(x-3)+(2x+2)-(3x-4)=(7x-4)+(3x+1)-64

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Solution for (x-3)+(2x+2)-(3x-4)=(7x-4)+(3x+1)-64 equation: 



(x-3)+(2x+2)-(3x-4)=(7x-4)+(3x+1)-64

We move all terms to the left:

(x-3)+(2x+2)-(3x-4)-((7x-4)+(3x+1)-64)=0

We get rid of parentheses

x+2x-3x-((7x-4)+(3x+1)-64)-3+2+4=0



We calculate terms in parentheses: -((7x-4)+(3x+1)-64), so:

(7x-4)+(3x+1)-64

We get rid of parentheses

7x+3x-4+1-64

We add all the numbers together, and all the variables

10x-67

Back to the equation:

-(10x-67)



We add all the numbers together, and all the variables

-(10x-67)+3=0

We get rid of parentheses

-10x+67+3=0

We add all the numbers together, and all the variables

-10x+70=0

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

-10x=-70

x=-70/-10

x=+7

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