# RD Sharma Solutions Class 7 Linear Equations In One Variable Exercise 8.3

## RD Sharma Solutions Class 7 Chapter 8 Exercise 8.3

### RD Sharma Class 7 Solutions Chapter 8 Ex 8.3 PDF Free Download

#### Exercise 8.3

Q1. 6x + 5 = 2x + 17

SOLUTION :

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

$\frac{4x}{4}$ = $\frac{12}{4}$

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.

Q2. 2(5x – 3)-3(2x – 1)= 9

SOLUTION :

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.

Q3 .  $\frac{x}{2}$ = $\frac{x}{3}$+ 1

SOLUTION :

$\frac{x}{2}$ = $\frac{x}{3}$ + 1

Transposing $\frac{x}{2}$ to LHS , we get

$\frac{x}{2}$$\frac{x}{3}$ = 1

$\frac{3x-2x}{6}$ = 1

$\frac{x}{6}$ = 1

Multiplying both sides by 6, we get

$\frac{x}{6}$ $\times$6 = 1$\times$6

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.

Q4. $\frac{x}{2}$ + $\frac{3}{2}$ = $\frac{2x}{5}$ – 1

SOLUTION :

$\frac{x}{2}$ + $\frac{3}{2}$ = $\frac{2x}{5}$ -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

$\frac{-25}{2}$ + $\frac{3}{2}$ = $\frac{2(-25)}{5}$ – 1

$\frac{-22}{2}$  = -10 -1

-11 = -11

LHS = RHS

Hence, verified.

Q5. $\frac{3}{4}$(x – 1) = x – 3

SOLUTION :

$\frac{3}{4}$(x – 1) = x – 3

On expanding the brackets on both sides, we get

=$\frac{3x}{4}$$\frac{3}{4}$ = x – 3

Transposing  $\frac{3x}{4}$  to RHS and 3 to LHS, we get

= 3 – $\frac{3}{4}$  = x – $\frac{3x}{4}$

= $\frac{12-3}{4}$  = $\frac{4x-3x}{4}$

= $\frac{9}{4}$  = $\frac{x}{4}$

Multiplying both sides by 4, we get

= x = 9

Verification :

Substituting x = 9 on both sides, we get

$\frac{3}{4}$ (9 – 1) = 9 – 3

$\frac{3}{4}$ \) $\times$ 8 = 6

6 = 6

LHS = RHS

Hence, verified.

Q6. 3(x – 3) = 5(2x + 1)

SOLUTION :

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

= –$\frac{7x}{7}$ = $\frac{14}{7}$

=x = -2

Verification :

Substituting x = -2 on both sides, we get

3( -2 – 3) = 5{2(-2) +1}

3(-5) = 5(-3)

-15 = -15

LHS = RHS

Hence, verified.

Q7. 3x – 2(2x – 5) = 2(x + 3) – 8

SOLUTION :

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(4) – 2{2(4) – 5} = 2(4 + 3) – 8

12 – 2(8 – 5) = 14 – 8

12 – 6 = 6

6 = 6

LHS = RHS

Hence, verified.

Q8. x – $\frac{x}{4}$$\frac{1}{2}$ = 3 + $\frac{x}{4}$

SOLUTION :

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.

Q9. $\frac{6x-2}{9}$ + $\frac{3x+5}{18}$ = $\frac{1}{3}$

SOLUTION :

$\frac{6x-2}{9}$ + $\frac{3x+5}{18}$ = $\frac{1}{3}$

= $\frac{6x(2)-2(2)+ 3x+5}{18}$ = $\frac{1}{3}$

= $\frac{12x-4+ 3x+5}{18}$ = $\frac{1}{3}$

$\frac{15x+1}{18}$  = $\frac{1}{3}$

Multiplying both sides by 18, we get

= $\frac{15x+1}{18}$ $\times$ 18  = $\frac{1}{3}$  $\times$ 18

= 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

$\frac{6(\frac{1}{3})-2}{9}$ + $\frac{3(\frac{1}{3})+5}{18}$ = $\frac{1}{3}$

$\frac{2-2}{9}$ + $\frac{1+5}{18}$ = $\frac{1}{3}$

0 + $\frac{6}{18}$ = $\frac{1}{3}$

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

LHS = RHS

Hence, verified.

Q10. m – $\frac{m – 1}{2}$ = 1 – $\frac{m – 2}{3}$

SOLUTION :

m – $\frac{m – 1}{2}$ = 1 – $\frac{m – 2}{3}$

= $\frac{2m -m + 1}{2}$ = $\frac{3 -m + 2}{3}$

= $\frac{m + 1}{2}$  = $\frac{5-m }{3}$

= $\frac{m + 1}{2}$ = $\frac{5}{3}$$\frac{m}{3}$

= $\frac{m }{2}$ + $\frac{ 1}{2}$ = $\frac{5}{3}$$\frac{m}{3}$

Transposing  $\frac{m}{3}$ 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{(\(\frac{7}{5}$$\frac{7}{5}\]”>) – 1}{2}$ = 1 – $\frac{(\(\frac{7}{5}$$\frac{7}{5}\]”> ) – 2}{3}$

$\frac{7}{5}-\frac{7-5}{10}=1-\frac{7-10}{15}$

$\frac{7}{5}-\frac{2}{10}=\frac{15+3}{15}$

$\frac{14-2}{10}=\frac{15+3}{15}$

$\frac{12}{10}=\frac{18}{15}$

$\frac{6}{5}=\frac{6}{5}$

LHS = RHS

Hence, verified.

Q11. $\frac{5x-1}{3}$$\frac{2x-2}{3}$ = 1

SOLUTION :

$\frac{5x-1}{3}$$\frac{2x-2}{3}$  = 1

$\frac{5x-1-2x+2}{3}$ = 1

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

Multiplying both sides by 3, we get 3

$\frac{3x+1}{3}$ $\times$ 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

$\frac{5(\(\frac{2}{3}$$\frac{2}{3}\]”>)-1}{3}$$\frac{2(\(\frac{2}{3}$$\frac{2}{3}\]”>)-2}{3}$ = 1

= $\frac{5\left ( \frac{2}{3} \right )-1}{3}-\frac{2\left ( \frac{2}{3} \right )-2}{3}$

= $\frac{\left ( \frac{10}{3} \right )-1}{3}-\frac{\left ( \frac{4}{3} \right )-2}{3}$

= $\frac{\frac{10-3}{3}}{3}-\frac{ \frac{4-6}{3} }{3}$

= $\frac{10-3}{9}-\frac{ 4-6 }{9}$

= $\frac{7}{9}+\frac{ 2 }{9}$

= $\frac{ 9 }{9}$

= RHS

LHS = RHS Hence, verified.

Q12. 0.6x + $\frac{4}{5}$ = 0.28x + 1.16

SOLUTION :

0.6x + $\frac{4}{5}$ = 0.28x + 1.16

Transposing 0.28x to LHS and $\frac{4}{5}$ 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

0.6($\frac{9}{8}$) + $\frac{4}{5}$  = 0.28($\frac{9}{8}$) + 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

LHS = RHS

Hence, verified.

Q13. 0.5x + $\frac{x}{3}$ = 0.25x + 7

SOLUTION :

0.5x + $\frac{x}{3}$ = 0.25x + 7

$\frac{5}{10}$x + $\frac{x}{3}$ = $\frac{25x}{100}$ + 7

$\frac{x}{2}$ + $\frac{x}{3}$ = $\frac{x}{4}$ + 7

Transposing $\frac{x}{4}$ to LHS , we get

$\frac{x}{2}$ + $\frac{x}{3}$$\frac{x}{4}$ = 7

$\frac{6x+4x-3x}{12}$ = 7

$\frac{7x}{12}$  = 7

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(12) + (12)3 = 0.25(12) + 7

6 + 4 = 3 + 7

10 = 10

LHS = RHS

Hence, verified.