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Solving Equations Having Variables on One side and Numbers

on the Other Side

Let us solve some more examples of linear equations having linear expression on one side and
rational numbers on the other side.

x 5 3
Example 4: Solve the following linear equation and verify the solution: + = − .
3 2 2
x 5 3 x 3 5 5
Solution: We have, + = − ⇒ = −− (Transposing to RHS)
3 2 2 3 2 2 2
x 8 x
⇒ = − ⇒ = –4
3 2 3
⇒ x = – 4 × 3 = –12 (Multiplying both sides by 3)
Thus, x = –12 is the required solution for the given equation.

Verification: Substitute x = –12 in the LHS of the given equation, we get
12 5 5 −+ 5 3
8
4
LHS = − + =− + = =− = RHS
3 2 2 2 2
Clearly, LHS = RHS. Hence, the solution is verified.
Example 5: Solve the following equations and verify your results.
15 y
( a) 3.5y + 1 = 8 (b) 4 – 7x = 9 (c) 1.8 = 36 ( d) 16 + 4p = 24
.
Solution: (a) 3.5y + 1 = 8
3.5y = 8 – 1 (Transposing 1 to RHS)
⇒ 3.5y = 7

7
⇒ y = (Transposing 3.5 to RHS)
.
35
7
= × 10 = 2.
35
So, y = 2 is the required solution.

Verification: LHS = 3.5y + 1 = 3.5 × 2 + 1 = 7 + 1 = 8
RHS = 8 ⇒ LHS = RHS, hence verified.

15
( b) – 7x = 9
4
⇒ –7x = 9 – 15 (Transposing 15 to RHS)
4 4
−
36 15 21 21 1
⇒ –7x = = ⇒ x = × (Transposing –7 to RHS)
4 4 4 − 7
−3
⇒ x = 4
−3
So, x = is the required solution.
4
Verification: LHS = 15 − 7 = 15 − 7 ×  − 3 = 15 + 21 = 15 + 21 = 36 = 9 = RHS
x


4 4  4  4 4 4 4
⇒ LHS = RHS, hence verified.
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