5(2-3b)-4(6+2b)=28-(b-2)

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Solution for 5(2-3b)-4(6+2b)=28-(b-2) equation: 



5(2-3b)-4(6+2b)=28-(b-2)

We move all terms to the left:

5(2-3b)-4(6+2b)-(28-(b-2))=0

We add all the numbers together, and all the variables

5(-3b+2)-4(2b+6)-(28-(b-2))=0

We multiply parentheses

-15b-8b-(28-(b-2))+10-24=0



We calculate terms in parentheses: -(28-(b-2)), so:

28-(b-2)

determiningTheFunctionDomain
-(b-2)+28

We get rid of parentheses

-b+2+28

We add all the numbers together, and all the variables

-1b+30

Back to the equation:

-(-1b+30)



We add all the numbers together, and all the variables

-23b-(-1b+30)-14=0

We get rid of parentheses

-23b+1b-30-14=0

We add all the numbers together, and all the variables

-22b-44=0

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

-22b=44

b=44/-22

b=-2

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