1/5(b+10)-7=1/5(b-9)

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Solution for 1/5(b+10)-7=1/5(b-9) equation: 



1/5(b+10)-7=1/5(b-9)

We move all terms to the left:

1/5(b+10)-7-(1/5(b-9))=0



Domain of the equation: 5(b+10)!=0

b∈R





Domain of the equation: 5(b-9))!=0

b∈R



We calculate fractions


(5bb/(5(b+10)*5(b-9)))+(-5bb/(5(b+10)*5(b-9)))-7=0



We calculate terms in parentheses: +(5bb/(5(b+10)*5(b-9))), so:

5bb/(5(b+10)*5(b-9))

We multiply all the terms by the denominator


5bb

Back to the equation:

+(5bb)





We calculate terms in parentheses: +(-5bb/(5(b+10)*5(b-9))), so:

-5bb/(5(b+10)*5(b-9))

We multiply all the terms by the denominator


-5bb

Back to the equation:

+(-5bb)



We get rid of parentheses

5bb-5bb-7=0

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

5bb-5bb=7

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