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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))!=0We add all the numbers together, and all the variables
h∈R
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:We get rid of parentheses
-6hh/(3(h+4)*3(-1h+5))
We multiply all the terms by the denominator
-6hh
Back to the equation:
+(-6hh)
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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