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.