(x-4)/7-(x+4)/5=(x+3)/7

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Solution for (x-4)/7-(x+4)/5=(x+3)/7 equation: 



(x-4)/7-(x+4)/5=(x+3)/7

We move all terms to the left:

(x-4)/7-(x+4)/5-((x+3)/7)=0

We calculate fractions


(5x-20)/()+(-x)/()+(-((x+3)*5)/()=0



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

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

We multiply all the terms by the denominator


-((x+3)*5)



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

(x+3)*5

We multiply parentheses

5x+15

Back to the equation:

-(5x+15)



We get rid of parentheses

-5x-15

Back to the equation:

+(-5x-15)



We add all the numbers together, and all the variables

(5x-20)/()+(-1x)/()+(-5x-15)=0

We get rid of parentheses

(5x-20)/()+(-1x)/()-5x-15=0

We multiply all the terms by the denominator


(5x-20)+(-1x)-5x*()-15*()=0

We add all the numbers together, and all the variables

(5x-20)+(-1x)-5x*()=0

We get rid of parentheses

5x-1x-5x*()-20=0

We add all the numbers together, and all the variables

4x-5x*()-20=0

We move all terms containing x to the left, all other terms to the right

4x-5x*()=20

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