1/(x-3)^2-4/x-3-5=0

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Solution for 1/(x-3)^2-4/x-3-5=0 equation: 



1/(x-3)^2-4/x-3-5=0



Domain of the equation: (x-3)^2!=0

x∈R





Domain of the equation: x!=0

x∈R



We add all the numbers together, and all the variables

1/(x-3)^2-4/x-8=0

We calculate fractions


x/((x-3)^2*x)+(-4*(x-3)^2)/((x-3)^2*x)-8=0



We calculate terms in parentheses: +(-4*(x-3)^2)/((x-3)^2*x), so:

-4*(x-3)^2)/((x-3)^2*x

We multiply all the terms by the denominator


-4*(x-3)^2)

We multiply parentheses

-4x-

We add all the numbers together, and all the variables

-4x

Back to the equation:

+(-4x)



We get rid of parentheses

x/((x-3)^2*x)-4x-8=0

We multiply all the terms by the denominator


x-4x*((x-3)^2*x)-8*((x-3)^2*x)=0

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