Question

How do you simplify $\frac{{\displaystyle \frac{4t^2-16}{8}}}{{\displaystyle \frac{t-2}{6}}}$?

Answer 1

Simplify the fraction $\frac{{\displaystyle \frac{4t^2-16}{8}}}{{\displaystyle \frac{t-2}{6}}}$:

The given fraction is $\frac{{\displaystyle \frac{4t^2-16}{8}}}{{\displaystyle \frac{t-2}{6}}}$.

To divide any number by a fraction, multiply the number by the reciprocal of the fraction.

Then,

$\begin{array}{rcl}\frac{{\displaystyle \frac{4t^2-16}{8}}}{{\displaystyle \frac{t-2}{6}}}& =& \frac{4t^2-16}{8}\times \frac{6}{t-2}\left[\frac{{\displaystyle \frac{a}{b}}}{{\displaystyle \frac{c}{d}}}=\frac{a}{b}·\frac{d}{c}\right]\\ & =& \frac{4(t^2-4)}{8}\times \frac{6}{t-2}\\ & =& \frac{t^2-4}{2}\times \frac{6}{t-2}\\ & =& \frac{t^2-2^2}{2}\times \frac{6}{t-2}\left[a^2-b^2=\left(a+b\right)\left(a-b\right)\right]\\ & =& \frac{\cancel{(t-2)}(t+2)}{2}\times \frac{6}{\cancel{t-2}}\\ & =& \frac{6\left(t+2\right)}{2}\\ & =& 3(t+2)\\ & =& 3t+6\end{array}$

Hence, the simplified form is $\mathbf{3}\mathit{t}\mathbf{+}\mathbf{6}$.