6(2u-3)+2u+2=4(u-1)-2u

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Solution for 6(2u-3)+2u+2=4(u-1)-2u equation: 



6(2u-3)+2u+2=4(u-1)-2u

We move all terms to the left:

6(2u-3)+2u+2-(4(u-1)-2u)=0

We add all the numbers together, and all the variables

2u+6(2u-3)-(4(u-1)-2u)+2=0

We multiply parentheses

2u+12u-(4(u-1)-2u)-18+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

2u-4

Back to the equation:

-(2u-4)



We add all the numbers together, and all the variables

14u-(2u-4)-16=0

We get rid of parentheses

14u-2u+4-16=0

We add all the numbers together, and all the variables

12u-12=0

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

12u=12

u=12/12

u=1

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