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Linear Equations with Variables on Both Sides

In the equation 3x – 4 = 11, there are two expressions: 3x – 4 and 11. RHS of the equation is 11,
which is just a number. In equations, we often find that RHS is just a number. But this need not
be always so. The RHS of an equation may be an expression containing the variable.

For example, the equation 2x – 11 = 4 – 3x has the expression (2x – 11) on the left and (4 – 3x) on
the right of the equality (=) sign. That is the equation has variable on both sides.
A suitable way of solving an equation having variable on both
sides is to transpose all the terms containing the variable to In an algebraic expression
one side and the constant terms to the other side. Note: one of the two expressions
of the equation must contain
Example 13: Solve: 5 – 3(5x + 2) = 4(7 – 3x) + 1 the variable.
Solution: Given, 5 – 3(5x + 2) = 4(7 – 3x) + 1

⇒ 5 – 15x – 6 = 28 – 12x + 1 ⇒ –15x – 1 = –12x + 29
⇒ –15x + 12x = 29 + 1 (Transposing 12x to LHS and 1 to RHS)

⇒ –3x = 30 ⇒ x = –10

Verification: Substitute x = –10 into 5 – 3(5x + 2) = 4(7 – 3x) + 1
5 – 3{5(–10) + 2} = 4{7 – 3(–10)} + 1

⇒ 5 – 3(–48) = 4(7 + 30) + 1 ⇒ 5 – 3(–48) = 4(37) + 1
⇒ 5 + 144 = 148 + 1 ⇒ 149 = 149

Therefore, LHS = RHS (Hence verified)

Reducing an Equation to a Simpler Form

To deal with linear equations whose terms involving fractions, we can simplify them by multiplying
the whole equation by the LCM of the denominators.

We then get the equation in simpler form, which can be solved using the methods discussed before.
2t − 1 2t + 2 4
Example 14: Verify: − = − t
5 3 5
2t − 1 2t + 2 4
Solution: We have, − = − t
5 3 5
2t − 1 2t + 2 4 − 5t
⇒ − =
5 3 5
(
(
t
(
15 2t − 1) 15 2t + 2) 15 4 − 5 )
⇒ − = (LCM of 3 and 5 is 15)
5 3 5
⇒ 3(2t – 1) – 5(2t + 2) = 3(4 – 5t) ⇒ 6t – 3 – 10t – 10 = 12 – 15t
⇒ 6t – 10t + 15t = 12 + 10 + 3
[Transposing the variable term to LHS and constant terms to RHS]

25
⇒ 11t = 25 ⇒ t = [Transposing 11 to RHS]
11

43 Linear Equations in One Variable