POTW 6 Posted

Not as much fun as the previous ones. When does a system of simultaneous equations have no solution?

Substituting 4+y for x in equation 1 yields 16=0 with all the y's cancelling, so my initial guess was that there is no solution to these two equations.
This is confirmed by realizing that equation one, written with x replaced by y+4, approaches the value of 2 in the limit of y = infinite but never actually reaches it.

Two parallel lines, both with slope 1. The first with x intercept at 0, the second with x intercept at 4. Never the twain shall meet.

The immediate answer is, "You can't cancel the xy denominator if xy = 0". So now all we have to do is establish that this is, in fact, the case.

Substituting x = y + 4, we have (2y^2 + 8y + 16)/(y^2 + 4y) = 2. If, and only if, y^2 + 4y 0, then this means that:

2y^2 + 8y + 16 = 2y^2 + 8y
ie 16 = 0
Since 16 0, it follows that y^2 = 4y = 0, or that y is either 0 or -4. If y = 0, clearly xy = 0. If y = -4, then x = 0, and still xy = 0.

Conclusion: cannot cancel the factor of xy, as it is equal to 0. There is no solution to this system of equations.

Hmm, my attempt at "does not equal" there came up blank. That is supposed to say "Since 16 does not equal 0"...

GAZZA: the problem is much simpler than you are trying to make it. They're just two parallel lines. They don't intersect, therefore they have no pair of x and y values in common, and 16 does not equal 0.

I'm not clear on how x/y + y/x = 2 is a line equation. It is discontinuous at x = 0, which is not a line. And I find algebraic proofs more convincing than geometric proofs - YMMV.

It's a line that lacks the single point (0,0). And you can easily do an algebraic proof by trying to find the solution to y=x and y=x-4.

GAZZA@7: The discontinuity at x = 0 is what we call a removable discontinuity. You can show that the limit of y as x approaches 0 is 0 and invoke L'Hopital's rule to say that y / x = (dy/dx)/(dx/dx) = 1/1 = 1.

It's the same argument that lets us say that sin(x)/x = 1 for x = 0. The ratio is only defined in the limiting sense, but since the function otherwise behaves as if its value is 1 at x = 0, we can treat it as having that value. Again, by L'Hopital's rule, sin(x)/x as x approaches 0 is equal to cos(0)/1 = 1.

Your argument in #4 also falls apart because your claimed solutions of (4,0) and (0,-4) don't work. You have a fraction with a nonzero numerator (16 either way) and a zero denominator supposedly equal to a finite number, which isn't true.

Gazza: In general, x/y+y/x=k is the same (apart from what happens when x=0 or y=0; see below) as x^2+y^2-kxy=0. For k>2 this is the equation of a pair of straight lines through the origin (factorize x^2-kxy+y^2 as (x-ay)(x-y/a) to find which straight lines). As k decreases towards 2, the angle between the lines decreases. At k=2 they coincide -- the quadratic factorizes as (x-y)^2 -- and we get the single line x=y. For k<2 there are no solutions to the quadratic other than (x,y)=(0,0).

Putting the equation in the form x/y+y/x=k rather than x^2+y^2=kxy means that instead of the equation being satisfied at (0,0) it's indeterminate there. As Eric says, the breakdown at (0,0) is "removable"; it's just an artefact of the perverse way we've framed the equation.

As for the problem itself, it's a lightly obfuscated version of this: "Consider the equations x-y=0 and x-y=4. Solving them, we find 0=4." whose problem is simply that considering a system of equations is no guarantee that it has any solutions; the correct step after "we find 0=4" is not "... and therefore 0=4" but "... which is a contradiction, and therefore these equations have no solution".