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Example 6: Solve the following simple equations using balancing method.
x
(a) x – 4 = 13 (b) y + 2 = 9 (c) = 14 (d) 9x = 18
7
(e) 3x – 4 = 5

Solution: (a) x – 4 = 13
x – 4 + 4 = 13 + 4 (Adding 4 on both sides) Let's Check: Substituting x = 17 in
x = 17 LHS, we get
LHS = x – 4 = 17 – 4 = 13 = RHS
(b) y + 2 = 9 Thus, LHS = RHS
y + 2 – 2 = 9 – 2 (Subtracting 2 from both sides)

y + 0 = 7 Let's Check: Substituting y = 7 in
LHS, we get
y = 7 LHS = y + 2 = 7 + 2 = 9 = RHS
x Hence, LHS = RHS
(c) = 14
7
x Let's Check: Substituting x = 98 in
LHS, we get
7 × 7 = 14 × 7 (Multiplying both sides by 7) x 98
x = 98 LHS = 7 = 7 = 14 = RHS
(d) 9x = 18 Hence, LHS = RHS
9x = 18 (Dividing both sides by 9)

9 9 Let's Check: Substituting x = 2 in
LHS, we get
x = 2 LHS = 9x = 9 × 2 = 18 = RHS
(e) 3x – 4 = 5 Hence, LHS = RHS
3x – 4 + 4 = 5 + 4 (Adding 4 on both sides)
Let's Check: Substituting x = 3 in
3x = 9 LHS, we get

3x 9 LHS = 3x – 4 = 3 × 3 – 4 = 9 – 4 = 5
= (Dividing both sides by 3) = RHS
3 3 Hence, LHS = RHS
x = 3
Pihu to Grandfather: Now, I can find my father’s age from the equation formed from the statement
that you said. If I consider my father’s present age be x years.

Therefore, 2x + 3 = 73 will be the equation formed from the statement ‘my age is 3 years more than
twice of your father’s age, which is 73 years.’

Now, 2x + 3 – 3 = 73 – 3 (Subtracting 3 from both sides)
⇒ 2x = 70
2x 70
⇒ = (Dividing both sides by 2)
2 2
x = 35

Thus, my father’s age is 35 years.

Teacher’s
Tip Tell the students that while solving a simple equation, cancel all the numerals and free the variable from
numerals.

135 Simple Equations