2/5(x-3)=1/3(2x+10)

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Solution for 2/5(x-3)=1/3(2x+10) equation: 



2/5(x-3)=1/3(2x+10)

We move all terms to the left:

2/5(x-3)-(1/3(2x+10))=0



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

x∈R





Domain of the equation: 3(2x+10))!=0

x∈R



We calculate fractions


(6x2/(5(x-3)*3(2x+10)))+(-5xx/(5(x-3)*3(2x+10)))=0



We calculate terms in parentheses: +(6x2/(5(x-3)*3(2x+10))), so:

6x2/(5(x-3)*3(2x+10))

We multiply all the terms by the denominator


6x2

We add all the numbers together, and all the variables

6x^2

Back to the equation:

+(6x^2)





We calculate terms in parentheses: +(-5xx/(5(x-3)*3(2x+10))), so:

-5xx/(5(x-3)*3(2x+10))

We multiply all the terms by the denominator


-5xx

Back to the equation:

+(-5xx)



We get rid of parentheses

6x^2-5xx=0

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