5(5b-4)+7b=3(7b+2)+7

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



5(5b-4)+7b=3(7b+2)+7

We move all terms to the left:

5(5b-4)+7b-(3(7b+2)+7)=0

We add all the numbers together, and all the variables

7b+5(5b-4)-(3(7b+2)+7)=0

We multiply parentheses

7b+25b-(3(7b+2)+7)-20=0



We calculate terms in parentheses: -(3(7b+2)+7), so:

3(7b+2)+7

We multiply parentheses

21b+6+7

We add all the numbers together, and all the variables

21b+13

Back to the equation:

-(21b+13)



We add all the numbers together, and all the variables

32b-(21b+13)-20=0

We get rid of parentheses

32b-21b-13-20=0

We add all the numbers together, and all the variables

11b-33=0

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

11b=33

b=33/11

b=3

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