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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

7u+6u-(4(u-3)+2u)+9+7=0



We calculate terms in parentheses: -(4(u-3)+2u), so:

4(u-3)+2u

We add all the numbers together, and all the variables

2u+4(u-3)

We multiply parentheses

2u+4u-12

We add all the numbers together, and all the variables

6u-12

Back to the equation:

-(6u-12)



We add all the numbers together, and all the variables

13u-(6u-12)+16=0

We get rid of parentheses

13u-6u+12+16=0

We add all the numbers together, and all the variables

7u+28=0

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

7u=-28

u=-28/7

u=-4

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