1/5(x+5)=1/7(2x+3)

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Solution for 1/5(x+5)=1/7(2x+3) equation: 



1/5(x+5)=1/7(2x+3)

We move all terms to the left:

1/5(x+5)-(1/7(2x+3))=0



Domain of the equation: 5(x+5)!=0

x∈R





Domain of the equation: 7(2x+3))!=0

x∈R



We calculate fractions


(7x2/(5(x+5)*7(2x+3)))+(-5xx/(5(x+5)*7(2x+3)))=0



We calculate terms in parentheses: +(7x2/(5(x+5)*7(2x+3))), so:

7x2/(5(x+5)*7(2x+3))

We multiply all the terms by the denominator


7x2

We add all the numbers together, and all the variables

7x^2

Back to the equation:

+(7x^2)





We calculate terms in parentheses: +(-5xx/(5(x+5)*7(2x+3))), so:

-5xx/(5(x+5)*7(2x+3))

We multiply all the terms by the denominator


-5xx

Back to the equation:

+(-5xx)



We get rid of parentheses

7x^2-5xx=0

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