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## Homework Statement

z[itex]^{3}[/itex]x+z-2y-1=0

xz+y-x[itex]^{2}[/itex]+5=0

Define z as a function of x, find z'.

## Homework Equations

I guess the two equations above...

## The Attempt at a Solution

Well, I just differentiated the first one with respect to x and got:

3z[itex]^{2}[/itex]xz'+z[itex]^{3}[/itex]3+1=0

z' = [itex]\frac{-1-z^{3}{3z[itex]^{2}[/itex]x}[/itex]

If I differentiate the second equation, I get something completely different...

z+xz'-2x=0

z' = [itex]\frac{2x-z}{x}[/itex]

I'm sorry if I typed some of my work above wrong, but the main reason I'm stuck on this question is:

**I don't even see why I need the two equations to solve this problem**

Thanks in advance to you guys.

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