4x+2(40-x)=128

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Solution for 4x+2(40-x)=128 equation: 



4x+2(40-x)=128

We move all terms to the left:

4x+2(40-x)-(128)=0

We add all the numbers together, and all the variables

4x+2(-1x+40)-128=0

We multiply parentheses

4x-2x+80-128=0

We add all the numbers together, and all the variables

2x-48=0

We move all terms containing x to the left, all other terms to the right

2x=48

x=48/2

x=24

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