Solve the 2 Indeterminate Equations
| } | (1) | ||
| (2) |
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 |
If x has any positive value less than 4, y and z are unreal.
By inspection, '0, 0,, 0' are one set of values.
Subtracting, we get x · (
1 |
y |
1 |
z |
∴ x = yz ·
y − z |
z − y |
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 |
∴ 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 |
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.