Is there a way to solve for x without using distribution in #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)#

by dividing by 2,

#=> 2(x-1)=x+3#

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

#=> 2(x-1)=(x-1)+4#

by subtracting #(x-1)#

#=> x-1=4#

by adding 1

#=> x=5#

I hope that this was helpful.