1/3(h+4)-6=2/3(5-h)

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Solution for 1/3(h+4)-6=2/3(5-h) equation: 



1/3(h+4)-6=2/3(5-h)

We move all terms to the left:

1/3(h+4)-6-(2/3(5-h))=0



Domain of the equation: 3(h+4)!=0

h∈R





Domain of the equation: 3(5-h))!=0

h∈R



We add all the numbers together, and all the variables

1/3(h+4)-(2/3(-1h+5))-6=0

We calculate fractions


(3h(-)/(3(h+4)*3(-1h+5)))+(-6hh/(3(h+4)*3(-1h+5)))-6=0



We calculate terms in parentheses: +(3h(-)/(3(h+4)*3(-1h+5))), so:

3h(-)/(3(h+4)*3(-1h+5))

We add all the numbers together, and all the variables

3h0/(3(h+4)*3(-1h+5))

We multiply all the terms by the denominator


3h0

We add all the numbers together, and all the variables

3h

Back to the equation:

+(3h)





We calculate terms in parentheses: +(-6hh/(3(h+4)*3(-1h+5))), so:

-6hh/(3(h+4)*3(-1h+5))

We multiply all the terms by the denominator


-6hh

Back to the equation:

+(-6hh)



We get rid of parentheses

3h-6hh-6=0

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

3h-6hh=6

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