$\begin{array} { l }\frac{ 5 }{ 2x-2 }=\frac{ 5 }{ {x}^{2}-1 }+\frac{ 5 }{ 3x-3 },& \begin{array} { l }x≠1,& x≠-1\end{array}\end{array}$
Move the expression to the left-hand side and change its sign$\frac{ 5 }{ 2x-2 }-\frac{ 5 }{ {x}^{2}-1 }-\frac{ 5 }{ 3x-3 }=0$
Factor out $2$ from the expression$\frac{ 5 }{ 2\left( x-1 \right) }-\frac{ 5 }{ {x}^{2}-1 }-\frac{ 5 }{ 3x-3 }=0$
Use ${a}^{2}-{b}^{2}=\left( a-b \right)\left( a+b \right)$ to factor the expression$\frac{ 5 }{ 2\left( x-1 \right) }-\frac{ 5 }{ \left( x-1 \right) \times \left( x+1 \right) }-\frac{ 5 }{ 3x-3 }=0$
Factor out $3$ from the expression$\frac{ 5 }{ 2\left( x-1 \right) }-\frac{ 5 }{ \left( x-1 \right) \times \left( x+1 \right) }-\frac{ 5 }{ 3\left( x-1 \right) }=0$
Write all numerators above the least common denominator $6\left( x+1 \right) \times \left( x-1 \right)$$\frac{ 15\left( x+1 \right)-30-10\left( x+1 \right) }{ 6\left( x+1 \right) \times \left( x-1 \right) }=0$
Distribute $15$ through the parentheses$\frac{ 15x+15-30-10\left( x+1 \right) }{ 6\left( x+1 \right) \times \left( x-1 \right) }=0$
Distribute $-10$ through the parentheses$\frac{ 15x+15-30-10x-10 }{ 6\left( x+1 \right) \times \left( x-1 \right) }=0$
Collect like terms$\frac{ 5x+15-30-10 }{ 6\left( x+1 \right) \times \left( x-1 \right) }=0$
Calculate the difference$\frac{ 5x-25 }{ 6\left( x+1 \right) \times \left( x-1 \right) }=0$
When the quotient of expressions equals $0$, the numerator has to be $0$$5x-25=0$
Move the constant to the right-hand side and change its sign$5x=25$
Divide both sides of the equation by $5$$\begin{array} { l }x=5,& \begin{array} { l }x≠1,& x≠-1\end{array}\end{array}$
Check if the solution is in the defined range$x=5$