Solve : $1 + x = x y = x^{2} y$

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Solution

$\Rightarrow 1 + x = x y$ and $x y = x^{2} y$

$x y (x - 1) = 0$

either $x y = 0$ or $x = 1$

if $x = 1$ if $x y = 0$

$1 + x = x y$ either $x = 0$ or $y = 0$

$1 + 1 = y$ $x = 0$ $y = 0$

$y = 2$ $1 + x = x y$ $1 + x = x (0)$

$1 + 0 = 0$ $x = - 1$
Not possible

So, $x = 1 \Rightarrow y = 2$

and $x = - 1 \Rightarrow y = 0$

two solution.

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