3(3x+1)+2(4-x)=^x-4(3x-2)

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



3(3x+1)+2(4-x)=^x-4(3x-2)

We move all terms to the left:

3(3x+1)+2(4-x)-(^x-4(3x-2))=0

We add all the numbers together, and all the variables

3(3x+1)+2(-1x+4)-(^x-4(3x-2))=0

We multiply parentheses

9x-2x-(^x-4(3x-2))+3+8=0



We calculate terms in parentheses: -(^x-4(3x-2)), so:

^x-4(3x-2)

We multiply parentheses

^x-12x+8

We add all the numbers together, and all the variables

-12x+^x+8

Back to the equation:

-(-12x+^x+8)



We add all the numbers together, and all the variables

7x-(-12x+^x+8)+11=0

We get rid of parentheses

7x+12x-^x-8+11=0

We add all the numbers together, and all the variables

19x-^x+3=0

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

19x-^x=-3

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