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HALP.. See if you can solve this fast.

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Solve:

This system of equations has one solution. What is z in this problem?

$$\frac{xy}{x + y} = 1, \quad \frac{xz}{x + z} = 2, \quad \frac{yz}{y + z} = 3$$

Aug 16, 2022

#1
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Hi Guest!

$$xy=x+y \\ xz=2(x+z) \\ yz=3(y+z)$$

$$x=\frac{y}{y-1}, \text{ From the first equation}$$

Now, substituting in the second:

$$\frac{y}{y-1}z=2(\frac{y}{y-1}+z)$$

Then, solving for z:

$$yz=2y+2(y-1)z \\ \implies z(y-2(y-1))=2y \\ z=\frac{2y}{2-y}$$

Now, substituting this in the third equation:

$$y(\frac{2y}{2-y})=3(y+\frac{2y}{2-y}) \\ \implies 2y^2=3(y)(2-y)+3(2y) \\ \implies 2y^2=6y-3y^2+6y \\ \implies 5y^2-12y=0 \implies y(5y-12)=0 \implies y_1=0 \text{ or } y_2=\frac{12}{5}$$

Case 1: $$y=0 \implies z=0,x=0$$     (Rejected).

Case 2: $$y=\frac{12}{5} \implies z=-12 \implies x=\frac{12}{7}$$

We reject case 1, as it will not satisfy the initial equations given. That is, the denominators will be 0.

Therefore, $$(x,y,z):=(\frac{12}{7},\frac{12}{5},-12)$$

Thus, $$z=-12$$

I hope this helps :)!

Aug 17, 2022
edited by Guest  Aug 17, 2022