If the value of $x = - 2$ , $y = 3$ , $z = 1$ , then find the value of the following :
$\frac{x y z + x z^{3}}{2 x z - y^{2}}$

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Solution

Given that :
$x = - 2$ , $y = 3$ , $z = 1$
And we have to find the value of the expression :
$\frac{x y z + x z^{3}}{2 x z - y^{2}}$
Finding the value of the expression at the given values of variables :
So
$\frac{x y z + x z^{3}}{2 x z - y^{2}} = \frac{(- 2) (3) (1) + (- 2) {(1)}^{3}}{2 (- 2) (1) - {(3)}^{2}} = \frac{- 6 + (- 2) (1)}{2 (- 2) - 9} = \frac{- 6 - 2}{- 4 - 9} = \frac{- 8}{- 13} = \frac{- 8}{- 13} = \frac{8}{13}$
Therefore value of the given expression at $x = - 2$ , $y = 3$ , $z = 1$ is :
$\frac{x y z + x z^{3}}{2 x z - y^{2}} = \frac{8}{13}$

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