(x+4)/(x-3)-(2x+5)/2x=0

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Solution for (x+4)/(x-3)-(2x+5)/2x=0 equation: 



(x+4)/(x-3)-(2x+5)/2x=0



Domain of the equation: (x-3)!=0

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

x!=3

x∈R





Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R



We calculate fractions


(2x^2+8x)/(2x^2-6x)+(-(2x+5)*(x-3))/(2x^2-6x)=0



We calculate terms in parentheses: +(-(2x+5)*(x-3))/(2x^2-6x), so:

-(2x+5)*(x-3))/(2x^2-6x

We add all the numbers together, and all the variables

-6x-(2x+5)*(x-3))/(2x^2

We multiply all the terms by the denominator


-6x*(2x^2-(2x+5)*(x-3))

Back to the equation:

+(-6x*(2x^2-(2x+5)*(x-3)))



We multiply all the terms by the denominator


(2x^2+8x)+((-6x*(2x^2-(2x+5)*(x-3))))*(2x^2-6x)=0



We calculate terms in parentheses: +((-6x*(2x^2-(2x+5)*(x-3))))*(2x^2-6x), so:

(-6x*(2x^2-(2x+5)*(x-3))))*(2x^2-6x

We add all the numbers together, and all the variables

-6x+(-6x*(2x^2-(2x+5)*(x-3))))*(2x^2

Back to the equation:

+(-6x+(-6x*(2x^2-(2x+5)*(x-3))))*(2x^2)



We get rid of parentheses

2x^2+8x+(-6x+(-6x*(2x^2-(2x+5)*(x-3))))*2x^2=0

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