2(b-5)+7b-8=-10-2(b+7)

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Solution for 2(b-5)+7b-8=-10-2(b+7) equation: 



2(b-5)+7b-8=-10-2(b+7)

We move all terms to the left:

2(b-5)+7b-8-(-10-2(b+7))=0

We add all the numbers together, and all the variables

7b+2(b-5)-(-10-2(b+7))-8=0

We multiply parentheses

7b+2b-(-10-2(b+7))-10-8=0



We calculate terms in parentheses: -(-10-2(b+7)), so:

-10-2(b+7)

determiningTheFunctionDomain
-2(b+7)-10

We multiply parentheses

-2b-14-10

We add all the numbers together, and all the variables

-2b-24

Back to the equation:

-(-2b-24)



We add all the numbers together, and all the variables

9b-(-2b-24)-18=0

We get rid of parentheses

9b+2b+24-18=0

We add all the numbers together, and all the variables

11b+6=0

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

11b=-6

b=-6/11

b=-6/11

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