Solving Systems of Equations with Ordered Triples

In this article, we will explore how to solve a complex system of equations using the method of ordered triples. We will walk through the process step-by-step, providing detailed explanations and insights to help you understand the underlying mathematical principles at play.

Introduction to the Problem

Consider the following system of equations:

[begin{cases} xyxyz 120 yzxyz 96 xzxyz 72 end{cases}]

From these equations, we introduce a new variable, (s x y z). This will allow us to simplify the original system and solve it more effectively.

Introducing a New Variable

Using the new variable (s x y z), we rewrite the original system in terms of (s):

(x y s 120) (y z s 96) (x z s 72)

We can further express these equations as fractions to isolate the products:

(x y frac{120}{s}) (y z frac{96}{s}) (x z frac{72}{s})

Expressing x, y, and z in Terms of s

We now express (x), (y), and (z) in terms of (s). Starting with the first equation:

(x y frac{120}{s})

From the second equation, we have:

(y z frac{96}{s})

Finally, from the third equation:

(x z frac{72}{s})

To find (z) in terms of (s) and (x), we start with the expression for (x y):

(y frac{120}{s x})

Substituting (y) into the second equation:

(left(frac{120}{s x} - xright) z frac{96}{s})

This simplifies to:

(z frac{96}{s} - frac{120}{s} x frac{-24}{s} x)

Substituting (y) into the third equation to find (x):

(x left(frac{-24}{s} xright) frac{72}{s})

This simplifies to:

(2x - frac{24}{s} x frac{72}{s})

(2x frac{72}{s} frac{24}{s} x)

(2x frac{96}{s} x)

(x frac{48}{s})

Substituting (x) back to find (y) and (z):

(y frac{120}{s} - frac{48}{s} frac{72}{s})

(z frac{-24}{s} cdot frac{48}{s} frac{24}{s})

Thus, we have:

(x frac{48}{s}) (y frac{72}{s}) (z frac{24}{s})

Finally, we express (xyz):

(x y z frac{48}{s} cdot frac{72}{s} cdot frac{24}{s} frac{144}{s})

Given that (s x y z), we have:

(s frac{144}{s})

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Multiplying both sides by (s):

(s^2 144)

(s 12)

(since (s > 0))

Substituting (s 12) back into the expressions for (x), (y), and (z):

(x frac{48}{12} 4)

(y frac{72}{12} 6)

(z frac{24}{12} 2)

Hence, the solution is (boxed{4, 6, 2}).

Further Analysis

Adding the three equations results in:

(2x2y2zxyz 288)

Dividing by (2):

(xyz^2 144)

(xyz) is either (12) or (-12).

Considering (xyz 12), we find:

(xy frac{120}{xyz} 10) (yz 8) (xz 6)

Subtracting each of these equations from (xyz 12) yields:

(x 4) (y 6) (z 2)

A similar argument shows that if (xyz -12), then:

(x -4) (y -6) (z -2)

Hence, the solutions are (4, 6, 2) and (-4, -6, -2).