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

• MHB
• solakis1
In summary, a mathematician found that there are only integer solutions to a system of equations in three variables, where x, y, and z are all integers.
solakis1
Solve the following :A) $$\displaystyle x^2-y\sqrt xy=126$$
$$\displaystyle y^2-x\sqrt xy=-63$$

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

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

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

## 1. What is a system of equalities?

A system of equalities is a set of equations with multiple variables that must be solved simultaneously in order to find a solution that satisfies all of the equations.

## 2. How do I solve a system of equalities?

To solve a system of equalities, you must use algebraic techniques such as substitution, elimination, or graphing to find the values of the variables that satisfy all of the equations in the system.

## 3. Can a system of equalities have more than one solution?

Yes, a system of equalities can have one, infinite, or no solutions. This depends on the specific equations and variables in the system.

## 4. What is the difference between a system of equalities and a system of inequalities?

A system of equalities consists of equations, while a system of inequalities consists of inequalities (such as <, >, ≤, ≥) that must be satisfied simultaneously.

## 5. How can I check if my solution to a system of equalities is correct?

To check if your solution is correct, plug the values of the variables into each equation in the system and see if they satisfy all of the equations. If they do, then your solution is correct.

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