Hi, please help:

\({{x+2}\over x^2-2x}-{{2}\over x-2}=-1\)

i need to have X solved...i have no idea...thank you kindly..

Guest Jul 23, 2018

#1**+1 **

**solve for X \dfrac{x+2}{x^2-2x} - \dfrac{2}{x-2}=-1**

\(\dfrac{x+2}{x^2-2x} - \dfrac{2}{x-2}=-1\)

\(\begin{array}{|rcll|} \hline \dfrac{x+2}{x^2-2x} - \dfrac{2}{x-2} &=& -1 \quad | \quad x-2 \ne 0 \Rightarrow x \ne 2 \\\\ && \quad | \quad x^2-2x \ne 0 \Rightarrow x \ne 2 \\\\ \dfrac{(x+2)(x-2)-2(x^2-2x)}{(x^2-2x)(x-2)} &=& -1 \\\\ (x+2)(x-2)-2(x^2-2x) &=& -(x^2-2x)(x-2) \\ x^2-4-2x^2+4x &=& -(x^2-2x)(x-2) \\ -x^2+4x-4 &=& -(x^2-2x)(x-2) \quad | \quad \cdot (-1) \\ x^2-4x+4 &=& (x^2-2x)(x-2) \\ (x-2)(x-2) &=& (x^2-2x)(x-2) \quad | \quad : (x-2) \\ x-2 &=& x^2-2x \\ x^2-2x &=& x-2 \\ x^2-3x+2 &=& 0 \\ (x-1)(x-2) &=& 0 \\\\ x = 1 && x \ne 2 \\ \hline \end{array}\)

heureka Jul 23, 2018

#3**0 **

Heureka,

thank you very much, although I fail to understand something:

why can't x be equal to 2?

Guest Jul 23, 2018

#3**0 **

Heureka,

thank you very much, although I fail to understand something:

why can't x be equal to 2?

Guest Jul 23, 2018