Solve for: (x+2)/(x-2)-(8)/(x^2-2x)=(2)/(x)

Expression: $\frac{ x+2 }{ x-2 }-\frac{ 8 }{ {x}^{2}-2x }=\frac{ 2 }{ x }$

Determine the defined range

$\begin{array} { l }\frac{ x+2 }{ x-2 }-\frac{ 8 }{ {x}^{2}-2x }=\frac{ 2 }{ x },& \begin{array} { l }x≠2,& x≠0\end{array}\end{array}$

Move the expression to the left-hand side and change its sign

$\frac{ x+2 }{ x-2 }-\frac{ 8 }{ {x}^{2}-2x }-\frac{ 2 }{ x }=0$

Factor out $x$ from the expression

$\frac{ x+2 }{ x-2 }-\frac{ 8 }{ x \times \left( x-2 \right) }-\frac{ 2 }{ x }=0$

Write all numerators above the least common denominator $x \times \left( x-2 \right)$

$\frac{ x \times \left( x+2 \right)-8-2\left( x-2 \right) }{ x \times \left( x-2 \right) }=0$

Distribute $x$ through the parentheses

$\frac{ {x}^{2}+2x-8-2\left( x-2 \right) }{ x \times \left( x-2 \right) }=0$

Distribute $-2$ through the parentheses

$\frac{ {x}^{2}+2x-8-2x+4 }{ x \times \left( x-2 \right) }=0$

Since two opposites add up to $0$, remove them from the expression

$\frac{ {x}^{2}-8+4 }{ x \times \left( x-2 \right) }=0$

Calculate the sum

$\frac{ {x}^{2}-4 }{ x \times \left( x-2 \right) }=0$

Use ${a}^{2}-{b}^{2}=\left( a-b \right)\left( a+b \right)$ to factor the expression

$\frac{ \left( x-2 \right) \times \left( x+2 \right) }{ x \times \left( x-2 \right) }=0$

Cancel out the common factor $x-2$

$\frac{ x+2 }{ x }=0$

When the quotient of expressions equals $0$, the numerator has to be $0$

$x+2=0$

Move the constant to the right-hand side and change its sign

$\begin{array} { l }x=-2,& \begin{array} { l }x≠2,& x≠0\end{array}\end{array}$

Check if the solution is in the defined range

$x=-2$