Solving System of Equalities: x^2-y\sqrt xy & y^2-x\sqrt xy =3

[solakis1]

Solve the following :A) \(\displaystyle x^2-y\sqrt xy=126\)
\(\displaystyle y^2-x\sqrt xy=-63\)

B) \(\displaystyle \frac{x} {y}+\frac{y}{z}+\frac{z}{x}=\frac{y}{x}+\frac{z}{y}+\frac{x}{z}=x+y+z=3\)

 

[topsquark]

B:
[sp]
I really wish you would say if you are looking for just integer solutions. I'm assuming that this is the case, as I'll point out below.

First, given the symmetry of the fractions it turns out that all we need to consider is
\(\displaystyle \begin{cases} \dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x} = 3 \\ x + y + z = 3 \end{cases} \)

Now, if we have x = y = z we get 3x = 3 from the bottom equation, so x = y = z = 1.

If we look at y = x with z indeterminate:
\(\displaystyle \begin{cases} \dfrac{x}{x} + \dfrac{x}{z} + \dfrac{z}{x} = 3\\ 2x + z = 3 \end{cases} \)

Putting z = 3 - 2x into the top equation we get
\(\displaystyle 9x^2 - 18x + 9 = 0\)

or x = 1, y = 1, z = 3 - 2(1) = 1. So we don't get a new solution. (This also holds for x = z, and y = z cases.)

The general argument is a bit more complicated but certainly do-able.

We have two equations in three variables. Let me use the following case: I'm going to let y = nx, with z undetermined, and use n as a parameter for the solutions.

\(\displaystyle \begin{cases} \dfrac{x}{nx} + \dfrac{z}{xn} + \dfrac{z}{x} = 3 \\ x + nx + z = 3 \end{cases}\)

The steps are the same as the n = 1 case above. The algebra isn't much fun but in the end it's just a quadratic equation. To get to the point I will just give the answer:
\(\displaystyle x = 3 \cdot \dfrac{(2n^2 + 5n - 1) \pm (n - 1) \sqrt{1 - 4n}}{2 (6n^3 + 6n^2 + 3n - 1)}\)

Now, we have to be careful. We have to leave out the values of n for where the denominator is 0 and we have to make sure that the argument of the square root is positive. This is fairly easily done and a simple check with a graphing calculator will show that there are indeed solutions.

Which brings me back to my original comment. There are an infinite number of irrational solutions. This is why I'm guessing we only want integer solutions.

With that thought I can finish the problem. The solution for x will only be rational for n = 1 because the square root factor is always irrational for any n. We've already found the n = 1 solution.

Thus x = y = z = 1 is the only possible integer solution.
[/sp]
-Dan

 

[solakis1]

[sp] hint:. put\(\displaystyle \frac{x}{y}=w ,\frac{y}{z}=e ,\frac{z}{x}=f\)[/sp]