Question

If $x=2$, $y=3$and $z=-1$ find the value of :

$\frac{2x-3y+4z}{3x-z}$

Answer 1

We have $\frac{2x-3y+4z}{3x-z}$

Substituting $x=2$$y=3$ and $z=-1$ in the given expression , we get

$$\begin{gathered} \frac{2x-3y+4z}{3x-z}=\frac{2\times 2-3\times 3+4\left(-1\right)}{3\times 2-\left(-1\right)} \\ =\frac{4-9-4}{6+1} \\ =\frac{4-13}{7} \\ =-\frac{9}{7} \\ =-1\frac{2}{7} \end{gathered}$$

Hence,the value of given expression will be $-1\frac{2}{7}$.