(k+3)/2=(2k-3)/7

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Solution for (k+3)/2=(2k-3)/7 equation: 



(k+3)/2=(2k-3)/7

We move all terms to the left:

(k+3)/2-((2k-3)/7)=0

We calculate fractions


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



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

-((2k-3)*2)/(

We multiply all the terms by the denominator


-((2k-3)*2)



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

(2k-3)*2

We multiply parentheses

4k-6

Back to the equation:

-(4k-6)



We get rid of parentheses

-4k+6

Back to the equation:

+(-4k+6)



We get rid of parentheses

k/()-4k+6=0

We multiply all the terms by the denominator


k-4k*()+6*()=0

We add all the numbers together, and all the variables

k-4k*()=0

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