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Mathematics

Solve the following equation and check your answer:

x−(2x−3x−47)=4x−273−3x-\Big(2x - \dfrac{3x - 4}{7}\Big) = \dfrac{4x - 27}{3} - 3

Linear Eqns One Variable

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Answer

We have:

=x−(2x−3x−47)=4x−273−3⇒x−2x+3x−47=4x−27−93⇒−x+3x−47=4x−363⇒−7x+3x−47=4x−363⇒−4x−47=4x−363⇒3(−4x−4)=7(4x−36)[By cross multiplication]⇒−12x−12=28x−252⇒−12x−28x=12−252[Transposing -12 to RHS and +28x to LHS]⇒−40x=−240⇒x=−240−40\phantom{=} x-\Big(2x - \dfrac{3x - 4}{7}\Big) = \dfrac{4x - 27}{3} - 3 \\[1em] \Rightarrow x - 2x + \dfrac{3x - 4}{7} = \dfrac{4x - 27 - 9}{3} \\[1em] \Rightarrow -x + \dfrac{3x - 4}{7} = \dfrac{4x - 36}{3} \\[1em] \Rightarrow \dfrac{-7x + 3x - 4}{7} = \dfrac{4x - 36}{3} \\[1em] \Rightarrow \dfrac{-4x - 4}{7} = \dfrac{4x - 36}{3} \\[1em] \Rightarrow 3(-4x - 4) = 7(4x - 36) \quad \text{[By cross multiplication]} \\[1em] \Rightarrow -12x - 12 = 28x - 252 \\[1em] \Rightarrow -12x - 28x = 12 - 252 \quad \text{[Transposing -12 to RHS and +28x to LHS]} \\[1em] \Rightarrow -40x = -240 \\[1em] \Rightarrow x = \dfrac{-240}{-40}

∴ x = 6

Check:

LHS=x−(2x−3x−47)LHS=6−(12−2)LHS=−4RHS=4x−273−3RHS=24−273−3RHS=−1−3RHS=−4\text{LHS} = x - \left(2x - \dfrac{3x - 4}{7}\right) \\[1em] \phantom{\text{LHS}} = 6 - \left(12 - 2\right) \\[1em] \phantom{\text{LHS}} = -4 \\[2em] \text{RHS} = \dfrac{4x - 27}{3} - 3 \\[1em] \phantom{\text{RHS}} = \dfrac{24 - 27}{3} - 3 \\[1em] \phantom{\text{RHS}} = -1 - 3 \\[1em] \phantom{\text{RHS}} = -4

Hence, LHS = RHS.

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