# RD Sharma Solutions Class 7 Linear Equations In One Variable Exercise 8.1

## RD Sharma Solutions Class 7 Chapter 8 Exercise 8.1

### RD Sharma Class 7 Solutions Chapter 8 Ex 8.1 PDF Download

#### Exercise 8.1

Q 1. Verify by substitution that :

(i). x = 4 is the root of 3x – 5 = 7

(ii). x = 3 is the root of 5 + 3x = 14

(iii). x = 2 is the root of 3x – 2 = 8x – 12

(iv). x = 4 is the root of 3x/2 = 6

(v). y = 2 is the root of y – 3 = 2y – 5

(vi). x = 8 is the root of $\frac{1}{2}x+7$ = 11

SOLUTION :

(i). x = 4 is the root of 3x — 5 = 7.

Now, substituting x = 4 in place of ‘x’ in the given equation 3x — 5 = 7,

3(4) –  5 = 7

12 – 5 = 7

7 = 7

Since, LHS = RHS

Hence, x = 4 is the root of 3x — 5 = 7.

(ii). x = 3 is the root of 5 + 3x = 14.

Now, substituting x = 3 in place of ‘x’ in the given equation 5 + 3x = 14,

5 + 3(3) = 14

5 + 9 = 14

14 = 14

Since, LHS = RHS

Hence, x = 3 is the root of 5 + 3x = 14.

(iii). x = 2 is the root of 3x – 2 = 8x – 12.

Now, substituting x = 2 in place of ‘x’ in the given equation 3x – 2 = 8x – 12,

3(2) – 2 = 8(2) – 12

6 – 2 = 16 – 12

4 =  4

Since , LHS = RHS

Hence, x = 2 is the root of 3x – 2 = 8x – 12.

(iv). x = 4 is the root of 3x/2 = 6.

Now, substituting x = 4 in place of ‘x’ in the given equation 3x/2 = 6,

(3×4)/2 = 6

12/2 = 6

6 = 6

Since, LHS = RHS

Hence, x = 4 is the root of 3x/2 = 6.

(v). y = 2 is the root of y – 3 = 2y – 5.

Now, substituting y = 2 in place of ‘y’ in the given equation y – 3 = 2y – 5,

2 – 3 = 2(2) – 5

-1 = 4 – 5

-1 = -1

Since,  LHS = RHS

Hence, y = 2 is the root of y – 3 = 2y –  5.

(vi). x = 8 is the root of  $\frac{1}{2}x+7$  = 11.

Now, substituting x = 8 in place of ‘x’ in the given equation $\frac{1}{2}x+7$  = 11,

$\frac{1}{2}(8)+7$ =11

4 + 7 = 11

11 = 11

Since,  LHS = RHS

Hence, x = 8 is the root of $\frac{1}{2}x+7$  = 11.

Q 2. Solve each of the following equations by trial and error method :

(i). x + 3 = 12

(ii). x – 7 = 10

(iii). 4x = 28

(iv). $\frac{x}{2}+7$ = 11

(v). 2x + 4 = 3x

(vi). $\frac{x}{4}$ = 12

(vii). $\frac{15}{x}$ = 3

(viii). $\frac{x}{18}$ = 20

SOLUTION :

(i). x + 3 = 12

Here, LHS = x + 3 and RHS = 12

 x LHS RHS Is LHS = RHS 1 1 + 3 = 4 12 No 2 2 + 3 = 5 12 No 3 3 + 3 = 6 12 No 4 4 + 3 = 7 12 No 5 5 + 3 = 8 12 No 6 6 + 3 = 9 12 No 7 7 + 3 = 10 12 No 8 8 + 3 = 11 12 No 9 9 + 3 = 12 12 Yes

Therefore, if x = 9, LHS = RHS.

Hence, x = 9 is the solution to this equation.

(ii). x – 7 = 10

Here, LHS = x – 7 and RHS = 10.

 x LHS RHS Is LHS = RHS 9 9 -7 = 2 10 No 10 10 – 7 = 3 10 No 11 11 – 7 =4 10 No 12 12 – 7 =5 10 No 13 13 – 7 = 6 10 No 14 14 – 7 = 7 10 No 15 15 – 7 = 8 10 No 16 16 – 7 = 9 10 No 17 17 – 7 = 10 10 Yes

Therefore, if x = 17, LHS = RHS.

Hence, x = 17 is the solution to this equation.

(iii). 4x = 28

Here, LHS = 4x and RHS = 28.

 x LHS RHS Is LHS = RHS 1 4 x 1 = 4 28 No 2 4 x 2 = 8 28 No 3 4 x 3 = 12 28 No 4 4 x 4 = 16 28 No 5 4 x 5 = 20 28 No 6 4 x 6 = 24 28 No 7 4 x 7 = 28 28 Yes

Therefore, if x = 7, LHS  = RHS

Hence, x = 7 is the solution to this equation.

(iv). $\frac{x}{2}+7$  = 11

Here, LHS = $\frac{x}{2}+7$  and RHS = 11.

Since RHS is a natural number, $\frac{x}{2}$  must also be a natural number, so we must substitute values of x that are multiples of 2.

 x LHS RHS Is LHS = RHS 2 $\frac{2}{2}+7$=8 11 No 4 $\frac{4}{2}+7$=9 11 No 6 $\frac{6}{2}+7$=10 11 No 8 $\frac{8}{2}+7$=11 11 Yes

Therefore, if x = 8, LHS = RHS.

Hence, x = 8 is the solution to this equation.

(v). 2x + 4 = 3x

Here, LHS = 2x + 4 and RHS = 3x.

 x LHS RHS Is LHS = RHS 1 2(1) + 4 =6 3(1) = 3 No 2 2(2) + 4 = 8 3(2) = 6 No 3 2(3) + 4 =10 3(3) = 9 No 4 2(4) + 5 = 12 3(4)= 12 Yes

Therefore, if x = 4, LHS = RHS.

Hence, x = 4 is the solution to this equation.

(vi). $\frac{x}{4}$ = 12

Here, LHS = $\frac{x}{4}$ and RHS = 12.

Since RHS is a natural number, $\frac{x}{4}$ must also be a natural number, so we must substitute values of x that are multiples of 4.

 X LHS RHS Is LHS = RHS 16 $\frac{16}{4}$=4 12 NO 20 $\frac{20}{4}$=5 12 NO 24 $\frac{24}{4}$=6 12 NO 28 $\frac{28}{4}$=7 12 NO 32 $\frac{32}{4}$=8 12 NO 36 $\frac{36}{4}$=9 12 NO 40 $\frac{40}{4}$=10 12 NO 44 $\frac{44}{4}$=11 12 NO 48 $\frac{48}{4}$=12 12 Yes

Therefore, if x = 48, LHS = RHS.

Hence, x = 48 is the solution to this equation.

(vii). $\frac{15}{x}$  = 3

Here, LHS = $\frac{15}{x}$  and RHS = 3.

Since RHS is a natural number, $\frac{15}{x}$  must also be a natural number , so we must substitute values of x that are factors of 15.

 x LHS RHS Is LHS = RHS 1 $\frac{15}{1}$=15 3 No 3 $\frac{15}{3}$=5 3 No 5 $\frac{15}{5}$=3 3 Yes

Therefore, if x = 5, LHS = RHS.

Hence, x = 5 is the solution to this equation.

(viii). $\frac{x}{18}$ = 20

Here, LHS = $\frac{x}{18}$ and RHS = 20.

Since RHS is a natural number, $\frac{x}{18}$ must also be a natural number, so we must substitute values of x that are multiples of 18.

 X LHS RHS Is LHS = RHS 324 $\frac{324}{18}$=18 20 No 342 $\frac{342}{18}$=19 20 No 360 $\frac{360}{18}$==20 20 Yes

Therefore, if x = 360, LHS = RHS.

Hence, x = 360 is the solution to this equation.