(b-10)+11(b-3)=3b+5

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Solution for (b-10)+11(b-3)=3b+5 equation: 



(b-10)+11(b-3)=3b+5

We move all terms to the left:

(b-10)+11(b-3)-(3b+5)=0

We multiply parentheses

(b-10)+11b-(3b+5)-33=0

We get rid of parentheses

b+11b-3b-10-5-33=0

We add all the numbers together, and all the variables

9b-48=0

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

9b=48

b=48/9

b=5+1/3

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