# Calculate: (2)/(x^2-9)=(4)/(x^2-5x+6)-(2)/(x^2-4)

## Expression: $\frac{ 2 }{ {x}^{2}-9 }=\frac{ 4 }{ {x}^{2}-5x+6 }-\frac{ 2 }{ {x}^{2}-4 }$

Determine the defined range

$\begin{array} { l }\frac{ 2 }{ {x}^{2}-9 }=\frac{ 4 }{ {x}^{2}-5x+6 }-\frac{ 2 }{ {x}^{2}-4 },& \begin{array} { l }x≠-3,& x≠3,& x≠2,& x≠-2\end{array}\end{array}$

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

$\frac{ 2 }{ {x}^{2}-9 }-\frac{ 4 }{ {x}^{2}-5x+6 }+\frac{ 2 }{ {x}^{2}-4 }=0$

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

$\frac{ 2 }{ \left( x-3 \right) \times \left( x+3 \right) }-\frac{ 4 }{ {x}^{2}-5x+6 }+\frac{ 2 }{ {x}^{2}-4 }=0$

Write $-5x$ as a difference

$\frac{ 2 }{ \left( x-3 \right) \times \left( x+3 \right) }-\frac{ 4 }{ {x}^{2}-2x-3x+6 }+\frac{ 2 }{ {x}^{2}-4 }=0$

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

$\frac{ 2 }{ \left( x-3 \right) \times \left( x+3 \right) }-\frac{ 4 }{ {x}^{2}-2x-3x+6 }+\frac{ 2 }{ \left( x-2 \right) \times \left( x+2 \right) }=0$

Factor out $x$ from the expression

$\frac{ 2 }{ \left( x-3 \right) \times \left( x+3 \right) }-\frac{ 4 }{ x \times \left( x-2 \right)-3x+6 }+\frac{ 2 }{ \left( x-2 \right) \times \left( x+2 \right) }=0$

Factor out $-3$ from the expression

$\frac{ 2 }{ \left( x-3 \right) \times \left( x+3 \right) }-\frac{ 4 }{ x \times \left( x-2 \right)-3\left( x-2 \right) }+\frac{ 2 }{ \left( x-2 \right) \times \left( x+2 \right) }=0$

Factor out $x-2$ from the expression

$\frac{ 2 }{ \left( x-3 \right) \times \left( x+3 \right) }-\frac{ 4 }{ \left( x-2 \right) \times \left( x-3 \right) }+\frac{ 2 }{ \left( x-2 \right) \times \left( x+2 \right) }=0$

Write all numerators above the least common denominator $\left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right)$

$\frac{ 2\left( x-2 \right) \times \left( x+2 \right)-4\left( x+3 \right) \times \left( x+2 \right)+2\left( x+3 \right) \times \left( x-3 \right) }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

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

$\frac{ 2\left( {x}^{2}-4 \right)-4\left( x+3 \right) \times \left( x+2 \right)+2\left( x+3 \right) \times \left( x-3 \right) }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

Distribute $-4$ through the parentheses

$\frac{ 2\left( {x}^{2}-4 \right)+\left( -4x-12 \right) \times \left( x+2 \right)+2\left( x+3 \right) \times \left( x-3 \right) }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

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

$\frac{ 2\left( {x}^{2}-4 \right)+\left( -4x-12 \right) \times \left( x+2 \right)+2\left( {x}^{2}-9 \right) }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

Distribute $2$ through the parentheses

$\frac{ 2{x}^{2}-8+\left( -4x-12 \right) \times \left( x+2 \right)+2\left( {x}^{2}-9 \right) }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

Simplify the expression

$\frac{ 2{x}^{2}-8-4{x}^{2}-8x-12x-24+2\left( {x}^{2}-9 \right) }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

Distribute $2$ through the parentheses

$\frac{ 2{x}^{2}-8-4{x}^{2}-8x-12x-24+2{x}^{2}-18 }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

Collect like terms

$\frac{ 0-8-8x-12x-24-18 }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

Calculate the sum of the negative numbers

$\frac{ 0-50-8x-12x }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

Collect like terms

$\frac{ 0-50-20x }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

Removing $0$ doesn't change the value, so remove it from the expression

$\frac{ -50-20x }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

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

$-50-20x=0$

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

$-20x=50$

Divide both sides of the equation by $-20$

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

Check if the solution is in the defined range

\begin{align*}&x=-\frac{ 5 }{ 2 } \\&\begin{array} { l }x=-2 \frac{ 1 }{ 2 },& x=-2.5\end{array}\end{align*}

Random Posts
Random Articles