5(2u-3)+(4-u)3-2(u+7)=(2u-4)+(6-u)

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Solution for 5(2u-3)+(4-u)3-2(u+7)=(2u-4)+(6-u) equation: 



5(2u-3)+(4-u)3-2(u+7)=(2u-4)+(6-u)

We move all terms to the left:

5(2u-3)+(4-u)3-2(u+7)-((2u-4)+(6-u))=0

We add all the numbers together, and all the variables

5(2u-3)+(-1u+4)3-2(u+7)-((2u-4)+(-1u+6))=0

We multiply parentheses

10u-3u-2u-((2u-4)+(-1u+6))-15+12-14=0



We calculate terms in parentheses: -((2u-4)+(-1u+6)), so:

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

We get rid of parentheses

2u-1u-4+6

We add all the numbers together, and all the variables

u+2

Back to the equation:

-(u+2)



We add all the numbers together, and all the variables

5u-(u+2)-17=0

We get rid of parentheses

5u-u-2-17=0

We add all the numbers together, and all the variables

4u-19=0

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

4u=19

u=19/4

u=4+3/4

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