Calculate: (5)/(2x-2)=(5)/(x^2-1)+(5)/(3x-3)

Expression: $\frac{ 5 }{ 2x-2 }=\frac{ 5 }{ {x}^{2}-1 }+\frac{ 5 }{ 3x-3 }$

Determine the defined range

$\begin{array} { l }\frac{ 5 }{ 2x-2 }=\frac{ 5 }{ {x}^{2}-1 }+\frac{ 5 }{ 3x-3 },& \begin{array} { l }x≠1,& x≠-1\end{array}\end{array}$

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

$x=5$

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