3b+4(1/2b)+(2b+5)=33

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Solution for 3b+4(1/2b)+(2b+5)=33 equation: 



3b+4(1/2b)+(2b+5)=33

We move all terms to the left:

3b+4(1/2b)+(2b+5)-(33)=0



Domain of the equation: 2b)!=0

b!=0/1

b!=0

b∈R



We add all the numbers together, and all the variables

3b+4(+1/2b)+(2b+5)-33=0

We multiply parentheses

3b+4b+(2b+5)-33=0

We get rid of parentheses

3b+4b+2b+5-33=0

We add all the numbers together, and all the variables

9b-28=0

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

9b=28

b=28/9

b=3+1/9

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