10(1+3b)+2=132

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Solution for 10(1+3b)+2=132 equation: 



10(1+3b)+2=132

We move all terms to the left:

10(1+3b)+2-(132)=0

We add all the numbers together, and all the variables

10(3b+1)+2-132=0

We add all the numbers together, and all the variables

10(3b+1)-130=0

We multiply parentheses

30b+10-130=0

We add all the numbers together, and all the variables

30b-120=0

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

30b=120

b=120/30

b=4

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