## RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8A.

**Other Exercises**

- RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8A
- RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8B
- RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8C

**Solve :**

**Question 1.**

**Solution:**

8x + 3 = 27 + 2x

=> 8x – 2x

=> 27 – 3

=> 6x = 24

=> x = \(\\ \frac { 24 }{ 6 } \) = 4

x = 4

**Question 2.**

**Solution:**

5x + 7 = 2x – 8

=> 5x – 2x = – 8 – 7

=> 3x = – 15

=> x = \(\\ \frac { -15 }{ 3 } \) = – 5

x = – 5

**Question 3.**

**Solution:**

2z – 1 = 14 – z

=> 2z + z = 14 + 1

=> 3z = 15

=> z = \(\\ \frac { 15 }{ 3 } \) = 5

z = 5

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**Question 4.**

**Solution:**

9x + 5 = 4(x – 2) +8

=> 9x + 5 = 4x – 8 + 8

=> 9x – 4x = – 8 + 8 – 5

=> 5x = – 5

=> x = \(\\ \frac { -5 }{ 5 } \) = – 1

x = – 1

**Question 5.**

**Solution:**

\(\\ \frac { 7y }{ 5 } \) = y – 4

Multiplying both sides by 5,

\(\\ \frac { 7y }{ 5 } \) x 5 = 5(y – 4)

=> 5 (y-4)

=> 7y = 5y – 20

=> 7y – 5y = – 20

=> 2y = – 20

=> y = \(\\ \frac { -20 }{ 2 } \) = – 10

Hence y = – 10 Ans.

**Question 6.**

**Solution:**

3x + \(\\ \frac { 2 }{ 3 } \) = 2x + 1

=> 3x – 2x = 1 – \(\\ \frac { 2 }{ 3 } \)

=> x = \(\\ \frac { 3-2 }{ 3 } \) = \(\\ \frac { 1 }{ 3 } \)

Hence x = \(\\ \frac { 1 }{ 3 } \) Ans.

**Question 7.**

**Solution:**

15(y – 4) – 2(y – 9) + 5(y + 6) = 0

15y – 60 – 2y + 18 + 5y + 30 = 0

=> 15y – 2y + 5y = 60 – 18 – 30

=> 18y = 12

=> y = \(\\ \frac { 12 }{ 18 } \) = \(\\ \frac { 2 }{ 3 } \)

=> y = \(\\ \frac { 2 }{ 3 } \)

**Question 8.**

**Solution:**

3(5x – 7) – 2(9x – 11) = 4(8x – 13) – 17

=> 15x – 21 – 18x + 22 = 32x – 52 – 17

=> 15x – 18x – 32x = – 52 – 17 + 21 – 22

=> 15x – 50x = – 70

=> – 35x = – 70

=> x = \(\\ \frac { -70 }{ -35 } \) = 2

x = 2

**Question 9.**

**Solution:**

\(\\ \frac { x-5 }{ 2 } \) – \(\\ \frac { x-3 }{ 5 } \) = \(\\ \frac { 1 }{ 2 } \)

Multiplying each term by 10, the L.C.M. of 2 and 5

**Question 10.**

**Solution:**

\(\\ \frac { 3t-2 }{ 4 } \) – \(\\ \frac { 2t+3 }{ 3 } \) = \(\\ \frac { 2 }{ 3 } \) – t

**Question 11.**

**Solution:**

\(\frac { 2x+7 }{ 5 } +\frac { 3x+11 }{ 2 } =\frac { 2x+8 }{ 3 } -5\)

Multiplying by 30, the L.C.M. of 5, 2 and 3.

**Question 12.**

**Solution:**

\(\frac { 5x-4 }{ 6 } =4x+1-\frac { 3x+10 }{ 2 } \)

Multiplying by 6, the L.C.M. of 6 and 2

**Question 13.**

**Solution:**

\(5x-\frac { 1 }{ 3 } \left( x+1 \right) =6\left( x+\frac { 1 }{ 30 } \right) \)

=> 5x – \(\\ \frac { x+1 }{ 3 } \) = 6x + \(\\ \frac { 1 }{ 5 } \)

Multiplying by 15, the L.C.M. of 3 and 5

**Question 14.**

**Solution:**

\(4-\frac { 2\left( z-4 \right) }{ 3 } =\frac { 1 }{ 2 } \left( 2z+5 \right)\)

**Question 15.**

**Solution:**

\(\frac { 3\left( y-5 \right) }{ 4 } -4y=3-\frac { \left( y-3 \right) }{ 2 } \)

Multiplying by 4, the L.C.M. of 4 and 2

**Question 16.**

**Solution:**

\(\\ \frac { 8x-3 }{ 3x } \) = \(\\ \frac { 2 }{ 1 } \)

By cross multiplication,

8x – 3 = 6x

=> 8x – 6x = 3

=> 2x = 3

=> x = \(\\ \frac { 3 }{ 2 } \)

x = \(\\ \frac { 3 }{ 2 } \)

**Question 17.**

**Solution:**

\(\\ \frac { 9x }{ 7-6x } \) = \(\\ \frac { 15 }{ 1 } \)

By cross multiplication,

9x = 105 – 90x

=> 9x + 90x = 105

=> 99x = 105

=> x = \(\\ \frac { 105 }{ 99 } \) = \(\\ \frac { 35 }{ 33 } \)

x = \(\\ \frac { 35 }{ 33 } \)

**Question 18.**

**Solution:**

\(\\ \frac { 3x }{ 5x+2 } \) = \(\\ \frac { -4 }{ 1 } \)

By cross multiplication,

3x × 1 = – 4×(5x + 2)

=> 3x = – 20x – 8

=> 3x + 20x = – 8

=> 23x = – 8

=> x = \(\\ \frac { -8 }{ 23 } \)

Hence x = \(\\ \frac { -8 }{ 23 } \)

**Question 19.**

**Solution:**

\(\\ \frac { 6y-5 }{ 2y } \) = \(\\ \frac { 7 }{ 9 } \)

By cross multiplication,

9(6y – 5) = 7 × 2y

=> 54y – 45 = 14y

=> 54y – 14y = 45

=> 40y = 45

=> y = \(\\ \frac { 45 }{ 40 } \) = \(\\ \frac { 9 }{ 8 } \)

Hence y = \(\\ \frac { 9 }{ 8 } \) Ans.

**Question 20.**

**Solution:**

\(\\ \frac { 2-9z }{ 17-4z } \) = \(\\ \frac { 4 }{ 5 } \)

By cross multiplication,

5 (2 – 9z) = 4(17 – 4z)

=> 10 – 45z = 68 – 16z

=> – 45z + 16z = 68 – 10

=> – 29 = 58

=> z = \(\\ \frac { 58 }{ -29 } \) = – 2

Hence z = – 2 Ans.

**Question 21.**

**Solution:**

\(\\ \frac { 4x+7}{ 9-3x } \) = \(\\ \frac { 1 }{ 4 } \)

By cross multiplication,

4(4x + 7) = 1 (9 – 3x)

=> 16x + 28 = 9 – 3x

=> 16x + 3x = 9 -28

=> 19x = – 19

=> x = \(\\ \frac { -19 }{ 19 } \) = – 1

Hence x = – 1 Ans.

**Question 22.**

**Solution:**

\(\\ \frac { 7y+4}{ y+2 } \) = \(\\ \frac { -4 }{ 3 } \)

By cross multiplication,

3 (7y + 4) = – 4 (y + 2)

=> 21y + 12 = – 4y – 8

=> 21y + 4y = – 8 – 12

=> 25y = – 20

=> y = \(\\ \frac { -20 }{ 25 } \) = \(\\ \frac { -4 }{ 5 } \)

y = \(\\ \frac { -4 }{ 5 } \)

**Question 23.**

**Solution:**

\(\\ \frac { 15(2-y)-5(y+6) }{ 1-3y } \) = \(\\ \frac { 10 }{ 1 } \)

By cross multiplication,

15 (2 – y) – 5(y + 6) = 10 (1 – 3y)

=> 30 – 15y – 5y – 30 = 10 – 30y

=> – 15y – 5y + 30y = 10 – 30 + 30

=> 30y – 20y = 10

=> 10y = 10

y = \(\\ \frac { 10 }{ 10 } \) = 1

Hence y = 1 Ans.

**Question 24.**

**Solution:**

\(\\ \frac { 2x-(7-5x) }{ 9x-(3x+4x) } \) = \(\\ \frac { 7 }{ 6 } \)

**Question 25.**

**Solution:**

\(m-\frac { \left( m-1 \right) }{ 2 } =1-\frac { \left( m-2 \right) }{ 3 } \)

**Question 26.**

**Solution:**

\(\\ \frac { 3x+5 }{ 4x+2 } \) = \(\\ \frac { 3x+4 }{ 4x+7 } \)

By cross multiplication,

(3x + 5)(4x + 7) = (3x + 4)(4x + 2)

=> 12x² + 21x + 20x + 35

**Question 27.**

**Solution:**

\(\\ \frac { 9x-7 }{ 3x+5 } \) = \(\\ \frac { 3x-4 }{ x+6 } \)

**Question 28.**

**Solution:**

\(\\ \frac { 2-7x }{ 1-5x } \) = \(\\ \frac { 3+7x }{ 4+5x } \)

Hope given RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8A are helpful to complete your math homework.

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