(h-5)/14=(h-3)/(2h+3)

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Solution for (h-5)/14=(h-3)/(2h+3) equation: 



(h-5)/14=(h-3)/(2h+3)

We move all terms to the left:

(h-5)/14-((h-3)/(2h+3))=0



Domain of the equation: (2h+3))!=0

h∈R



We calculate fractions


((h-5)*(2h+3)))/28h+(-((h-3)*14)/28h=0

We calculate fractions


(((h-5)*(2h+3)))*28h)/(28h+(*28h)+(-((h-3)*14)*28h)/(28h+(*28h)=0



We calculate terms in parentheses: +(-((h-3)*14)*28h)/(28h+(*28h), so:

-((h-3)*14)*28h)/(28h+(*28h

We multiply all the terms by the denominator


-((h-3)*14)*28h)+((*28h)*(28h

We add all the numbers together, and all the variables

-((h-3)*14)*28h)+((+*28h)*(28h

Back to the equation:

+(-((h-3)*14)*28h)+((+*28h)*(28h)



We add all the numbers together, and all the variables

(((h-5)*(2h+3)))*28h)/(28h+(+*28h)+(-((h-3)*14)*28h)+((+*28h)*28h=0

We get rid of parentheses

(((h-5)*(2h+3)))*28h)/(28h+*28h+(-((h-3)*14)*28h)+((+*28h)*28h=0

We multiply all the terms by the denominator


(((h-5)*(2h+3)))*28h)+(*28h)*(28h+((-((h-3)*14)*28h))*(28h+(((+*28h)*28h)*(28h=0

We add all the numbers together, and all the variables

(((h-5)*(2h+3)))*28h)+(+*28h)*(28h+((-((h-3)*14)*28h))*(28h+(((+*28h)*28h)*(28h=0

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