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

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



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

We move all terms to the left:

2-(3-3(x+1)-(3x+2(3+2x)))=0

We add all the numbers together, and all the variables

-(3-3(x+1)-(3x+2(2x+3)))+2=0



We calculate terms in parentheses: -(3-3(x+1)-(3x+2(2x+3))), so:

3-3(x+1)-(3x+2(2x+3))

determiningTheFunctionDomain
-3(x+1)-(3x+2(2x+3))+3

We multiply parentheses

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



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

3x+2(2x+3)

We multiply parentheses

3x+4x+6

We add all the numbers together, and all the variables

7x+6

Back to the equation:

-(7x+6)



We add all the numbers together, and all the variables

-3x-(7x+6)

We get rid of parentheses

-3x-7x-6

We add all the numbers together, and all the variables

-10x-6

Back to the equation:

-(-10x-6)



We get rid of parentheses

10x+6+2=0

We add all the numbers together, and all the variables

10x+8=0

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

10x=-8

x=-8/10

x=-4/5

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