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

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



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

We move all terms to the left:

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

We calculate fractions


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



We calculate terms in parentheses: +(-((3x+1)*5)/(), so:

-((3x+1)*5)/(

We multiply all the terms by the denominator


-((3x+1)*5)



We calculate terms in parentheses: -((3x+1)*5), so:

(3x+1)*5

We multiply parentheses

15x+5

Back to the equation:

-(15x+5)



We get rid of parentheses

-15x-5

Back to the equation:

+(-15x-5)



We get rid of parentheses

2x/()-15x-5=0

We multiply all the terms by the denominator


2x-15x*()-5*()=0

We add all the numbers together, and all the variables

2x-15x*()=0

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