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

NCERT Solutions Class 8 Mathematics 
Chapter – 2 (Linear Equations in One Variable) 

The NCERT Solutions in English Language for Class 8 Mathematics Chapter – 2 Linear Equations in One Variable Exercise 2.3 has been provided here to help the students in solving the questions from this exercise. 

Chapter 2: Linear Equations in One Variable

Exercise – 2.3 

Solve the following equations and check your results.

1. 3x = 2x + 18

Solution –
3x = 2x + 18
⇒ 3x – 2x = 18
⇒ x = 18
Putting the value of x in RHS and LHS,
we get, 3 × 18 = (2 × 18) +18
⇒ 54 = 54
⇒ LHS = RHS

2. 5t – 3 = 3t – 5

Solution –
5t – 3 = 3t – 5
⇒ 5t – 3t = -5 + 3
⇒ 2t = -2
⇒ t = -1
Putting the value of t in RHS and LHS,
we get,
5× (-1) – 3 = 3× (-1) – 5
⇒ -5 – 3 = -3 – 5
⇒ -8 = -8
⇒ LHS = RHS

3. 5x + 9 = 5 + 3x

Solution –
5x + 9 = 5 + 3x
⇒ 5x – 3x = 5 – 9
⇒ 2x = -4
⇒ x = -2
Putting the value of x in RHS and LHS,
we get,
5× (-2) + 9 = 5 + 3× (-2)
⇒ -10 + 9 = 5 + (-6)
⇒ -1 = -1
⇒ LHS = RHS

4. 4z + 3 = 6 + 2z

Solution –
4z + 3 = 6 + 2z
⇒ 4z – 2z = 6 – 3
⇒ 2z = 3
⇒ z = \frac{3}{2}
Putting the value of z in RHS and LHS,
we get,
(4 × \frac{3}{2}) + 3 = 6 + (2 × \frac{3}{2})
⇒ 6 + 3 = 6 + 3
⇒ 9 = 9
⇒ LHS = RHS

5. 2x – 1 = 14 – x

Solution –
2x – 1 = 14 – x
⇒ 2x + x = 14 + 1
⇒ 3x = 15
⇒ x = 5
Putting the value of x in RHS and LHS,
we get,
(2 × 5) – 1 = 14 – 5
⇒ 10 – 1 = 9
⇒ 9 = 9
⇒ LHS = RHS

6. 8x + 4 = 3 (x – 1) + 7

Solution –
8x + 4 = 3 (x – 1) + 7
⇒ 8x + 4 = 3x – 3 + 7
⇒ 8x + 4 = 3x + 4
⇒ 8x – 3x = 4 – 4
⇒ 5x = 0
⇒ x = 0
Putting the value of x in RHS and LHS,
we get,
(8 × 0) + 4 = 3 (0 – 1) + 7
⇒ 0 + 4 = 0 – 3 + 7
⇒ 4 = 4
⇒ LHS = RHS

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

Solution –
x = \frac{4}{5} (x + 10)
⇒ x = \frac{4x}{5} + \frac{40}{5}  

⇒ x – \frac{4x}{5} = 8

⇒ \frac{5x-4x}{5} = 8
⇒ x = 8 × 5
⇒ x = 40
Putting the value of x in RHS and LHS,
we get,
40 = \frac{4}{5} (40 + 10)
⇒ 40 = \frac{4}{5} × 50
⇒ 40 = 4 × 10
⇒ 40 = 40
⇒ LHS = RHS

8. \mathbf{\frac{2x}{3}} + 1 = \mathbf{\frac{7x}{15}} + 3

Solution –
\frac{2x}{3} + 1 = \frac{7x}{15} + 3

⇒ \frac{2x}{3} – \frac{7x}{15} = 3 – 1

⇒ \frac{10x-7x}{15} = 2

⇒ 3x = 2 × 15
⇒ 3x = 30

⇒ x = \frac{30}{3}

⇒ x = 10
Putting the value of x in RHS and LHS,
we get,
\frac{2\times 10}{3} + 1 = \frac{7\times 10}{15} +3

⇒ \frac{20}{3} +1 = \frac{7\times 2}{3} +3

⇒ \frac{20 +3}{3} = \frac{14 +9}{3}

⇒ \frac{23}{3} = \frac{23}{3}
⇒ LHS = RHS

9. 2y + \mathbf{\frac{5}{3}} = \mathbf{\frac{26}{3}} – y

Solution –
2y + \frac{5}{3} = \frac{26}{3} – y

⇒ 2y + y = \frac{26}{3}  – \frac{5}{3}  

⇒ 3y = \frac{26-5}{3}

⇒ 3y = \frac{21}{3}

⇒ 3y = 7

⇒ y = \frac{7}{3}

Putting the value of y in RHS and LHS,
we get,

⇒ (2 × \frac{7}{3}) + 5/3 = \frac{26}{3} – \frac{7}{3}

⇒ \frac{14}{3} + \frac{5}{3} = \frac{26}{3} – \frac{7}{3}

⇒ \frac{14+5}{3} = \frac{26-7}{3}

⇒ \frac{19}{3} = \frac{19}{3}  

⇒ LHS = RHS

10. 3m = 5m – \mathbf{\frac{8}{5}}

Solution –
3m = 5m – \frac{8}{5}

⇒ 5m – 3m = \frac{8}{5}

⇒ 2m = \frac{8}{5}

⇒ 2m × 5 = 8

⇒ 10m = 8

⇒ m = \frac{8}{10}

⇒ m = \frac{4}{5}

Putting the value of m in RHS and LHS,
we get,
⇒ 3 × \frac{4}{5} = 5 × \frac{4}{5} – \frac{8}{5}

⇒ \frac{12}{5} = 4 – \frac{8}{5}

⇒ \frac{12}{5} = \frac{20-8}{5}

⇒ \frac{12}{5} = \frac{12}{5}

⇒ LHS = RHS

 

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