x = 1/(3+a) and y = 1/(3–a), where a is any number. (So if a = 0, then x = y = 1/3.)

The way I came about this was really crazy. I said, let x and y be the zeros of a quadratic function written in terms of z, such that (z–x)(z–y) = 0. Multiplying this out gives z² – (x+y)z + xy = 0.

Rewriting the given equation with a common denominator, you get (y+x)/(xy) = 6, so (y+x) = 6xy. Let r = xy, so that x+y = 6r. Substituting into the prior quadratic equation, we get z² – 6rz + r = 0.

Using the Quadratic Formula, you get z = 3r ± 2r*sqrt(2). But remembering that x and y are the zeros of that quadratic equation, and the solutions found for z are said zeros, you can arbitrarily assign each of these solutions as x and y, respectively. Thus, if I say x = 3r + 2r*sqrt(2), then I can set that equation equal to zero and factor out an r to get x – r(3 + 2sqrt(2)) = 0.

Remembering that r = xy, we can rewrite this as x – xy(3 + 2sqrt(2)) = 0. Factoring out an x, we get x[1 – y(3+2sqrt(2)] = 0. Since x cannot equal zero, we get y = 1/(3 + 2sqrt(2)). Solving the similar equation y = 3r – 2r*sqrt(2), you would obtain x = 1/(3 – 2sqrt(2)).

Once I got these values for x and y, I realized that the 2sqrt(2) terms would add up to zero. If those terms were left as arbitrary values, they would still add up to zero, which is how I generated the final solution above.

Jason, that really is a crazy way! It's easy to see that what you got works, since "flip 1/(3+a)" plus "flip 1/(3-a)" equals (3+a) + (3-a) = 6. Having now seen this, lots of other representations of the solutions suggest themselves to me: x = 1/(4+a) and y = 1/(2-a), x = 1/(0.22 + a) and y = 1/(5.88 - a), etc.

Of course, one can also just solve for y in terms of x: Multiplying both sides by xy gives y + x = 6xy, then subtract y to get x = 6xy - y = y(6x - 1), now divide by 6x - 1 to get y = x/(6x - 1).

x = 1/(3+a) and y = 1/(3–a), where a is any number. (So if a = 0, then x = y = 1/3.)

ReplyDeleteThe way I came about this was really crazy. I said, let x and y be the zeros of a quadratic function written in terms of z, such that (z–x)(z–y) = 0. Multiplying this out gives z² – (x+y)z + xy = 0.

Rewriting the given equation with a common denominator, you get (y+x)/(xy) = 6, so (y+x) = 6xy. Let r = xy, so that x+y = 6r. Substituting into the prior quadratic equation, we get z² – 6rz + r = 0.

Using the Quadratic Formula, you get z = 3r ± 2r*sqrt(2). But remembering that x and y are the zeros of that quadratic equation, and the solutions found for z are said zeros, you can arbitrarily assign each of these solutions as x and y, respectively. Thus, if I say x = 3r + 2r*sqrt(2), then I can set that equation equal to zero and factor out an r to get x – r(3 + 2sqrt(2)) = 0.

Remembering that r = xy, we can rewrite this as x – xy(3 + 2sqrt(2)) = 0. Factoring out an x, we get x[1 – y(3+2sqrt(2)] = 0. Since x cannot equal zero, we get y = 1/(3 + 2sqrt(2)). Solving the similar equation y = 3r – 2r*sqrt(2), you would obtain x = 1/(3 – 2sqrt(2)).

Once I got these values for x and y, I realized that the 2sqrt(2) terms would add up to zero. If those terms were left as arbitrary values, they would still add up to zero, which is how I generated the final solution above.

Jason, that really is a crazy way! It's easy to see that what you got works, since "flip 1/(3+a)" plus "flip 1/(3-a)" equals (3+a) + (3-a) = 6. Having now seen this, lots of other representations of the solutions suggest themselves to me: x = 1/(4+a) and y = 1/(2-a), x = 1/(0.22 + a) and y = 1/(5.88 - a), etc.

ReplyDeleteOf course, one can also just solve for y in terms of x: Multiplying both sides by xy gives y + x = 6xy, then subtract y to get x = 6xy - y = y(6x - 1), now divide by 6x - 1 to get y = x/(6x - 1).