-3(3u-4)+u=6+2(5u-2)

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Solution for -3(3u-4)+u=6+2(5u-2) equation: 



-3(3u-4)+u=6+2(5u-2)

We move all terms to the left:

-3(3u-4)+u-(6+2(5u-2))=0

We add all the numbers together, and all the variables

u-3(3u-4)-(6+2(5u-2))=0

We multiply parentheses

u-9u-(6+2(5u-2))+12=0



We calculate terms in parentheses: -(6+2(5u-2)), so:

6+2(5u-2)

determiningTheFunctionDomain
2(5u-2)+6

We multiply parentheses

10u-4+6

We add all the numbers together, and all the variables

10u+2

Back to the equation:

-(10u+2)



We add all the numbers together, and all the variables

-8u-(10u+2)+12=0

We get rid of parentheses

-8u-10u-2+12=0

We add all the numbers together, and all the variables

-18u+10=0

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

-18u=-10

u=-10/-18

u=5/9

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