Chapter 2 Linear Equations in One Variable Ex 2.1 Solutions
Question - 1 : - Solve the equation: x – 2 = 7.
Answer - 1 : -
Given: x – 2 = 7
⇒ x – 2 + 2 = 7 + 2 (adding 2 on both sides)
⇒ x = 9 (Required solution)
Question - 2 : - Solve the equation: 6 = z + 2
Answer - 2 : -
We have 6 = z + 2
⇒ 6 – 2 = z + 2 – 2 (subtracting 2 from each side)
⇒ 4 = z
Thus, z = 4 is the required solution.
Question - 3 : - Solve the equation: y + 3 = 10.
Answer - 3 : -
⇒ y + 3 – 3 = 10 – 3 (subtracting 3 from each side)
⇒ y = 7 (Required solution)
Question - 4 : - Solve the equations: 37 + x = 177
Answer - 4 : -
Question - 5 : - Solve the equation 6x = 12.
Answer - 5 : -
We have 6x = 12
⇒ 6x ÷ 6 = 12 ÷ 6 (dividing each side by 6)
⇒ x = 2
Thus, x = 2 is the required solution.
Question - 6 : - Solve the equation 1.6 = y/15
Answer - 6 : -
1.6 = y/1.5
y/1.5 = 1.6
y = 1.6 × 1.5
y = 2.4
Question - 7 : - Solve the equation 17 + 6p = 9.
Answer - 7 : -
We have, 17 + 6p = 9
⇒ 17 – 17 + 6p = 9 – 17 (subtracting 17 from both sides)
⇒ 6p = -8
⇒ 6p ÷ 6 = -8 ÷ 6 (dividing both sides by 6)
⇒ p = −86
⇒ p = −43
Thus, p = −43 is the required solution.
Question - 8 : - Solve the equation t5 = 10.
Answer - 8 : -
Given t5 = 10
⇒ t5 × 5 = 10 × 5 (multiplying both sides by 5)
⇒ t = 50
Thus, t = 50 is the required solution.
Question - 9 : - Solve the equation x/3 + 1 = 715
Answer - 9 : - 
Question - 10 : - Solve the equation 7x – 9 = 16.
Answer - 10 : -
We have 7x – 9 = 16
⇒ 7x – 9 + 9 = 16 + 9 (adding 9 to both sides)
⇒ 7x = 25
⇒ 7x ÷ 7 = 25 ÷ 7 (dividing both sides by 7)
⇒ x = 257
Thus, x = 257 is the required solution.