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.