+35 The system of equations has exactly one solution. What is in this solution?
\(\frac{xy}{x + y} = 1, \quad \frac{xz}{x + z} = 2, \quad \frac{yz}{y + z} = 4\)
(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
+177 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!
