1/5(50-10b)+8=1/3(9b-30)-7

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Solution for 1/5(50-10b)+8=1/3(9b-30)-7 equation: 



1/5(50-10b)+8=1/3(9b-30)-7

We move all terms to the left:

1/5(50-10b)+8-(1/3(9b-30)-7)=0



Domain of the equation: 5(50-10b)!=0

b∈R





Domain of the equation: 3(9b-30)-7)!=0

b∈R



We add all the numbers together, and all the variables

1/5(-10b+50)-(1/3(9b-30)-7)+8=0

We calculate fractions


(3b9/(5(-10b+50)*3(9b-30))+(-5b0/(5(-10b+50)*3(9b-30))+8=0

We can not solve this equation

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