You can Download KSEEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Questions and Answers helps you to revise the complete syllabus.

## KSEEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3

Solve the following equations and check your results:

Question 1.

3x = 2x + 18

Solution:

3x = 2x + 18

⇒ 3x – 2x = 18

⇒ x = 18

Check: LHS = 3 × 18 = 54

R.H.S = 2 × 18 + 18 = 36 + 18 = 54

54 = 54

L.H.S = R.H.S

Question 2.

5t – 3 = 3t – 5

Solution:

5t – 3 = 3t – 5

⇒ 5t – 3t = -5 + 3

⇒ 2t = -2

⇒ t = -1

Check: L.H.S = 5(-1) – 3 = -5 – 3 = -8

R.H.S = 3(-1) – 5 = -3 – 5 = -8

-8 = -8

L.H.S = R.H.S

Question 3.

5x + 9 = 5 + 3x

Solution:

5x + 9 = 5 + 3x

⇒ 5x – 3x = 5 – 9

⇒ 2x = -4

⇒ x = -2

Check: L.H.S = 5(-2) + 9 = -10 + 9 = -1

R.H.S = 5 + 3(-2) = 5 – 6 = -1

-1 = -1

L.H.S = R.H.S

Question 4.

4z + 3 = 6 + 2z

Solution:

4z + 3 = 6 + 2z

⇒ 4z – 2z = 6 – 3

⇒ 2z = 3

⇒ z = \(\frac{3}{2}\)

L.H.S = 4(\(\frac{3}{2}\)) + 3 = 6 + 3 = 9

R.H.S = 6 + 2(\(\frac{3}{2}\)) = 6 + 3 = 9

9 = 9

L.H.S = R.H.S

Question 5.

2x – 1 = 14 – x

Solution:

2x – 1 = 14 – x

⇒ 2x + x = 14 + 1

⇒ 3x = 15

⇒ x = \(\frac{15}{3}\) = 5

Check: L.H.S = 2(5) – 1 = 10 – 1 = 9

R.H.S = 14 – 5 = 9

9 = 9

L.H.S = R.H.S

Question 6.

8x + 4 = 3(x – 1) + 7

Solution:

8x + 4 = 3(x – 1) + 7

⇒ 8x + 4 = 3x – 3 + 7

⇒ 8x – 3x = -3 + 7 – 4

⇒ 5x = -7 + 7

⇒ 5x = 0

⇒ x = 0

Check: L.H.S = 8(0) + 4 = 0 + 4 = 4

R.H.S = 3(0 – 1) + 7 = -3 + 7 = 4

4 = 4

L.H.S = R.H.S

Question 7.

x = \(\frac{4}{5}\) (x + 10)

Solution:

x = \(\frac{4}{5}\) (x + 10)

⇒ 5x = 4(x + 10)

⇒ 5x = 4x + 40

⇒ 5x – 4x = 40

⇒ x = 40

Check: LHS = 5 × 40 = 200

RHS = 4(40 + 10) = 4(50) = 200

200 = 200

L.H.S = R.H.S

Question 8.

\(\frac{2 x}{3}+1=\frac{7 x}{15}+3\)

Solution:

Check:

Question 9.

\(2 y+\frac{5}{3}=\frac{26}{3}-y\)

Solution:

\(y=\frac{21}{3 \times 3}=\frac{7}{3}\)

Check:

Question 10.

3m = 5m – \(\frac{8}{5}\)

Solution:

Check: