First, we have to find the LCD. In this case, the LCD will be [latex]xy[/latex]. We then multiply each expression by the appropriate form of 1 to obtain [latex]xy[/latex] as the denominator for each fraction.

Now that the expressions have the same denominator, we simply add the numerators to find the sum.

Analysis of the Solution

Multiplying by [latex]\dfrac{y}{y}[/latex] or [latex]\dfrac{x}{x}[/latex] does not change the value of the original expression because any number divided by itself is 1, and multiplying an expression by 1 gives the original expression.

[latex]\begin{array}{cc}\dfrac{6}{{\left(x+2\right)}^{2}}-\dfrac{2}{\left(x+2\right)\left(x - 2\right)}\hfill & \text{Factor}.\hfill \\ \dfrac{6}{{\left(x+2\right)}^{2}}\cdot \dfrac{x - 2}{x - 2}-\dfrac{2}{\left(x+2\right)\left(x - 2\right)}\cdot \dfrac{x+2}{x+2}\hfill & \text{Multiply each fraction to get the LCD as the denominator}.\hfill \\ \dfrac{6\left(x - 2\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}-\dfrac{2\left(x+2\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Multiply}.\hfill \\ \dfrac{6x - 12-\left(2x+4\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Apply distributive property}.\hfill \\ \dfrac{4x - 16}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Subtract}.\hfill \\ \dfrac{4\left(x - 4\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Simplify}.\hfill \end{array}[/latex]