Question

z=$\frac{1}{x}$+y

Answer 1

Answer:

refer the attachment

Answer 2

Answer:

Solve for x

$z= \frac{ 1 }{ x } +y \\ zx=1+xy \\ zx-xy=1 \\ \left(z-y\right)x=1 \\ \frac{\left(z-y\right)x}{z-y}=\frac{1}{z-y} \\ x=\frac{1}{z-y} \\ x=\frac{1}{z-y}\text{, }x\neq 0$

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