HBSE 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.3

Haryana State Board HBSE 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.3 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions 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
or, 3x – 2x = 18
∴ x = 18.
Check:
L.H.S. = 3 × 18 = 54
R.H.S. = 2 × 18 + 18 = 54
∴ L.H.S. = R.H.S.

Question 2.
5t – 3 = 3t – 5
Solution:
5t – 3 = 3t – 5
or, 5t – 3t = -5 + 3
or, 2t = -2
or, t = -1
∴ t = -1
Check:
L.H.S = 5(-1) – 3
= -5 – 3 = -8
R.H.S = 3(-1) – 5
= -3 – 5 = -8
∴ L.H.S = R.H.S

HBSE 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.3

Question 3.
5x + 9 = 5 + 3x
Solution:
5x + 9 = 5 + 3x
or, 5x – 3x = 5 – 9
or, 2x = -4
or, x = -2
∴ x = -2
Check:
L.H.S. = 5 (-2) + 9
= – 10 + 9 = -1
R.H.S. = 5 +3 (-2)
= 5 – 6 = -1
∴ L.H.S. = R.H.S.

Question 4.
4z + 3 = 6 + 2z
Solution:
4z – 2z = 6 – 3
or, 2z = 3
or, z = \(\frac{3}{2}\)
Check:
L.H.S. = 4z + 3
= 4 × \(\frac{3}{2}\) + 3 = 9
R.H.S. = 6 + 2 × \(\frac{3}{2}\) = 9
∴ L.H.S. = R.H.S.

HBSE 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.3

Question 5.
2x – 1 = 14 – x
Solution:
2x – 1 = 14 – x
or, 2x + x = 14 + 1
or, 3x = 15
or, x = 5
∴ x = 5
Check:
L.H.S. = 2 × 5 – 1
= 10 – 1 = 9
R.H.S. = 14 – 5 = 9
∴ L.H.S. = R.H.S.

Question 6.
8x + 4 = 3 (x – 1) + 7
Solution:
8x + 4 = 3 (x – 1) + 7
or, 8x + 4 = 3x – 3 + 7
or, 8x – 3x = -3 + 7 – 4
or, 5x = 0
∴ x = 0
Check:
L.H.S. = 8 × 0 + 4 = 4
R.H.S. = 3(0 – 1) + 7
= -3 + 7 = 4
∴ L.H.S. = R.H.S.

HBSE 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.3

Question 7.
x = \(\frac{4}{5}\)(x + 10)
Solution:
x = \(\frac{4}{5}\)(x + 10)
or, 5x = 4(x + 10)
or, 5x = 4x + 40
or, 5x – 4x =40
∴ x = 40
Check:
L.H.S. = 40
R.H.S. = \(\frac{4}{5}\)(40 + 10)
= \(\frac{4}{5}\) × 50 = 40
∴ L.H.S. = R.H.S.

Question 8.
\(\frac{2x}{3}\) + 1 = \(\frac{7x}{15}\) + 3
Solution:
HBSE 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.3 1

HBSE 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.3

Question 9.
2y + \(\frac{5}{3}\) = \(\frac{26}{3}\) – y.
Solution:
2y + \(\frac{5}{3}\) = \(\frac{26}{3}\) – y
or, 2y + y = \(\frac{26}{3}\) – \(\frac{5}{3}\)
or, 3y = \(\frac{26-5}{3}\) = \(\frac{21}{3}\)
or, 3y = 7
∴ y = \(\frac{7}{3}\)

Question 10.
3m = 5m – \(\frac{8}{5}\)
Solution:
3m = 5m – \(\frac{-8}{5}\)
or, 3m – 5m = \(\frac{-8}{5}\)
or, -2m =\(\frac{-8}{5}\)
or, 2m = \(\frac{8}{5}\)
∴ m = \(\frac{8}{5 \times 2}\) = \(\frac{4}{5}\)
Check:
L.H.S = 3 × \(\frac{4}{5}\) = \(\frac{12}{5}\)
R.H.S = 5 × \(\frac{4}{5}\) – \(\frac{8}{5}\)
= 4 – \(\frac{8}{5}\) = \(\frac{12}{5}\)
∴ L.H.S = R.H.S

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