## 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**

- RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9A
- RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9B
- RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9C

**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

(Adding 7 to both sides)

=> 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

(Adding 2 on both sides)

=> 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

(Adding 5 on both sides)

=> 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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