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.
$$\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$$
Solution:
$$\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$$
Transposing $$-\frac{1}{5}$$ to R.H.S. and $$\frac{\mathrm{x}}{3}$$ to L.H.S
$$\frac{x}{2}-\frac{x}{3}=\frac{1}{4}+\frac{1}{5}$$
$$\frac{3 x-2 x}{6}=\frac{5+4}{20}$$
$$\frac{x}{6}=\frac{9}{20}$$
By cross-multiplication, we get
20(x) = 9 × 6
20x = 54
Dividing both sides by 20
$$\frac{20 \mathrm{x}}{20}=\frac{54}{20}$$
x = $$\frac{27}{10}$$

Question 2.
$$\frac{\mathrm{n}}{2}-\frac{3 n}{4}+\frac{5 \mathrm{n}}{6}=21$$
Solution:
$$\frac{\mathrm{n}}{2}-\frac{3 n}{4}+\frac{5 \mathrm{n}}{6}=21$$
L.C.M of 2, 4 and 6 = 12
Multiplying each term by 12, we get
$$12 \times \frac{\mathrm{n}}{2}-12 \times \frac{3 \mathrm{n}}{4}+12 \times \frac{5 \mathrm{n}}{6}=21 \times 12$$
6n – 9n + 10n = 252
7n = 252
Dividing both sides by 7
$$\frac{7 \mathrm{n}}{7}=\frac{252}{7}$$
n = 36

Question 3.
$$x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}$$
Solution:
$$x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}$$
Transposing 7 to R.H.S. and $$\frac{-5 x}{2}$$ to L.H.S
$$x-\frac{8 x}{3}+\frac{5 x}{2}=\frac{17}{6}-7$$
L.C.M of 3, 2 and 6 = 6
Multiplying each term by 6, we get
$$6 \times x-6 \times \frac{8 x}{3}+6 \times \frac{5 x}{2}=6 \times \frac{17}{6}-6 \times 7$$
6x – 16x + 15x = 17 – 42
5x = -25
Dividing both sides by 5
$$\frac{5 x}{5}=-\frac{25}{5}$$
x = -5

Question 4.
$$\frac{x-5}{3}=\frac{x-3}{5}$$
Solution:
$$\frac{x-5}{3}=\frac{x-3}{5}$$
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
$$\frac{2 \mathrm{x}}{2}$$ = $$\frac{16}{2}$$
x = 8

Question 5.
$$\frac{3 \mathrm{t}-2}{4}-\frac{2 \mathrm{t}+3}{3}=\frac{2}{3}-\mathrm{t}$$
Solution:
$$\frac{3 \mathrm{t}-2}{4}-\frac{2 \mathrm{t}+3}{3}=\frac{2}{3}-\mathrm{t}$$
Transposing -t to L.H.S., we get
$$\frac{3 \mathrm{t}-2}{4}-\frac{2 \mathrm{t}+3}{3}+\mathrm{t}=\frac{2}{3}$$
L.C.M of 4 and 3 is 12
Multiplying each term by 12
$$12 \frac{(3 \mathrm{t}-2)}{4}-12 \frac{(2 \mathrm{t}+3)}{3}+12 \mathrm{t}=12 \times \frac{2}{3}$$
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
$$\frac{13 t}{13}=\frac{26}{13}$$
t = 2

Question 6.
$$m-\frac{m-1}{2}=1-\frac{m-2}{3}$$
Solution:
$$m-\frac{m-1}{2}=1-\frac{m-2}{3}$$
Transposing $$\left(\frac{m-2}{3}\right)$$ to L.H.S. we get
$$\mathrm{m}-\frac{\mathrm{m}-1}{2}+\frac{\mathrm{m}-2}{3}=1$$
L.C.M of 2 and 3 is 6
Multiplying each term by 6, we get
6m – $$\frac{6(m-1)}{2}+\frac{6(m-2)}{3}$$ = 1 × 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
$$\frac{5 m}{5}=\frac{7}{5}$$
m = $$\frac{7}{5}$$

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
$$\frac{-7 t}{-7}=\frac{14}{-7}$$
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.
$$\frac{18 y}{18}=\frac{12}{18}$$
y = $$\frac{2}{3}$$

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
$$\frac{-35 z}{-35}=\frac{-70}{-35}$$
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
$$\frac{0.5 \mathrm{f}}{0.5}=\frac{0.3}{0.5}$$
f = $$\frac{3}{5}$$ = 0.6
f = 0.6

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