NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5

These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 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 2.5

Solve the following linear equations.

Question 1.

Question 2.

Question 3.

Question 4.
x−5/3=x−3/5
Solution:
x−5/3=x−3/5
By cross-multiplication we get
5(x – 5) = 3(x – 3)
5x – 25 = 3x – 9
Transposing -25 to R.H.S and 3x to L.H.S.
5x – 3x = 25 – 9
2x = 16
Dividing both sides by 2
2x/2 = 16/2
x = 8

Question 5.

3(3t – 2) – 4(2t + 3) + 12t = 8
9t – 6 – 8t – 12 + 12t = 8
13t – 18 = 8
Transposing -18 to R.H.S., we get
13t = 8 + 18
13t = 26
Dividing both sides by 13
13t/13=26/13
t = 2

Question 6.

6m – 3(m – 1) + 2(m – 2) = 6
6m – 3m + 3 + 2m – 4 = 6
5m – 1 = 6
Transposing -1 to the R.H.S., we get
5m = 6 + 1
5m = 7
Dividing both sides by 5, we get
5m/5=7/5
m = 7/5

Simplify and solve the following linear equations.

Question 7.
3(t – 3) = 5 (2t + 1)
Solution:
3(t – 3) = 5 (2t + 1)
Opening the brackets, we get
3t – 9 = 10t + 5
Transposing -9 to R.H.S. and 10t to L.H.S.
3t – 10t = 5 + 9
-7t = 14
Dividing both sides by -7, we get
−7t/−7=14/−7
t = -2

Question 8.
15(y – 4) – 2(y – 9) + 5 (y + 6) = 0
Solution:
15(y – 4) – 2(y – 9) + 5(y + 6) = 0
Opening the brackets, we get
15y – 60 – 2y + 18 + 5y + 30 = 0
15y – 2y + 5y – 60 + 18 + 30 = 0
18y – 12 = 0
Transposing -12 to R.H.S.
18y = 12
Dividing both sides by 18, we get.
18y/18=12/18
y = 2/3

Question 9.
3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Solution:
3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Opening the brackets, we get
15z – 21 – 18z + 22 = 32z – 52 – 17
15z – 18z + 22 – 21 = 32z – 69
-3z + 1 = 32z – 69
Transposing 1 to R.H.S. and 32z to L.H.S.
-3z – 32z = – 69 – 1
-35z = -70
Dividing both sides by -35, we get
−35z/−35=−70/−35
z = 2

Question 10.
0.25(4f – 3) = 0.05(10f – 9)
Solution:
0.25(4f – 3) = 0.05(10f – 9)
Opening the brackets, we get
0.25(4f) – 0.25 × 3 = 0.05 × 10f – 0.05 × 9
f – 0.75 = 0.5f – 0.45
f – 0.5f = 0.75 – 0.45
0.5f = 0.3
Dividing both sides by 0.5, we get
0.5f/0.5=0.3/0.5
f = 3/5 = 0.6
f = 0.6