10x+(6+4x)=36-(2x-2)

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Solution for 10x+(6+4x)=36-(2x-2) equation: 



10x+(6+4x)=36-(2x-2)

We move all terms to the left:

10x+(6+4x)-(36-(2x-2))=0

We add all the numbers together, and all the variables

10x+(4x+6)-(36-(2x-2))=0

We get rid of parentheses

10x+4x-(36-(2x-2))+6=0



We calculate terms in parentheses: -(36-(2x-2)), so:

36-(2x-2)

determiningTheFunctionDomain
-(2x-2)+36

We get rid of parentheses

-2x+2+36

We add all the numbers together, and all the variables

-2x+38

Back to the equation:

-(-2x+38)



We add all the numbers together, and all the variables

14x-(-2x+38)+6=0

We get rid of parentheses

14x+2x-38+6=0

We add all the numbers together, and all the variables

16x-32=0

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

16x=32

x=32/16

x=2

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