15=((3x+2)+(2x-2))/2

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Solution for 15=((3x+2)+(2x-2))/2 equation: 



15=((3x+2)+(2x-2))/2

We move all terms to the left:

15-(((3x+2)+(2x-2))/2)=0

We multiply all the terms by the denominator


-(((3x+2)+(2x-2))+15*2)=0



We calculate terms in parentheses: -(((3x+2)+(2x-2))+15*2), so:

((3x+2)+(2x-2))+15*2

We add all the numbers together, and all the variables

((3x+2)+(2x-2))+30



We calculate terms in parentheses: +((3x+2)+(2x-2)), so:

(3x+2)+(2x-2)

We get rid of parentheses

3x+2x+2-2

We add all the numbers together, and all the variables

5x

Back to the equation:

+(5x)



Back to the equation:

-(5x+30)



We get rid of parentheses

-5x-30=0

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

-5x=30

x=30/-5

x=-6

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