## RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9B

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9B.

Other Exercises

Solve each of the following equations and verify the answer in each case :

Question 1.
Solution:
x + 5 = 12
=> x + 5 – 5 = 12 – 5
(Subtracting 5 from both sides)
=> x = 7
.’. x = 7 is the solution of the given equation.
Check : Substituting x = 7 in the given equation, we get
L.H.S. = 7 + 5 = 12 and R.H.S. = 12
∴When x = 7, we have L.H.S. = R.H.S.

Question 2.
Solution:
x + 3 = – 2
=>x + 3 – 3 = – 2 – 3
(Subtracting 3 from both sides)
=> x = – 5
∴ x = – 5 is the solution of the given equation.
Check : Substituting x = – 5 in the given equation, we get:
L.H.S. = – 5 + 3 = – 2 and R.H.S. = – 2
When x = – 5,
∴we have L.H.S. = R.H.S.

Question 3.
Solution:
x – 7 = 6
=>x – 7 + 7 = 6 + 7
=> x – 13
So, x = 13 is the solution of the given equation.
Check : Substituting x – 13 in the given equation, we get
L.H.S.= 13 – 7 = 6 and R.H.S. = 6
∴When x = 13, we have L.H.S. = R.H.S.

Question 4.
Solution:
x – 2 = – 5
=> x – 2 + 2 = – 5 + 2
=> x = – 3
So, x = – 3 is the solution of the given equation.
Check : Substituting x = – 3 in the given equation, we get
L.H.S. = – 3 – 2 = – 5 and R.H.S. = – 5 When x = – 3,
we have
L.H.S. = R.H.S.

Question 5.
Solution:
3x – 5 = 13
=>3x – 5 + 5 = 13 + 5
=> 3x = 18
=>$$\\ \frac { 3x }{ 3 }$$ = $$\\ \frac { 18 }{ 3 }$$
(Dividing both sides by 3)
=> x = 6
x = 6 is the solution of the given equation.
Check : Substituting x = 6 in the given equation, we get
L.H.S. = 3 x 6 – 5 = 18 – 5 = 13 and R.H.S. = 13
∴ When x = 6, we have L.H.S. = R.H.S

Question 6.
Solution:
4x + 7 = 15
=> 4x + 7 – 7 = 15 – 7
(Subtracting 7 from both sides)
=> 4x = 8
=> $$\\ \frac { 4x }{ 4 }$$ = $$\\ \frac { 8 }{ 4 }$$
(Dividing both sides by 4)
=> x = 2
x = 2 is the solution of the given equation.
Check : Substituting x = 2 in the given equation, we get
L.H.S. = 4 x 2 + 7 = 8 + 7 = 15 and R.H.S. = 15
∴When x = 2, we have L.H.S. = R.H.S.

Question 7.
Solution:
$$\\ \frac { x }{ 5 }$$ = 12
=> $$\\ \frac { x }{ 5 }$$ x 5 = 12 x 5
(Multiplying both sides by 5)
=> x = 60
x = 60 is the solution of the given equation.
Check : Substituting x = 60 in the given equation, we get
L.H.S. = $$\\ \frac { 60 }{ 5 }$$ = 12 and R.H.S. = 12
When x = 60, we have
∴L.H.S. = R.H.S.

Question 8.
Solution:
$$\\ \frac { 3x }{ 5 }$$ = 15
=> $$\\ \frac { 3x }{ 5 }$$ x $$\\ \frac { 5 }{ 3 }$$ = 15 x $$\\ \frac { 5 }{ 3 }$$

Question 9.
Solution:
5x – 3 = x + 17
=> 5x – x = 17 + 3

So. x = 5 is a solution of the given
equation.
Check : Substituting x = 5 in the given
equation, we get
L.H.S. = 5 x 5 – 3 = 25 – 3 = 22
R.H.S. 5 + 17 = 22
∴When x = 5, we have L.H.S. = R.H.S.

Question 10.
Solution:
2x – $$\\ \frac { 1 }{ 2 }$$ = 3
=> 2x = 3 + $$\\ \frac { 1 }{ 2 }$$

Question 11.
Solution:
3(x + 6) = 24
=> $$\\ \frac { 3(x+6) }{ 3 }$$ = $$\\ \frac { 24 }{ 3 }$$
(Dividing both sides by 3)
x + 6 = 8
=> x = 8 – 6
(Transposing 6 to R.H.S.)
=> x = 2
x = 2 is a solution of the given equation.
Check : Substituting the value of x = 2
in the given equation, we get
L.H.S. = 3(2 + 6 ) = 3 x 8 = 24
and RH.S. = 24
∴When x = 2, we have L.H.S. = R.H.S.

Question 12.
Solution:
6x + 5 = 2x + 17
=> 6x – 2x = 17 – 5
(Transposing 2 x to L.H.S. and 5 to R.H.S.)
=> 4x = 12
=> $$\\ \frac { 4x }{ 4 }$$ = $$\\ \frac { 12 }{ 4 }$$
(Dividing both sides by 4)
=> x = 3
x = 3 is a solution of the given
equation.
Check : Substituting x = 3 in the given
equation, we get
L.H.S.= 6 x 3 + 5 = 18 + 5 = 23
R.H.S.= 2 x 3 + 17 = 6 + 17 = 23
∴When x = 3, we have L.H.S. = R.H.S.

Question 13.
Solution:
$$\\ \frac { x }{ 4 }$$ – 8 = 1
=> $$\\ \frac { x }{ 4 }$$ = 1 + 8

R.H.S = 1
∴When x = 36,we have L.H.S. = R.H.S.

Question 14.
Solution:
$$\\ \frac { x }{ 2 }$$ = $$\\ \frac { x }{ 2 }$$ + 1
=> $$\\ \frac { x }{ 2 }$$ – $$\\ \frac { x }{ 3 }$$ = 1

Question 15.
Solution:
3(x + 2) – 2(x – 1) = 7
=> 3x + 6 – 2x + 2 = 7
(Removing brackets)
3x – 2x + 6 + 2 = 7
x + 8 = 7
x = 7 – 8
(Transposing 8 to R.H.S.)
x = – 1 is a solution of the given
equation.
Check : Substituting x = – 1 in the given
equation, we get
L.H.S. = 3 ( – 1 + 2) – 2( – 1 – 1)
= 3 x 1 + ( – 2 x – 2)
= 3 + 4 = 7 and
R.H.S. = 7
When x = – 1, we have
L.H.S. = R.H.S.

Question 16.
Solution:
5 (x- 1) + 2 (x + 3) + 6 = 0
= 5 (x – 1) + 2 (x + 3) = – 6
(Transposing 6 to R.H.S.)
= 5x – 5 + 2x + 6 = – 6

Question 17.
Solution:
6(1 – 4 x) + 7 (2 + 5 x) – 53
=> 6 – 24x + 14 + 35 x = 53
(Removing brackets)
=> – 24 x + 35 x + 14 + 6 = 53
=> 11 x + 20 = 53
=> 11 x = 53 – 20
=> 11 x = 33
(Transposing 20 to R.H.S.)

Question 18.
Solution:
16 (3x – 5) – 10 (4x – 8) = 40
=> 48x – 80 – 40x + 80 = 40
(Removing brackets)
=> 48x – 40 x – 80 + 80 = 40
=> 8x = 40

Question 19.
Solution:
3 (x + 6) + 2 (x + 3) = 64
=> 3x + 18 + 2x + 6 = 64
(Removing brackets)
=> 3x + 2x + 18 + 6 = 64
=> 5x + 24 = 64
=> 5x = 64 – 24

Question 20.
Solution:
3(2 – 5x) – 2 (1 – 6x) = 1
=> 6 – 15x – 2 + 12x = 1
(Removing brackets)
=> 6 – 2 – 15x + 12x = 1
=> 4 – 3x = 1
– 3x = 1 – 4

Question 21.
Solution:
$$\\ \frac { n }{ 4 } -5$$ = $$\\ \frac { n }{ 6 }$$ + $$\\ \frac { 1 }{ 2 }$$
Multiplying each term by 12, the L.C.M. of 4, 6, 2, we get

Question 22.
Solution:
$$\\ \frac { 2m }{ 3 } +8$$ = $$\\ \frac { m }{ 2 } -1$$
Multiplying each term by 6, the L.C.M. of 2 and 3, we get
$$\\ \frac { 2m }{ 3 }$$ x 6 + 8 x 6 = $$\\ \frac { m }{ 2 }$$ x 6 – 1 x 6

Question 23.
Solution:
$$\\ \frac { 2x }{ 5 }$$ – $$\\ \frac { 3 }{ 2 }$$ = $$\\ \frac { x }{ 2 } +1$$
Multiplying each term by 10, the L.C.M. of 5 and 2, we get

Question 24.
Solution:
$$\\ \frac { x-3 }{ 5 }$$ – 2 = $$\\ \frac { 2x }{ 5 }$$
multiplying each term by 5, we get

Question 25.
Solution:
$$\\ \frac { 3x }{ 10 }$$ – 4 = 14
=> $$\\ \frac { 3x }{ 10 }$$ = 14 + 4

Question 26.
Solution:
$$\\ \frac { 3 }{ 4 } (x-1)$$ = (x – 3)

Hope given RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9B are helpful to complete your math homework.

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