Pillow-Problems: Problem #47

Problem #45 | Problem #50
For more information on this collection, see Pillow-Problems by Charles L. Dodgson (Lewis Carroll).

Problem:

Solve the 2 Indeterminate Equations

x y = x − z;	}	(1)
x z = x − y;		(2)

and find the limits, if any, between which the real values lie.

Answer:

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:

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

Subtracting, we get x · (

1

y

−

1

z

) = y − z;

∴ x = yz ·

y − z

z − y

= −yz, unless y = z, in which case x =

0

0

.

Now, by (1), x = xy − yz;

∴, when y ≠ z, 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 y ≠ z, 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.

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

By (1),

x

y

= x − y

∴ 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

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.