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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
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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

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