Question

If $p=4$$q=3$ and $r=-2$find the value of :

$\frac{3pq+2q{r}^2}{p+q-r}$

Answer 1

We have $\frac{3pq+2q{r}^2}{p+q-r}$

Substituting $p=4$$q=3$ and $r=-2$ in the given expression , we get

$$\begin{gathered} \frac{3pq+2q{r}^2}{p+q-r}=\frac{3\times 4\times 3+2\times 3\times {\left(-2\right)}^2}{4+3-\left(-2\right)} \\ =\frac{36+24}{7+2}\mathbf{} \\ =\frac{60}{9} \\ =\mathbf{}6\frac{6}{9}=6\frac{2}{3} \end{gathered}$$

Hence,the value of given expression will be $6\frac{2}{3}$.