\frac{5}{5} - 2$
= 1 $-$ 2
= $-$1
= RHS
LHS = RHS
Hence, verified.

Answer:

5(x − 2) + 3(x + 1) = 25
On expanding the brackets, we get
⇒ (5 $\times$ x) $-$ (5 $\times$ 2) + (3 $\times$ x) + (3$\times$1)  =  25
⇒ 5x $-$ 10 + 3x + 3 = 25
⇒ 5x  + 3x $-$ 10 + 3 = 25
⇒ 8x $-$ 7  = 25
Adding 7 to both sides, we get
⇒ 8x $-$ 7 + 7 = 25 +7
⇒ 8x  = 32
Dividing both sides by 8, we get
⇒ $\frac{8x}{8} = \frac{32}{8}$
⇒ x = 4

Verification:
Substituting x = 4  in LHS, we get

= 5(4 − 2) + 3(4 + 1)
= 5(2) + 3(5)
= 10 + 15
= 25
= RHS
LHS = RHS
Hence, verified.

Answer:

We have
⇒ 6x + 5 = 2x + 17
Transposing 2x to LHS and 5 to RHS, we get
⇒ 6x $-$ 2x = 17 $-$ 5
⇒ 4x = 12   
Dividing both sides by 4, we get  
$\Rightarrow \frac{4\text{x}}{4}=\frac{12}{4}$
⇒ x = 3
Verification:
Substituting x =3 in the given equation, we get
6$\times$3 + 5 = 2$\times$3 + 17
18 + 5 = 6 + 17
23 = 23
LHS = RHS
Hence, verified.

Answer:

We have
⇒2(5x − 3) − 3(2x − 1) = 9
Expanding the brackets, we get
⇒ $2\times 5x - 2\times 3 -3\times 2x + 3\times 1 = 9$
⇒ 10x − 6 − 6x + 3 = 9
⇒ 10x − 6x − 6 + 3 = 9
⇒ 4x  − 3 = 9
Adding 3 to both sides, we get    
⇒ 4x − 3 + 3= 9 + 3
⇒ 4x = 12
Dividing both sides by 4, we get
⇒ $\frac{4x}{4} = \frac{12}{4}$
⇒ Thus, x = 3.
Verification:
Substituting x =3 in LHS, we get
=2(5$\times$3 − 3) − 3(2$\times$3 − 1)
=2$\times$12 − 3 $\times$ 5
=24 − 15
= 9
LHS = RHS

Hence, verified.

Answer:

$\frac{x}{2} = \frac{x}{3} + 1$
Transposing $\frac{x}{3}$ to LHS, we get
$$\begin{gathered} \Rightarrow \frac{x}{2}-\frac{x}{3} = 1 \\ \Rightarrow \frac{3x-2x}{6} = 1 \end{gathered}$$ 
$\Rightarrow \frac{x}{6} = 1$      
Multiplying both sides by 6, we get                 
$\Rightarrow \frac{x}{6}\times 6 = 1\times 6$                            
$\Rightarrow$ x = 6
Verification:
Substituting x = 6 in the given equation, we get
$\frac{6}{2}= \frac{6}{3}+ 1$
3 = 2 + 1
3 = 3
LHS = RHS
Hence, verified.

Answer:

 $\frac{x}{2}+\frac{3}{2}=\frac{2x}{5}-1$   
Transposing $\frac{2x}{5}$ to LHS and $\frac{3}{2}$ to RHS, we get
=> $\frac{x}{2}-\frac{2x}{5} = -1 -\frac{3}{2}$
                
=> $\frac{5x-4x}{10}= \frac{-2-3}{2}$
=> $\frac{x}{10}= \frac{-5}{ 2 }$
Multiplying both sides by 10, we get                 
=> $\frac{x}{10}\times 10 = \frac{-5}{2}\times 10$                                    
=>  x = $-25$
Verification:
Substituting x = $-25$ in the given equation, we get
$\frac{-25}{2}+\frac{3}{2}= \frac{2\times \left(-25\right)}{5}-1$
$\frac{-22}{2}= -10 -1$
$-11= -11$
LHS = RHS
Hence, verified.

Answer:

 $\frac{3}{4}\left(x-1\right) = x-3$
On expanding the brackets on both sides, we get
=>$\frac{3}{4}x- \frac{3}{4}= x-3$
Transposing $\frac{3}{4}x$ to RHS and 3 to LHS, we get
=> $3-\frac{3}{4}= x-\frac{3}{4}x$                       
=> $\frac{12-3}{4}=\frac{4x-3x}{4}$
=> $\frac{9}{4}=\frac{x}{4}$
Multiplying both sides by 4, we get
=> x = 9                                       

Verification:
Substituting x = 9 on both sides, we get
$\frac{3}{4}\left(9-1\right)=9-3$
$\frac{3}{4}\times 8=6$
6 = 6
LHS = RHS
Hence, verified.

Answer:

6. 3(x − 3) = 5(2x + 1)
On expanding the brackets on both sides, we get
=> $3\times x - 3\times 3 = 5\times 2x + 5\times 1$           
=> 3x $-$ 9 = 10x + 5
Transposing 10x to LHS and 9 to RHS, we get
=> 3x $-$ 10x = 9 + 5                            
=> $-$7x = 14
Dividing both sides by 7, we get
=> $\frac{-7\mathrm{x}}{7 } =\frac{14}{7}$                                   
=> x = $-$2
Verification:
Substituting x = $-$2 on both sides, we get
$3\left(-2-3\right) = 5\left(2\left(-2\right) +1\right)$
$3\left(-5\right) = 5\left(-3\right)$
$-$15 = $-$15
LHS = RHS
Hence, verified.

Answer:

3x − 2 (2x − 5) = 2(x + 3) − 8
On expanding the brackets on both sides, we get
=> $3x-2\times 2x+2\times 5 = 2\times x + 2\times 3 -8$                 
=> $3x -4x + 10 = 2x + 6 -8$
=> $-x + 10 = 2x - 2$
Transposing x to RHS and 2 to LHS, we get
=> 10 + 2 = 2x + x                                                  
=> 3x = 12
Dividing both sides by 3, we get
=> $\frac{3x }{3} = \frac{12}{3}$                                                       
 => x = 4
Verification:
Substituting x = 4 on both sides, we get
$3\left(4\right) - 2\left(2\left(4\right)-5\right) = 2\left(4+3\right)-8$
12$-$2 (8 $-$ 5) = 14$-$8
12 $-$ 6 = 6
6 = 6
LHS = RHS
Hence, verified.

Answer:

$x - \frac{x}{4}-\frac{1}{2}= 3 + \frac{x}{4}$
Transposing $\frac{x}{4}$ to LHS and $-\frac{1}{2}$ to RHS, we get
=> $x - \frac{x}{4} -\frac{x}{4} = 3 + \frac{1}{2}$               
=> $\frac{4x-x - x}{4} = \frac{6 + 1}{2}$
=> $\frac{2x}{4} = \frac{7}{2}$
Multiplying both sides by 4, we get
=> $\frac{2x}{4}\times 4 = \frac{7}{2}\times 4$                         
 => 2x = 14
Dividing both sides by 2, we get
=> $\frac{2x}{2} = \frac{14}{2}$                                 
=> x = 7
Verification:
Substituting x = 7 on both sides, we get

$7 -\frac{7}{4}- \frac{1}{2} = 3 + \frac{7}{4}$
$\frac{28 - 7 - 2}{4} = \frac{12 + 7}{4}$
$\frac{19}{4} = \frac{19}{4}$
LHS = RHS
Hence, verified.

Answer:

$\frac{6x - 2}{9} + \frac{3x + 5}{18} = \frac{1}{3}$
=> $\frac{6x\times 2 - 2\times 2 + 3x + 5}{18} = \frac{1}{3}$
=> $\frac{12x - 4 + 3x + 5}{18} = \frac{1}{3}$
=> $\frac{15x + 1}{18} = \frac{1}{3}$
Multiplying both sides by 18, we get
=> $\frac{15x + 1}{18}\times \left(18\right) = \frac{1}{3}\times \left(18\right)$                 
=> 15x + 1 = 6
Transposing 1 to RHS, we get
=> 15x = 6$-$1                                           
=> 15x = 5
Dividing both sides by 15, we get
=> $\frac{15x}{15} = \frac{5}{15}$                                         
=> x = $\frac{1}{3}$
Verification:
Substituting x = $\frac{1}{3}$ on both sides, we get
$\frac{6\left({\displaystyle \frac{1}{3}}\right)-2}{9} + \frac{3\left({\displaystyle \frac{1}{3}}\right) +5}{18} = \frac{1}{3}$
$\frac{2 - 2}{9} + \frac{1 +5 }{18} = \frac{1}{3}$
$0 + \frac{6}{18} = \frac{1}{3}$
$\frac{1}{3} = \frac{1}{3}$
LHS = RHS
Hence, verified.

Answer:

$m-\frac{m-1}{2} = 1 - \frac{m-2}{3}$
=> $\frac{2m-m -(- 1) }{2} = \frac{3 -m -(-2)}{3}$
=> $\frac{m + 1}{2} = \frac{3 - m + 2}{3}$
=> $\frac{m+ 1 }{2} = \frac{5-m}{3}$
=> $\frac{m}{2} + \frac{1}{2} = \frac{5}{3} -\frac{m}{3}$
Transposing m/3 to LHS and 1/2 to RHS, we get
=> $\frac{m}{2} + \frac{m}{3} = \frac{5}{3} -\frac{1}{2}$                       
=> $\frac{3m+2m}{6} = \frac{10-3}{6}$
Multiplying both sides by 6, we get
=> $\frac{5m}{6}\times 6 = \frac{7}{6}\times 6$                               
 => 5m = 7
Dividing both sides by 5, we get
=> $\frac{5m}{5} = \frac{7}{5}$                                         
=> m = $\frac{7}{5}$
Verification:
Substituting m =$\frac{7}{5}$ on both sides, we get
$\frac{7}{5}-\frac{{\displaystyle \frac{7}{5}}-1}{2} = 1 -\frac{{\displaystyle \frac{7}{5}}-2}{3}$
$\frac{7}{5}-\frac{{\displaystyle \frac{7-5}{5}}}{2} = 1 - \frac{{\displaystyle \frac{7-10}{5}}}{3}$
$\frac{7}{5}-\frac{2}{5\times 2} = 1 - \frac{-3}{5\times 3}$
$\frac{7}{5}- \frac{1}{5} = 1 + \frac{1}{5}$
$\frac{6}{5} = \frac{6}{5}$
LHS = RHS
Hence, verified.

Answer:

$\frac{\left(5x -1\right)}{3}- \frac{\left(2x - 2\right)}{3} = 1$
 $$\begin{gathered} \Rightarrow \frac{5x - 1 - 2x - (-2)}{3}= 1 \\ \Rightarrow \frac{5x - 1 - 2x + 2}{3} = 1 \\ \Rightarrow \frac{5x - 2x + 2 -1}{3 } = 1 \\ \Rightarrow \frac{3x + 1}{3} = 1 \end{gathered}$$
Multiplying both sides by 3, we get
$\Rightarrow \left(\frac{3x + 1 }{3}\right)\times 3 = 1\times 3$                        
=> 3x + 1 = 3
Subtracting 1 from both sides, we get
=> 3x + 1 $-$ 1 = 3 $-$ 1                               
=> 3x = 2
Dividing both sides by 3, we  get
=> $\frac{3x }{3}= \frac{2}{3}$                                             
=> x = $\frac{2}{3}$
Verification:
Substituting x =$\frac{2}{3}$ in LHS, we get
$$\begin{gathered} =\frac{5\left({\displaystyle \frac{2}{3}}\right) - 1}{3}-\frac{2\left({\displaystyle \frac{2}{3}}\right) - 2}{3} \\ =\frac{{\displaystyle \frac{10}{3}}-1}{3} - \frac{{\displaystyle \frac{4}{3}}- 2}{3}=\frac{{\displaystyle \frac{10-3}{3}}}{3}-\frac{{\displaystyle \frac{4-6}{3}}}{3}=\frac{7}{3\times 3}-\left({\displaystyle \frac{-2}{3\times 3}}\right)=\frac{7}{9} + \frac{2}{9}= \frac{9}{9}= 1 = RHS \end{gathered}$$
LHS = RHS
Hence, verified.

Answer:

$0.6x + \frac{4}{5} = 0.28x + 1.16$
Transposing 0.28x to LHS and 4/5 to RHS, we get
=> 0.6x $-$ 0.28x = 1.16 $-$ $\frac{4}{5}$                   
=> 0.32x = 1.16 $-$ 0.8
 => 0.32x = 0.36
Dividing both sides by 0.32, we get
  => $\frac{0.32x}{0.32} = \frac{0.36}{0.32}$                               
 => x = $\frac{9}{8}$
Verification:
Substituting x = $\frac{9}{8}$ on both sides, we get
$$\begin{gathered} 0.6\left(\frac{9}{8}\right) + \frac{4}{5} = 0.28\left(\frac{9}{8}\right) + 1.16 \\ \frac{5.4}{8} + \frac{4}{5} = \frac{2.52}{8} + 1.16 \\ 0.675 + 0.8 = 0.315 + 1.16 \\ 1.475 = 1.475 \end{gathered}$$

    LHS = RHS
Hence, verified.

Answer:

$$\begin{gathered} 0.5x + \frac{x}{3} = 0.25x + 7 \\ \Rightarrow \frac{05}{10}\text{x}+\frac{\text{x}}{3}=\frac{25\text{x}}{100}+7 \\ \Rightarrow \frac{x}{2} + \frac{x}{3} = \frac{x}{4} + 7 \end{gathered}$$
Transposing x/4 to LHS, we get

$\Rightarrow \frac{x}{2} + \frac{x}{3} -\frac{x}{4} = 7$                     
$$\begin{gathered} \Rightarrow \frac{6x + 4x - 3x}{12 }= 7 \\ \Rightarrow \frac{7x}{12} = 7 \end{gathered}$$
Multiplying both sides by 12, we get
=> $\frac{7x}{12}\times 12 = 7 \times 12$                      
=> 7x = 84
Dividing both sides by 7, we get
=> $\frac{7x}{7} = \frac{84}{7}$                                   
=> x = 12
Verification:
Substituting x = 12 on both sides, we get
$0.5\left(12\right) + \frac{12}{3} = 0.25\left(12\right) + 7$
6 + 4 = 3 + 7
10 = 10
LHS =RHS
Hence, verified.

Answer:

Let the required number be 'x'. Then, 5 subtracted from 3 times x = 3x $-$ 5.
⇒ 3x $-$ 5 = 16
Adding 5 to both sides, we get
⇒ 3x $-$ 5 + 5 = 16 + 5                        
⇒ 3x= 21
Dividing both sides by 3, we get
⇒ $\frac{3x}{3} = \frac{21}{3}$                                     
⇒ x =7
Thus, the required number is 7.

Answer:

Let the required number be 'x'. Thus, when multiplied by 7, it gives 7x, and x increases by 78.
⇒ 7x = x + 78
Transposing x to LHS, we get
⇒ 7x $-$ x = 78                        
⇒ 6x = 78
Dividing both sides by 6, we get
⇒ $\frac{6x}{6} = \frac{78}{6}$                                     
⇒ x =13
Thus, the required number is 13.

Answer:

Let the first number be 'x'. Hence, the second number = x + 1 and the third number = x + 2.
⇒ Sum of first and second numbers = (x) + (x + 1).
ATQ:
⇒ (x) + (x + 1)  = 15 + (x + 2)
⇒ 2x + 1 = 17 + x
Transposing x to LHS and 1 to RHS, we get
⇒ 2x $-$ x = 17 $-$ 1         
⇒ x = 16
So, first number = x = 16
   Second number = x + 1 = 16 + 1 = 17
   Third number = x + 2 = 16 + 2 = 18
Thus, the required consecutive natural numbers are 16, 17 and 18.

Answer:

Let the smaller number be 'x'. So, the larger number = x + 7.
ATQ:
⇒ 6x + (x + 7) = 77
⇒ 6x + x + 7 = 77
⇒ 7x + 7 = 77
Subtracting 7 from both sides, we get
⇒ 7x + 7 $-$ 7 = 77 $-$ 7           
⇒ 7x = 70
Dividing both sides by 7, we get
⇒ $\frac{7x}{7} = \frac{70}{7}$                         
 x = 10
Thus, the smaller number = x = 10, and the larger number = x + 7 = 10 + 7 = 17.
The two required numbers are 10 and 17.

Answer:

Let the number thought of by the man be 'x'.
So, ATQ:
⇒ $\frac{x}{3} + 5 = 2x$
Transposing x/3 to RHS, we get
⇒ 5 = $2x - \frac{x}{3}$                           
⇒ 5 = $\frac{6x - x}{3}$
⇒ 5 = $\frac{5x}{3}$
Multiplying both sides by 3, we get
⇒ $5\times 3 = \frac{5x}{3}\times 3$                       
⇒ 15 = 5x
Dividing both sides by 5, we get
⇒ $\frac{15}{5}= \frac{5x}{5}$                             
⇒ x = 3
Thus, the number thought of by the man is 3.

Answer:

Let the required number be 'x'.
So, ATQ:
⇒ 3x + 5 = 50
Subtracting 5 from both sides, we get
⇒ 3x + 5 $-$ 5 = 50 $-$ 5                 
⇒ 3x = 45
Dividing both sides by 3, we get
⇒ $\frac{3x}{3} = \frac{45}{3}$                               
⇒ x = 15
Thus, the required number is 15.

Answer:

Let the present age of Shikha = 'x' years.
So, the present age of Shikha's brother Ravish = (x + 3) years.
So, sum of their ages  = x + (x+ 3)
⇒ x  + ( x + 3 ) = 37
⇒ 2x + 3 = 37
Subtracting 3 from both sides, we get
⇒ 2x + 3 $-$ 3 = 37 $-$ 3         
⇒ 2x = 34
Dividing both sides by 2, we get
⇒ $\frac{2x}{2 }= \frac{34}{2}$                       
⇒ x = 17
So, the present age of Shikha = 17 years, and the present age of Ravish = x+ 3 = 17 + 3 = 20 years.

Answer:

 Let the present age of Nilu  = 'x' years.
Therefore, the present age of Nilu's mother, Mrs. Jain = (x + 27) years.
So, after 8 years,
Nilu's age = (x + 8), and Mrs. Jain's age  = (x + 27 + 8) = (x + 35) years
⇒ x + 35 = 2(x + 8)
Expanding the brackets, we get
⇒ x + 35 = 2x + 16
Transposing x to RHS and 16 to LHS, we get
⇒ 35 $-$ 16 = 2x $-$ x                    
⇒ x = 19
So, the present age of Nilu  = x =  19 years, and the present age of Nilu's mother  = x+ 27 = 19 + 27 = 46 years.

Answer:

Let the present age of the son  = 'x' years.
Therefore, the present age of his father  = '4x'  years.
So, after 16 years,
Son's age = (x + 16) and father's age  = (4x + 16) years
ATQ:
⇒ 4x + 16 = 2(x + 16)
⇒ 4x + 16 = 2x + 32
Transposing 2x to LHS and 16 to RHS, we get
⇒ 4x - 2x = 32 - 16                    
⇒ 2x = 16
Dividing both sides by 2, we get
⇒ $\frac{2x}{2} = \frac{16}{2}$                             
⇒ x = 8
So, the present age of the son  = x =  8 years, and the present age of the father = 4x = 4(8) = 32 years.

Answer:

Let the age of the girl = 'x' years.
So, the age of her younger sister = (x $-$ 4) years.
Thus, the age of the brother = (x $-$ 4 $-$ 4) years = (x $-$ 8) years.
ATQ:
⇒ (x $-$ 4) + (x $-$ 8) = 16
⇒  x + x $-$ 4 $-$ 8 = 16
⇒ 2x $-$ 12 = 16
Adding 12 to both sides, we get
⇒ 2x $-$ 12 + 12 = 16 + 12           
⇒ 2x = 28
Dividing both sides by 2, we get
⇒ $\frac{2x}{2} = \frac{28}{2}$                             
⇒ x = 14
Thus, the age of the girl = x = 14 years, the age of the younger sister  = x $-$ 4 = 14 $-$ 4 = 10 years,
and the age of the younger brother = x $-$ 8 = 14 $-$  8 = 6 years.

Answer:

 Let the number of sea shells found by Sandy  = 'x'.
So, the number of sea shells found by Anita = (x + 5).
The number of sea shells found by Shella = 2 (x + 5 ).
According to the question,
⇒ x + 2 (x + 5)  = 16
⇒ x + 2x  + 10  = 16
⇒ 3x + 10 = 16
Subtracting 10 from both sides, we get
⇒ 3x + 10 $-$ 10 = 16 $-$ 10             
⇒ 3x = 6
Dividing both sides by 3, we get
⇒ $\frac{3x}{3} = \frac{6}{3}$                                 
⇒ x = 2
Thus, the number of sea shells found by Sandy = x = 2, the number of sea shells found by Anita = x + 5 = 2 + 5 = 7,
and the number of sea shells found by Shella  = 2(x + 5) = 2(2 + 5) = 2(7 ) = 14.

Answer:

Let the number of marbles with Pandy = 'x'.
So, the number of marbles with Andy = '2x'.
Thus, the number of marbles with Sandy = $\frac{1}{2}\left(x +2x\right)$ = $\frac{3x}{2}$.
According to the question,
$\frac{3x}{2}$ $-$ 115  = 110
Adding 115 to both sides, we get
$\frac{3x}{2}$ $-$ 115  + 115 = 110 + 115                   
$\frac{3x}{2}$ = 225
Multiplying both sides by 2, we get
$\frac{3x}{2}\times 2 = 225 \times 2$                                     
 3x = 450
Dividing both sides by 3, we get
$\frac{3x}{3} = \frac{450}{3}$                                             
  x = 150
So, Pandy has 150 marbles, Andy has 2x = 2(150) =  300 marbles, and Sandy has $\frac{3x}{2}$ = $\frac{3\times 150}{2}$ = 225 marbles.

Answer:

Let the number of 50 paise coins  = 'x'.
So, the money value contribution of 50 paise coins = 0.5x.
The number of 25 paise coins = '4x'.
The money value contribution of 25 paise coins = 0.25(4x) = x.
According to the question,
⇒ 0.5x + x = 30
⇒ 1.5x = 30
Dividing both sides by 1.5, we get
⇒ $\frac{1.5x}{1.5}= \frac{30}{1.5}$              
⇒ x = 20
Thus, the number of 50 paise coins  = 'x' = 20, and the number of 25 paise coins  = '4x' = 4 (20) = 80.

Answer:

Let the breadth of the rectangle = 'x' metres.
According to the question,
Length of the rectangle = '2x' metres
Perimeter of a rectangle = 2 (length + breadth)
So,       2 (2x + x) = 228
           =>  2 (3x) = 228
            => 6x = 228
Dividing both sides by 6, we get
           => $\frac{6x}{6} = \frac{228}{6}$                   
            => x = 38
So, the breadth of the rectangle = x = 38 metres, and the length of the rectangle  = 2x = 2(38) = 76 metres.

Answer:

Let the number of 25-paise coins in the purse be 'x'.
So, the value of money in the purse = 0.25x.
But 0.25x = 17.5.
Dividing both sides by 0.25, we get
      => $\frac{0.25x}{0.25} = \frac{17.5}{0.25}$       
      => x = 70
Thus, the number of 25-paise coins in the purse = 70.

Answer:

Let the number of students in the hostel be 'x'.
Quantity of rice consumed by each student = 400 gm.
So, daily rice consumption in the hostel mess = 400(x).
But, daily rice consumption = 50 kg = 50 $\times$ 1000 = 50000 gm  [since 1 kg = 1000 gm].
According to the question,
           400x = 50000
Dividing both sides by 400, we get
         =>  $\frac{400x}{400} =\frac{ 50000}{400}$                                           
         =>  x = 125
Thus, 125 students have their meals in the hostel mess.

Answer:

$$\begin{gathered} \mathrm{If} 3x+2=0, \mathrm{then} \\ 3x=-2 \left(\mathrm{Transposing} +2 \mathrm{to} \mathrm{RHS}\right) \\ \Rightarrow x=-\frac{2}{3} \end{gathered}$$

So, the zero of 3x + 2 is $-\frac{2}{3}$.

Note: A zero is that number, when put in place of the variable, makes the expression equal to zero.

Hence, the correct alternative is option (c).

Answer:

$$\begin{gathered} \mathrm{As}, 2x-\frac{3}{2}=5x+\frac{3}{4} \\ \Rightarrow 2x-5x=\frac{3}{2}+\frac{3}{4} \left(\mathrm{By} \mathrm{transposing} -\frac{3}{2} \mathrm{to} \mathrm{RHS} \mathrm{and} 5x \mathrm{to} \mathrm{LHS}\right) \\ \Rightarrow -3x=\frac{6}{4}+\frac{3}{4} \\ \Rightarrow -3x=\frac{6+3}{4} \\ \Rightarrow x=\frac{9}{4\times \left(-3\right)} \left(\mathrm{By} \mathrm{transposing} -3 \mathrm{to} \mathrm{RHS}\right) \\ \Rightarrow x=\frac{3}{4\times \left(-1\right)} \\ \Rightarrow x=\frac{3}{-4} \\ \therefore x=-\frac{3}{4} \end{gathered}$$

Hence, the correct alternative is option (b).

Answer:

$$\begin{gathered} \mathrm{As}, \frac{x}{2}-4=\frac{x}{3}-1 \\ \Rightarrow \frac{x}{2}-\frac{x}{3}=4-1 \left(\mathrm{By} \mathrm{transposing} \frac{x}{3} \mathrm{to} \mathrm{LHS} \mathrm{and} -4 \mathrm{to} \mathrm{RHS}\right) \\ \Rightarrow \frac{3x}{6}-\frac{2x}{6}=3 \\ \Rightarrow \frac{3x-2x}{6}=3 \\ \Rightarrow \frac{x}{6}=3 \\ \Rightarrow x=3\times 6 \left(\mathrm{By} \mathrm{transposing} 6 \mathrm{to} \mathrm{RHS}\right) \\ \therefore x=18 \end{gathered}$$

Hence, the correct alternative is option (c).

Answer:

$$\begin{gathered} \mathrm{As}, \frac{x+2}{x-2}=\frac{2}{3} \\ \Rightarrow 3\left(x+2\right)=2\left(x-2\right) \left(\mathrm{By} \mathrm{cross} \mathrm{multiplication}\right) \\ \Rightarrow 3x+6=2x-4 \\ \Rightarrow 3x-2x=-6+4 \left(\mathrm{By} \mathrm{transposing} 2x \mathrm{to} \mathrm{LHS} \mathrm{and} 6 \mathrm{to} \mathrm{RHS}\right) \\ \therefore x=-10 \end{gathered}$$

Hence, the correct alternative is option (a).

Answer:

$$\begin{gathered} \mathrm{As}, \frac{x}{6}+\frac{x}{4}=\frac{x}{2}+\frac{3}{4} \\ \Rightarrow \frac{x}{6}+\frac{x}{4}-\frac{x}{2}=\frac{3}{4} \left(\mathrm{By} \mathrm{transposing} \frac{x}{2} \mathrm{to} \mathrm{LHS}\right) \\ \Rightarrow \frac{2x}{12}+\frac{3x}{12}-\frac{6x}{12}=\frac{3}{4} \\ \Rightarrow \frac{2x+3x-6x}{12}=\frac{3}{4} \\ \Rightarrow \frac{-x}{12}=\frac{3}{4} \\ \Rightarrow -x\times 4=3\times 12 \left(\mathrm{By} \mathrm{cross} \mathrm{multiplication}\right) \\ \Rightarrow -4x=36 \\ \Rightarrow x=\frac{36}{-4} \\ \therefore x=-9 \end{gathered}$$

Hence, the correct alternative is option (c).

Answer:

$$\begin{gathered} \mathrm{As}, 2x+\frac{5}{3}=\frac{1}{4}x+4 \\ \Rightarrow 2x-\frac{1}{4}x=4-\frac{5}{3} \left(\mathrm{By} \mathrm{transposing} \frac{5}{3} \mathrm{to} \mathrm{RHS} \mathrm{and} \frac{1}{4}x \mathrm{to} \mathrm{LHS}\right) \\ \Rightarrow \frac{2x}{1}-\frac{x}{4}=\frac{4}{1}-\frac{5}{3} \\ \Rightarrow \frac{8x}{4}-\frac{x}{4}=\frac{12}{3}-\frac{5}{3} \\ \Rightarrow \frac{8x-x}{4}=\frac{12-5}{3} \\ \Rightarrow \frac{7x}{4}=\frac{7}{3} \\ \Rightarrow 7x\times 3=4\times 7 \left(\mathrm{By} \mathrm{cross} \mathrm{multiplication}\right) \\ \Rightarrow 21x=28 \\ \Rightarrow x=\frac{28}{21} \\ \therefore x=\frac{4}{3} \end{gathered}$$

Hence, the correct alternative is option (d).

Answer:

$$\begin{gathered} \mathrm{As}, \frac{x}{2}-\frac{x}{3}=5 \\ \Rightarrow \frac{3x}{6}-\frac{2x}{6}=5 \\ \Rightarrow \frac{3x-2x}{6}=5 \\ \Rightarrow \frac{x}{6}=5 \\ \Rightarrow x=5\times 6 \left(\mathrm{By} \mathrm{transposing} 6 \mathrm{to} \mathrm{RHS}\right) \\ \therefore x=30 \end{gathered}$$

Hence, the correct alternative is option (d).

Answer:

$$\begin{gathered} \mathrm{As}, \frac{x-2}{3}=\frac{2x-1}{3}-1 \\ \Rightarrow \frac{x-2}{3}-\frac{2x-1}{3}=-1 \left(\mathrm{By} \mathrm{transposing} \frac{2x-1}{3} \mathrm{to} \mathrm{LHS}\right) \\ \Rightarrow \frac{\left(x-2\right)-\left(2x-1\right)}{3}=-1 \\ \Rightarrow \frac{x-2-2x+1}{3}=-1 \\ \Rightarrow \frac{-x-1}{3}=-1 \\ \Rightarrow -x-1=-1\times 3 \left(\mathrm{By} \mathrm{transposing} 3 \mathrm{to} \mathrm{RHS}\right) \\ \Rightarrow -x-1=-3 \\ \Rightarrow -x=-3+1 \left(\mathrm{By} \mathrm{transposing} -1 \mathrm{to} \mathrm{RHS}\right) \\ \Rightarrow -x=-2 \\ \therefore x=2 \end{gathered}$$

Hence, the correct alternative is option (a).

Answer:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{two} \mathrm{consecutive} \mathrm{whole} \mathrm{numbers} \mathrm{be} x \mathrm{and} x+1. \\ \mathrm{As}, \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{two} \mathrm{consecutive} \mathrm{whole} \mathrm{numbers} \mathrm{is} 43. \\ \Rightarrow x+\left(x+1\right)=43 \\ \Rightarrow 2x+1=43 \\ \Rightarrow 2x=43-1 \left(\mathrm{By} \mathrm{transposing} 1 \mathrm{to} \mathrm{RHS}\right) \\ \Rightarrow 2x=42 \\ \Rightarrow x=\frac{42}{2} \left(\mathrm{By} \mathrm{transposing} 2 \mathrm{to} \mathrm{RHS}\right) \\ \therefore x=21 \end{gathered}$$

So, the smaller number is 21.

Hence, the correct alternative is option (a).

Answer:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{two} \mathrm{consecutive} \mathrm{odd} \mathrm{numbers} \mathrm{be} x \mathrm{and} x+2. \\ \mathrm{As}, \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{two} \mathrm{consecutive} \mathrm{odd} \mathrm{numbers} \mathrm{is} 36. \\ \Rightarrow x+\left(x+2\right)=36 \\ \Rightarrow 2x+2=36 \\ \Rightarrow 2x=36-2 \\ \Rightarrow 2x=34 \\ \Rightarrow x=\frac{34}{2} \\ \Rightarrow x=17 \\ \therefore x+2=17+2=19 \end{gathered}$$

So, the larger number is 19.

Hence, the correct alternative is option (c).

Answer:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{be} x. \\ \mathrm{As}, \mathrm{twice} \mathrm{the} \mathrm{number} \mathrm{when} \mathrm{increased} \mathrm{by} 7 \mathrm{gives} 25. \\ \Rightarrow 2x+7=25 \\ \Rightarrow 2x=25-7 \left(\mathrm{By} \mathrm{transposing} 7 \mathrm{to} \mathrm{RHS}\right) \\ \Rightarrow 2x=18 \\ \Rightarrow x=\frac{18}{2} \left(\mathrm{By} \mathrm{transposing} 2 \mathrm{to} \mathrm{RHS}\right) \\ \therefore x=9 \end{gathered}$$

So, the number is 9.

Hence, the correct alternative is option (b).

Answer:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{width} \mathrm{of} \mathrm{the} \mathrm{rectangle} \mathrm{be} x. \mathrm{Then}, \\ \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{rectangle}=3x \\ \mathrm{As}, \mathrm{perimeter} \mathrm{of} \mathrm{the} \mathrm{rectangle}=56 \mathrm{m} \\ \Rightarrow 2\times \left(\mathrm{Length}+\mathrm{Breadth}\right)=56 \\ \Rightarrow 2\times \left(3x+x\right)=56 \\ \Rightarrow 2\times 4x=56 \\ \Rightarrow 8x=56 \\ \Rightarrow x=\frac{56}{8} \\ \therefore x=7 \\ \mathrm{So}, \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{rectangle}=3x=3\times 7=21 \mathrm{m} \end{gathered}$$

Hence, the correct alternative is option (c).