These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.3

Solve the following equations and check your results.

Question 1.

3x = 2x + 18

Solution:

3x = 2x + 18

Transposing 2x from RHS to L.H.S, we get

3x – 2x = 18

x = 18

Check:

Put x = 18 in L.H.S. and R.H.S. of the equation.

L.H.S. = 3x = 3 × 18 = 54

R.H.S. = 2x + 18 = 2(18) + 18 = 36 +18 = 54

∴ L.H.S. = R.H.S

Question 2.

5t – 3 = 3t – 5

Solution:

5t – 3 = 3t – 5

Transposing (-3) to R.H.S. and 3t to L.H.S., we get

5t – 3t = -5 + 3

2t = -2

Dividing both sides by 2, we get

\(\frac{2 t}{2}=\frac{-2}{2}\)

t = -1

Check:

Put t = -1 in L.H.S. and RHS of the equation

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

R.H.S. = 3t – 5 = 3(-1) – 5 = -3 – 5 = -8

Hence, L.H.S. = R.H.S.

Question 3.

5x + 9 = 5 + 3x

Solution:

5x + 9 = 5 + 3x

Transposing 9 to R.H.S. and 3x to L.H.S.

5x – 3x = 5 – 9

2x = -4

Dividing both sides by 2, we get

\(\frac{2 x}{2}=\frac{-4}{2}\)

x = -2

Check:

Put x = -2 in L.H.S. and R.H.S. of the equation

L.H.S. = 5x + 9

= 5 (-2) + 9

= -10 + 9

= -1

R.H.S. = 5 + 3x

= 5 + 3(-2)

= 5 – 6 = -1

Hence, L.H.S. = R.H.S.

Question 4.

4z + 3 = 6 + 2z

Solution:

4z + 3 = 6 + 2z

Transposing 3 to R.H.S. and 2z to LHS

4z – 2z = 6 – 3

2z = 3

Dividing both sides by 2, we get

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

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

Check:

Put z = \(\frac{3}{2}\) in L.H.S. and R.H.S of the equation

L.H.S. = 4z + 3

= 4\(\left(\frac{3}{2}\right)\) + 3

= 2(3) + 3

= 6 + 3

= 9

R.H.S = 6 + 2z

= 6 + 2\(\left(\frac{3}{2}\right)\)

= 6 + 3

= 9

Hence, L.H.S. = R.H.S.

Question 5.

2x – 1 = 14 – x

Solution:

2x – 1 = 14 – x

Transposing -1 to RHS and -x to L.H.S.

2x + x = 14 + 1

3x = 15

Dividing both sides by 3, we get

\(\frac{3 x}{3}=\frac{15}{3}\)

x = 5

Check:

Put x = 5 in LHS and RHS of the equation

LHS = 2x – 1

= 2(5) – 1

= 10 – 1

= 9

R.H.S = 14 – x

= 14 – 5

= 9

Hence, 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 + 4 = 3x + 4

Transposing 4 to R.H.S. and 3x to LHS, we get

8x – 3x = 4 – 4

5x = 0

Dividing both sides by 5

\(\frac{5 x}{5}=\frac{0}{5}\)

x = 0

Check:

Put x = 0 in L.H.S. and R.H.S. of the equation

L.H.S. = 8x + 4

= 8(0) + 4

= 4

R.H.S = 3(x – 1) + 7

= 3(0 – 1) + 7

= -3 + 7

= 4

Hence, L.H.S. = R.H.S.

Question 7.

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

Solution:

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

Multiplying both sides by 5, we get

5x = 5 × \(\frac{4}{5}\) (x + 10)

5x = 4(x + 10)

5x = 4x + 40

Transposing 4x to L.H.S.

5x – 4x = 40

x = 40

Check:

Put x = 40 in L.H.S. and R.H.S. of the equation

L.H.S. = x = 40

R.H.S. = \(\frac{4}{5}\) (x + 10)

= \(\frac{4}{5}\) (40 + 10)

= \(\frac{4}{5}\) x 50

= 4 × 10

= 40

Hence, L.H.S. = R.H.S

Question 8.

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

Solution:

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

Transposing 1 to R.H.S. and \(\frac{7 \mathrm{x}}{15}\) to L.H.S.

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

\(\frac{10 x-7 x}{15}\) = 2

\(\frac{3 x}{15}\) = 2

Multiplying both sides by 15, we get

\(\frac{3 x}{15}\) × 15 = 2 × 15

3x = 30

Dividing both sides by 3

\(\frac{3 x}{3}=\frac{30}{3}\)

x = 10

Check:

Put x = 10 in L.H.S. and R.H.S. of the equation

Hence, L.H.S. = R.H.S.

Question 9.

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

Solution:

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

Transposing \(\frac{5}{3}\) to R.H.S. and -y to L.H.S.

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

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

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

Divide both sides by 3, we get

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

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

Check:

Put y = \(\frac{7}{3}\) in .L.H.S. and R.H.S of the equation

Hence, L.H.S. = R.H.S

Question 10.

3m = 5m – \(\frac{8}{5}\)

Solution:

3m = 5m – \(\frac{8}{5}\)

Transposing 5m to L.H.S. we get

3m – 5m = \(\frac{-8}{5}\)

-2m = \(\frac{-8}{5}\)

Dividing both sides by -2, we get

\(\frac{-2 m}{-2}=\frac{-8}{5} \div(-2)\)

m = \(\frac{8}{10}=\frac{4}{5}\)

Check:

Put m = \(\frac{4}{5}\) in L.H.S. and R.H.S. of the equation

Hence, L.H.S = R.H.S