6(b-3)+2(b+1)=3(b-5)

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Solution for 6(b-3)+2(b+1)=3(b-5) equation: 



6(b-3)+2(b+1)=3(b-5)

We move all terms to the left:

6(b-3)+2(b+1)-(3(b-5))=0

We multiply parentheses

6b+2b-(3(b-5))-18+2=0



We calculate terms in parentheses: -(3(b-5)), so:

3(b-5)

We multiply parentheses

3b-15

Back to the equation:

-(3b-15)



We add all the numbers together, and all the variables

8b-(3b-15)-16=0

We get rid of parentheses

8b-3b+15-16=0

We add all the numbers together, and all the variables

5b-1=0

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

5b=1

b=1/5

b=1/5

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