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5. 2x – 1 = 14 – x

Given 2x – 1 = 14 – x
$$\Rightarrow$$ 2x + x = 14 + 1 (Transposing x to LHS and 1 to RHS)

$$\Rightarrow$$ 3x = 15

$$\Rightarrow$$x = 15 ÷ 3 = 5

Hence x = 5 is the required solution.

Checking:

2x – 1 = 14 – x

Substituting x = 5 in the given equation,we have,

LHS= 2x – 1 = 2 × 5 – 1 = 10 – 1 = 9

RHS = 14 – x = 14 – 5 = 9

LHS = RHS

Hence x = 5 is the required solution.