## Intermediate Algebra (6th Edition)

Let \begin{array}{l}\require{cancel}Y_1= (4x)^{-2}-(3x)^{-1} ,\text{ and }\\Y_2= \dfrac{3-16x}{48x^2} ,\end{array} where $Y_1$ is the given in item $108$ and $Y_2$ is the answer to item $108.$ Using a graphing calculator, the red dotted graph is the graph of $Y_1$ and the blue solid graph is the graph of $Y_2.$ Since the graphs coincide, then $Y_1$ and $Y_2$ are equal. That is, $Y_2$ is the simplified form of $Y_1.$