2(b-5)+3(b-5)=8+7(b-4)

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Solution for 2(b-5)+3(b-5)=8+7(b-4) equation: 



2(b-5)+3(b-5)=8+7(b-4)

We move all terms to the left:

2(b-5)+3(b-5)-(8+7(b-4))=0

We multiply parentheses

2b+3b-(8+7(b-4))-10-15=0



We calculate terms in parentheses: -(8+7(b-4)), so:

8+7(b-4)

determiningTheFunctionDomain
7(b-4)+8

We multiply parentheses

7b-28+8

We add all the numbers together, and all the variables

7b-20

Back to the equation:

-(7b-20)



We add all the numbers together, and all the variables

5b-(7b-20)-25=0

We get rid of parentheses

5b-7b+20-25=0

We add all the numbers together, and all the variables

-2b-5=0

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

-2b=5

b=5/-2

b=-2+1/2

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