RD Sharma solution class 7 chapter 8 Linear Equations In One Variables Exercise 8.2

Exercise 8.2

Page-8.12

Question 1:

Solve each of the following equations and check your answers:
x − 3 = 5

Answer 1:

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 2:

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.

Question 3:

Solve each of the following equations and check your answers:
$x-\frac{3}{5}=\frac{7}{5}$

Answer 3:

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.

Question 4:

Solve each of the following equations and check your answers:
3x = 0

Answer 4:

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.

Question 5:

Solve each of the following equations and check your answes:
$\frac{x}{2}=0$

Answer 5:

 $\frac{x}{2}$ = 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.

Question 6:

Solve each of the following equations and check your answers:
$x-\frac{1}{3}=\frac{2}{3}$

Answer 6:

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.

Question 7:

Solve each of the following equations and check your answers:
$x+\frac{1}{2}=\frac{7}{2}$

Answer 7:

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.

Question 8:

Solve each of the following equations and check your answers:
10 − y = 6

Answer 8:

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.

Question 9:

Solve each of the following equations and check your answers:
7 + 4y = −5

Answer 9:

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.

Question 10:

Solve each of the following equations and check your answers:
$\frac{4}{5}-x=\frac{3}{5}$

Answer 10:

$\frac{4}{5}$ − 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.

Question 11:

Solve each of the following equations and check your answers:
$2y-\frac{1}{2}=-\frac{1}{3}$

Answer 11:

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.

Question 12:

Solve each of the following equations and check your answers:
$14=\frac{7x}{10}-8$

Answer 12:

 $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.

Question 13:

Solve each of the following equations and check your answers:
3 (x + 2) = 15

Answer 13:

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.

Question 14:

Solve each of the following equations and check your answers:
$\frac{x}{4}=\frac{7}{8}$

Answer 14:

 $\frac{x}{4} = \frac{7}{8}$
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.

Question 15:

Solve each of the following equations and check your answers:
$\frac{1}{3}-2x=0$

Answer 15:

 $\frac{1}{3} - 2x = 0$
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.

Question 16:

Solve each of the following equations and check your answers:
3(x + 6) = 24

Answer 16:

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.

Question 17:

Solve each of the following equations and check your answers:
3(x + 2) − 2(x − 1) = 7

Answer 17:

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.

Question 18:

Solve each of the following equations and check your answers:
8(2x − 5) − 6(3x − 7) = 1

Answer 18:

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.

Question 19:

Solve each of the following equations and check your answers:
6(1 − 4x) + 7(2 + 5x) = 53

Answer 19:

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.

Question 20:

Solve each of the following equations and check your answers:
5(2 − 3x) − 17(2x − 5) = 16

Answer 20:

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.

Question 21:

Solve each of the following equations and check your answers:
$\frac{x-3}{5}-2=-1$

Answer 21:

 $\frac{x - 3}{5}- 2 = -1$
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.

Question 22:

Solve each of the following equations and check your answers:
5(x − 2) + 3(x + 1) = 25

Answer 22:

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.