Solve for: (3)/(x-1)+(2x)/(x+1)=2

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

Determine the defined range

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

Write all numerators above the least common denominator $\left( x-1 \right) \times \left( x+1 \right)$

$\frac{ 3\left( x+1 \right)+2x \times \left( x-1 \right) }{ \left( x-1 \right) \times \left( x+1 \right) }=2$

Distribute $3$ through the parentheses

$\frac{ 3x+3+2x \times \left( x-1 \right) }{ \left( x-1 \right) \times \left( x+1 \right) }=2$

Distribute $2x$ through the parentheses

$\frac{ 3x+3+2{x}^{2}-2x }{ \left( x-1 \right) \times \left( x+1 \right) }=2$

Collect like terms

$\frac{ x+3+2{x}^{2} }{ \left( x-1 \right) \times \left( x+1 \right) }=2$

Multiply both sides of the equation by $\left( x-1 \right) \times \left( x+1 \right)$

$x+3+2{x}^{2}=2\left( x-1 \right) \times \left( x+1 \right)$

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

$x+3+2{x}^{2}=2\left( {x}^{2}-1 \right)$

Distribute $2$ through the parentheses

$x+3+2{x}^{2}=2{x}^{2}-2$

Cancel equal terms on both sides of the equation

$x+3=-2$

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

$x=-2-3$

Calculate the difference

$\begin{array} { l }x=-5,& \begin{array} { l }x≠1,& x≠-1\end{array}\end{array}$

Check if the solution is in the defined range

$x=-5$