4(b+3)-4=8(1/2b+1)

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Solution for 4(b+3)-4=8(1/2b+1) equation: 



4(b+3)-4=8(1/2b+1)

We move all terms to the left:

4(b+3)-4-(8(1/2b+1))=0



Domain of the equation: 2b+1))!=0

b∈R



We multiply parentheses

4b-(8(1/2b+1))+12-4=0

We multiply all the terms by the denominator


4b*2b+12*2b+1))-4*2b+1))+1))-(8(1=0

Wy multiply elements

8b^2+24b+1))-4*2b+1))+1))-(8(1=0

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