2/3(39b-18)-16b=1/2(8b-10)

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Solution for 2/3(39b-18)-16b=1/2(8b-10) equation: 



2/3(39b-18)-16b=1/2(8b-10)

We move all terms to the left:

2/3(39b-18)-16b-(1/2(8b-10))=0



Domain of the equation: 3(39b-18)!=0

b∈R





Domain of the equation: 2(8b-10))!=0

b∈R



We add all the numbers together, and all the variables

-16b+2/3(39b-18)-(1/2(8b-10))=0

We calculate fractions


-16b+(4b8/(3(39b-18)*2(8b-10)))+(-3b3/(3(39b-18)*2(8b-10)))=0



We calculate terms in parentheses: +(4b8/(3(39b-18)*2(8b-10))), so:

4b8/(3(39b-18)*2(8b-10))

We multiply all the terms by the denominator


4b8

We add all the numbers together, and all the variables

4b^8

We do not support ebpression: b^8

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