(c-23)/2=(2c+5)/7

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Solution for (c-23)/2=(2c+5)/7 equation: 



(c-23)/2=(2c+5)/7

We move all terms to the left:

(c-23)/2-((2c+5)/7)=0

We calculate fractions


c/()+(-((2c+5)*2)/()=0



We calculate terms in parentheses: +(-((2c+5)*2)/(), so:

-((2c+5)*2)/(

We multiply all the terms by the denominator


-((2c+5)*2)



We calculate terms in parentheses: -((2c+5)*2), so:

(2c+5)*2

We multiply parentheses

4c+10

Back to the equation:

-(4c+10)



We get rid of parentheses

-4c-10

Back to the equation:

+(-4c-10)



We get rid of parentheses

c/()-4c-10=0

We multiply all the terms by the denominator


c-4c*()-10*()=0

We add all the numbers together, and all the variables

c-4c*()=0

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