RD Sharma solution class 7 chapter 8 Linear Equations In One Variables Exercise 8.1

Exercise 8.1

Page-8.6

Question 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 $\frac{3x}{2}=6$
(v) y = 2 is the root of y − 3 = 2y − 5
(vi) x = 8 is the root of $\frac{1}{2}x+7=11$

Answer 1:

(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
  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
  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
   LHS = RHS
   Hence, x = 2 is the root of 3x − 2 = 8x − 12.

(iv) x = 4 is the root of $\frac{3x}{2} = 6$.
  Now, substituting x = 4 in place of 'x' in the given equation $\frac{3x}{2} = 6$,
    $$\begin{gathered} \frac{3 \times 4}{2}= 6 \\ \frac{12}{2} = 6 \\ 6 = 6 \end{gathered}$$  
   LHS = RHS
  Hence, x = 4 is the root of $\frac{3x}{2} = 6$.
(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
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 $\frac{1}{2}x + 7 = 11$,
  $\frac{1}{2}\times 8 + 7 = 11$
   4 + 7 = 11  
    11 = 11
  LHS = RHS
Hence, x = 8 is the root of $\frac{1}{2}x + 7 = 11$.

Question 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) $\frac{x}{18}=20$

Answer 2:

(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$\times$1=4	28	No
2	4$\times$2=8	28	No
3	4$\times$3=12	28	No
4	4$\times$4=16	28	No
5	4$\times$5=20	28	No
6	4$\times$6=24	28	No
7	4$\times$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)+4=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.