4(3u+3)+3u+2=4(u-1)+2u

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



4(3u+3)+3u+2=4(u-1)+2u

We move all terms to the left:

4(3u+3)+3u+2-(4(u-1)+2u)=0

We add all the numbers together, and all the variables

3u+4(3u+3)-(4(u-1)+2u)+2=0

We multiply parentheses

3u+12u-(4(u-1)+2u)+12+2=0



We calculate terms in parentheses: -(4(u-1)+2u), so:

4(u-1)+2u

We add all the numbers together, and all the variables

2u+4(u-1)

We multiply parentheses

2u+4u-4

We add all the numbers together, and all the variables

6u-4

Back to the equation:

-(6u-4)



We add all the numbers together, and all the variables

15u-(6u-4)+14=0

We get rid of parentheses

15u-6u+4+14=0

We add all the numbers together, and all the variables

9u+18=0

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

9u=-18

u=-18/9

u=-2

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