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# I need help

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The system of equations has exactly one solution. What is  in this solution?

$$\frac{xy}{x + y} = 1, \quad \frac{xz}{x + z} = 2, \quad \frac{yz}{y + z} = 4$$

Aug 20, 2023

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(x y)/(x + y) = 1
(x z)/(x + z) = 2
(y z)/(y + z) = 4, solve for x, y, z

Use elimination and substitution to get:

x = 8/5,  y = 8/3,  z=  - 8

Aug 20, 2023
#3
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This problem just requires some clever algebraic manipulation. I know others have already posted the answer to this system of equations, but I will show you some work as to how one might find one of the variables because it may not be too straightforward of a process. If I can avoid it, I try to resist having fractions in a problem involving system of equations, so I decided to eliminate the fractions from the original equation. There is probably a faster way, but it is typically hard to find.

 $$\frac{xy}{x + y} = 1 \\ xy = x + y$$ $$\frac{xz}{x + z} = 2 \\ xz = 2x + 2z$$ $$\frac{yz}{y + z} = 4 \\ yz = 4y + 4z$$

Afterwards, I solved for one of the variables in both equations. I chose to solve for y, but any choice should suffice. I used the technique of grouping all the terms with y and then factoring.

 $$xy = x + y \\ xy - y = x \\ y(x - 1) = x \\ y = \frac{x}{x - 1}$$ $$yz = 4y + 4z \\ yz - 4y = 4z \\ y(z - 4) = 4z \\ y = \frac{4z}{z - 4}$$

We have two equations solved for y, so we can find a relationship between x and z variables.

$$y = \frac{x}{x - 1}, y = \frac{4z}{z - 4} \\ \frac{x}{x - 1} = \frac{4z}{z - 4} \\ x(z - 4) = 4z(x - 1) \\ xz - 4x = 4xz - 4z \\ -3xz = 4x - 4z$$

Now, we can conveniently use the $$xz = 2x + 2z$$ equation to help solve this system of equations.

$$xz = 2x + 2z \\ 3xz = 6x + 6z$$

Now, we can use technique of elimination to find a relationship between x and z.

\begin{align*} 3xz &= &6x &+ 6z \\ -3xz &= &4x &- 4z \\ 0 &= &10x &+ 2z \end{align*} \\ 5x + z = 0 \\ z = -5x

Now, we can substitute this relationship for z into an equation with x and z variables and solve for x.

$$xz = 2x + 2z \\ x * -5x = 2x + 2 * -5x \\ -5x^2 = -8x \\ -5x^2 + 8x = 0 \\ -x(5x - 8) = 0 \\ x = 0 \text{ or } 5x - 8 = 0 \\ x = 0 \text{ or } x = \frac{8}{5}$$

Now, you might notice that we obtained x = 0 as a potential solution, but we can reject $$x = 0$$ immediately because if we were to substitute this x-value into the original equation $$\frac{xy}{x + y} = 1$$, there would be no solution, so $$x = 0$$ is not a candidate. At this point, I will not go any further because the process should be similar to find the other variables. Happy solving!

Aug 21, 2023