# Pillow-Problems: Problem #47

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## Problem:

47.

Solve the 2 Indeterminate Equations
 x y
= xz;
} (1)
 x z
= xy;
(2)
and find the limits, if any, between which the real values lie.

[12/90

47.

One set of values is 0, 0, 0.

A 2nd set is x = y = 0; z has any value.

A 3rd is x = z = 0; y has any value.

And the 4th set is x =
 k² k − 1
, y = z = k; where k has any value.

If x has any positive value less than 4, y and z are unreal.

## Solution:

47.

By inspection, '0, 0,, 0' are one set of values.

Subtracting, we get x · (
 1 y
 1 z
) = yz;

x = yz ·
 y − z z − y
= −yz, unless y = z, in which case x =
 0 0
.

Now, by (1), x = xyyz;

∴, when yz, x = xy + x;

xy = 0, unless x be infinite.

Similarly, by (2), xz = 0, unless x be infinite.

Hence, if x be finite, and if yz, either x or y = 0, and also either x or z = 0; i.e. either x = 0, or else y = z = 0. But the latter is excluded by our hypothesis. Hence x = 0. Hence yz = 0; i.e. either y or z = 0, and the other may take any value.

This gives us 2 more sets of values, viz.

x = y = 0; z has any value;
x = z = 0; y has any value.

We have now to ascertain what happens when y = z.

By (1),
 x y
= xy

y² = x · (y − 1); i.e. x =
 y² y − 1
.

Similarly, by (2), x =
 z² z − 1
.

This gies us a 4th set of values, viz. x =
 k² k − 1
, y = z = k; where k has any value.

Now y and z may evidently have any real values, but x is restricted by the equation

y² − xy + x = 0,

in which y cannot be real, unless (x² − 4x) > 0. Hence x may have any negative value, and any positive value that is not less than 4; but it cannot have any positive value, less than 4, without making y unreal.

Q.E.F.