1/5(h-4)-7=4/10(2h+2)

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Solution for 1/5(h-4)-7=4/10(2h+2) equation: 



1/5(h-4)-7=4/10(2h+2)

We move all terms to the left:

1/5(h-4)-7-(4/10(2h+2))=0



Domain of the equation: 5(h-4)!=0

h∈R





Domain of the equation: 10(2h+2))!=0

h∈R



We calculate fractions


(10h2/(5(h-4)*10(2h+2)))+(-20hh/(5(h-4)*10(2h+2)))-7=0



We calculate terms in parentheses: +(10h2/(5(h-4)*10(2h+2))), so:

10h2/(5(h-4)*10(2h+2))

We multiply all the terms by the denominator


10h2

We add all the numbers together, and all the variables

10h^2

Back to the equation:

+(10h^2)





We calculate terms in parentheses: +(-20hh/(5(h-4)*10(2h+2))), so:

-20hh/(5(h-4)*10(2h+2))

We multiply all the terms by the denominator


-20hh

Back to the equation:

+(-20hh)



We get rid of parentheses

10h^2-20hh-7=0

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

10h^2-20hh=7

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