(2x-3)/2=(3x-2)/5

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



(2x-3)/2=(3x-2)/5

We move all terms to the left:

(2x-3)/2-((3x-2)/5)=0

We calculate fractions


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



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

-((3x-2)*2)/(

We multiply all the terms by the denominator


-((3x-2)*2)



We calculate terms in parentheses: -((3x-2)*2), so:

(3x-2)*2

We multiply parentheses

6x-4

Back to the equation:

-(6x-4)



We get rid of parentheses

-6x+4

Back to the equation:

+(-6x+4)



We get rid of parentheses

2x/()-6x+4=0

We multiply all the terms by the denominator


2x-6x*()+4*()=0

We add all the numbers together, and all the variables

2x-6x*()=0

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