b-5+4b=(11b+4)+11(b-11)

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Solution for b-5+4b=(11b+4)+11(b-11) equation: 



b-5+4b=(11b+4)+11(b-11)

We move all terms to the left:

b-5+4b-((11b+4)+11(b-11))=0

We add all the numbers together, and all the variables

5b-((11b+4)+11(b-11))-5=0



We calculate terms in parentheses: -((11b+4)+11(b-11)), so:

(11b+4)+11(b-11)

We multiply parentheses

(11b+4)+11b-121

We get rid of parentheses

11b+11b+4-121

We add all the numbers together, and all the variables

22b-117

Back to the equation:

-(22b-117)



We get rid of parentheses

5b-22b+117-5=0

We add all the numbers together, and all the variables

-17b+112=0

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

-17b=-112

b=-112/-17

b=6+10/17

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