 # RS Aggarwal Solutions for Class 8 Chapter 8 - Linear Equations Exercise 8A

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## Download PDF of RS Aggarwal Solutions for Class 8 Chapter 8 – Linear Equations Exercise 8A    ### Access answers to Maths RS Aggarwal Solutions for Class 8 Chapter 8 – Linear Equations Exercise 8A

Solve:

1. 8x+3=27+2x

Solution:

Given 8x+3=27+2x

By transposing the above equation we can write as

8x-2x=27-3

6x=24

Again by transposing

x=24/6=4

2. 5x+7=2x-8

Solution:

Given 5x+7=2x-8,

By transposing the above equation we can write as

5x-2x=-7-8

3x=-15

Again by transposing

x=-15/3=-5

3. 2z-1=14-z

Solution:

Given 2z-1=14-z,

By transposing the above equation we can write as

2z+z=14+1

3z=15

Again by transposing

z=15/3=5

4. 9x+5=4(x-2)+8

Solution:

Given 9x+5=4(x-2) + 8,

By transposing the above equation we can write as

9x+5=4x-8+8

9x-4x=5

Again by transposing

5x=5

x=5/5=1

5. (7y)/5=y-4

Solution:

Given (7y)/5=y-4,

By transposing the above equation we can write as

7y=5(y-4)

7y=5y-20

Again by transposing

7y-5y=-20

2y=-20

y=-20/2=-10

6. 3x+2/3=2x+1

Solution:

Given 3x+2/3=2x+1

By transposing the above equation we can write as

3x+2=3(2x+1)

3x+2=6x+3

Again by transposing

3x-6x=3-2

-3x=1

x=-1/3

7. 15(y-4)-2(y-9)+5(y+6)=0

Solution:

Given 15(y-4)-2(y-9) + 5(y+6)=0

Now by rearranging we get

15y-60-2y+18+5y+30=0

18y-12=0

By transposing the above equation we can write as

18y=12

Again by transposing

y=12/18

y=6/9=2/3

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

Solution:

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

Now by rearranging we get

15x-21-18x+22=32x-52-17

-3x+1=32x-69

By transposing the above equation we can write as

35x=70

Again by transposing

x=70/35

x=2

9. (x-5)/2-(x-3)/5=1/2

Solution:

Given (x-5)/2-(x-3)/5=1/2

Now by taking L.C.M for 5 and 2 is 10

(5(x-5)-2(x-3))/10=1/2

By transposing the above equation we can write as

(5x-25-2x+6)=10/2

3x-19=2

Again by transposing

3x=19+2=21

x=21/3=7

10. (3t-2)/4-(2t+3)/3=2/3-t

Solution:

Given (3t-2)/4-(2t+3)/3=2/3-t

Now by taking L.C.M for 4 and 3 is 12

(3(3t-2)-4(2t+3))/12=2/3-t

By transposing the above equation we can write as

(9t-6-8t-12)/12=2/3-t

Again by transposing

t=26/13

t=2

11. (2x+7)/5-(3x+11)/2=(2x+8)/3-5

Solution:

Given (2x+7)/5-(3x+11)/2 = (2x+8)-15

Now by taking L.C.M for 2 and 5 is 10

15(2(2x+7)-5(3x+11))/10 = (2x+8)/3-5

By transposing the above equation we can write as

33x-20x=123-70

Again by transposing

x=-53/53

x=-1

12. (5x-4)/6=4x+1-(3x+10)/2

Solution:

Given (5x-4)/6=4x+1-(3x+10)/2

Now by taking L.C.M for 1 and 2 is 1

5x-4-6(4x+1) + 3(3x+10)/6=0

By cross multiplication we get

5x-4-6(4x+1) + 3(3x+10) = 0

-10x=-20

Again by transposing,

x=20/10=2

13. 5x-(1/3)(x+1)=6(x+(1/30))

Solution:

Given 5x-(1/3) (x+1) = 6(x + (1/30))

Taking L.C.M on both sides, we get

(15x-(x+1))/3=6(30x+1)/30

By cross multiplication,

10(14x-1) = 6(30x+1)

140x-180x=6+10

-40x=16

-x=2/5

14. 4-(2(z-4))/2=1/2(2z+5)

Solution:

Given 4-(2(z-4))/2=1/2(2z+5)

Now by taking L.C.M of 1 and 3 is 3

12-2(z-4)/3 = (2z+5)/2

By cross multiplication we get

2(12-2z+8) = 3(2z+5)

40-4z=6z+15

-10z=-25

z=25/10=5/2

### Access other exercises of RS Aggarwal Solutions for Class 8 Chapter 8 – Linear Equations

1. Exercise 8B Solutions 27 questions
2. Exercise 8C Solutions 18 questions

## RS Aggarwal Solutions for Class 8 Maths Chapter 8 – Linear Equations Exercise 8A

Exercise 8A of RS Aggarwal Solutions for Chapter 8, Linear Equations deals with the topics related to equations and rules for solving a linear equation. Some of the topics focused prior to exercise 8A include the following.

• Definition of equation
• Rules for solving a linear equation
• Transposing

The RS Aggarwal Solutions can help the students in practising and learning each and every concept as it provides solutions to all questions asked in the RS Aggarwal textbook. Those who aim to score high in the Maths of class 8 are advised to practice all the questions present in RS Aggarwal as many times as possible.