-3(3u-2)+u=2+5(2u-2)

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Solution for -3(3u-2)+u=2+5(2u-2) equation: 



-3(3u-2)+u=2+5(2u-2)

We move all terms to the left:

-3(3u-2)+u-(2+5(2u-2))=0

We add all the numbers together, and all the variables

u-3(3u-2)-(2+5(2u-2))=0

We multiply parentheses

u-9u-(2+5(2u-2))+6=0



We calculate terms in parentheses: -(2+5(2u-2)), so:

2+5(2u-2)

determiningTheFunctionDomain
5(2u-2)+2

We multiply parentheses

10u-10+2

We add all the numbers together, and all the variables

10u-8

Back to the equation:

-(10u-8)



We add all the numbers together, and all the variables

-8u-(10u-8)+6=0

We get rid of parentheses

-8u-10u+8+6=0

We add all the numbers together, and all the variables

-18u+14=0

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

-18u=-14

u=-14/-18

u=7/9

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