Is there a way to solve for x without using distribution in 4(x-1) = 2 (x+3)?

Nov 7, 2014

Yes, there is. It might not be easier than with distribution, I will give it a shot.

$4 \left(x - 1\right) = 2 \left(x + 3\right)$

by dividing by 2,

$\implies 2 \left(x - 1\right) = x + 3$

by rewriting $x + 3$ as $\left(x - 1\right) + 4$,

$\implies 2 \left(x - 1\right) = \left(x - 1\right) + 4$

by subtracting $\left(x - 1\right)$

$\implies x - 1 = 4$

$\implies x = 5$