## MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.1

Solve the following equations.

Question 1.

x – 2 = 7.

Solution:

We have, x – 2 = 7

Transposing -2 to R.H.S., we get

x = 7 + 2 ⇒ x = 9, which is the required solution.

Question 2.

y + 3 = 10.

Solution:

We have, y + 3 = 10

Transposing 3 to R.H.S., we get y = 10 – 3

⇒ y = 7, which is the required solution.

Question 3.

6 = z + 2.

Solution:

We have, 6 = z + 2

Transposing 2 to L.H.S., we get

6 – 2 = z

⇒ z = 4, which is the required solution.

Question 4.

\(\frac{3}{7}\) + x = \(\frac{17}{7}\).

Solution:

⇒ x = 2, which is the required solution.

Question 5.

6x = 12.

Solution:

We have, 6x = 12

Dividing both sides by 6, we get

\(\frac{6 x}{6}=\frac{12}{6}\)

⇒ x = 2, which is the required solution.

Question 6.

\(\frac{t}{5}=10\)

Solution:

We have, \(\frac{t}{5}\) = 10

Multiplying both sides by 5, we get

⇒ t = 50, which is the required solution.

Question 7.

\(\frac{2 x}{3}\) = 18

Solution:

We have,

\(\frac{2 x}{3}\) = 18

Dividing both sides by 2, we get

Now, multiplying both sides by 3, we get

⇒ x = 27, which is the required solution.

Question 8.

\(1.6=\frac{y}{1.5}\)

Solution:

We have, \(1.6=\frac{y}{1.5}\)

Multiplying both sides by 1.5, we get

⇒ y = 2.4, which is the required solution.

Question 9.

7x – 9 = 16.

Solution:

We have, 7x – 9 = 16

Transposing -9 to R.H.S., we get

7x = 16 + 9

⇒ 7x = 25

Now, dividing both sides by 7, we get

⇒ x = \(\frac{25}{7}\), which is the required solution.

Question 10.

14y – 8 = 13.

Solution:

We have, 14y -8 = 13

Transposing -8 to R.H.S., we get

14y = 13 + 8 ⇒ 14y = 21

Now, dividing both sides by 14, we get

Question 11.

17 + 6p = 9.

Solution:

We have, 17 + 6p = 9

Transposing 17 to R.H.S., we get 6p = 9 – 17

⇒ 6p = – 8

Now, dividing both sides by 6, we get

Question 12.

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

Solution:

Now, multiplying both sides by 3, we get

⇒ x = \(-\frac{8}{5}\), which is the required solution.