1/2(3-4b)-5(2b+3)=b-17

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Solution for 1/2(3-4b)-5(2b+3)=b-17 equation: 



1/2(3-4b)-5(2b+3)=b-17

We move all terms to the left:

1/2(3-4b)-5(2b+3)-(b-17)=0



Domain of the equation: 2(3-4b)!=0

b∈R



We add all the numbers together, and all the variables

1/2(-4b+3)-5(2b+3)-(b-17)=0

We multiply parentheses

1/2(-4b+3)-10b-(b-17)-15=0

We get rid of parentheses

1/2(-4b+3)-10b-b+17-15=0

We multiply all the terms by the denominator


-10b*2(-4b+3)-b*2(-4b+3)+17*2(-4b+3)-15*2(-4b+3)+1=0

Wy multiply elements

-20b^2(--2b^2(-+34b(--30b(-+1=0

We use the square of the difference formula

-20b^2(+2b^2(-34b(+30b(-1=0

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