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

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



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

We move all terms to the left:

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

We multiply parentheses

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



We calculate terms in parentheses: -(5(u-4)+u), so:

5(u-4)+u

We add all the numbers together, and all the variables

u+5(u-4)

We multiply parentheses

u+5u-20

We add all the numbers together, and all the variables

6u-20

Back to the equation:

-(6u-20)



We add all the numbers together, and all the variables

3u-(6u-20)+10=0

We get rid of parentheses

3u-6u+20+10=0

We add all the numbers together, and all the variables

-3u+30=0

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

-3u=-30

u=-30/-3

u=+10

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