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

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Answer 1 · 2014-11-07T14:50:26

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

$4 (x - 1) = 2 (x + 3)$ $4(x-1)=2(x+3)$

by dividing by 2,

$\Rightarrow 2 (x - 1) = x + 3$ $=> 2(x-1)=x+3$

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

$\Rightarrow 2 (x - 1) = (x - 1) + 4$ $=> 2(x-1)=(x-1)+4$

by subtracting $(x - 1)$ $(x-1)$

$\Rightarrow x - 1 = 4$ $=> x-1=4$

by adding 1

$\Rightarrow x = 5$ $=> x=5$

I hope that this was helpful.

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