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

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



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

We move all terms to the left:

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

We multiply parentheses

6r-(3(r+3)+7)-8=0



We calculate terms in parentheses: -(3(r+3)+7), so:

3(r+3)+7

We multiply parentheses

3r+9+7

We add all the numbers together, and all the variables

3r+16

Back to the equation:

-(3r+16)



We get rid of parentheses

6r-3r-16-8=0

We add all the numbers together, and all the variables

3r-24=0

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

3r=24

r=24/3

r=8

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