Wikipedia:Reference desk/Archives/Mathematics/2013 July 28

= July 28 =

Solve this pair of two equations
Solve for x and y using two equations given below.

x2 + y2 = 25

x3 + y3 = 91

I have been finding the value of x and y using above two relations for a number of days but I failed. At last I gave up and put this one here. Please suggest me different ways to solve this. Also suggest me some simple tricks (or handy methods or videos) for solving problems like this. Thank you! Publisher54321 (talk) 03:54, 28 July 2013 (UTC)
 * You could use Newton's method to solve it numerically, but it turns out that it might be easier to just graph the two. They happen to intersect at a point where both x and y are integers. I don't know of a way to reduce these two equations to one except by substitution - first solving for y from the first equation yields $$y=\sqrt{25-x^2}$$, which we can substitute into the second equation to obtain $$x^3 + (25-x^2)\sqrt{25-x^2}=91$$. We move the $$x^3$$ term to the right, square both sides, and solve.-Jasper Deng (talk) 04:00, 28 July 2013 (UTC)
 * By observation, x = 3, y = 4 and x = 4, y = 3 are both valid solutions. You can easily deduce that solutions must satisfy |x| &le; 5 and |y| &le; 5, plotting the curves will give you a good idea of the number of solutions.  EdChem (talk) 04:13, 28 July 2013 (UTC)
 * Indeed (although it's more correct to say they must be on the circle with r=5 and center at the origin than that) - the only way to solve it algebraically is to then try to factor that sixth-order polynomial via synthetic division - it might take a long time, but that's the only algebraic way I can think of.--Jasper Deng (talk) 04:18, 28 July 2013 (UTC)


 * Addendum: After plotting both equations, I see there are in fact four solutions. These first two points with integer coordinates aren't that hard, but there are two others that have to be found using the quartic formula.--Jasper Deng (talk) 05:45, 28 July 2013 (UTC)

The equations are symmetric: if (x,y) is a solution then so is (y,x). Therefore each equation can be written in terms of the product p=xy and the sum s=x+y, and knowing p and s you find x by solving the equation x2 &minus; sx + p = 0. The equations are

= 3s(s2 &minus; 2p &minus; 25) &minus; 2(s3 &minus; 3ps &minus; 91) = s3 &minus; 75s + 182 = (s &minus; 7)(s2 + 7 s &minus; 26)
 * 0 = s2 &minus; 2p &minus; 25 = s3 &minus; 3ps &minus; 91

Bo Jacoby (talk) 11:13, 28 July 2013 (UTC). The equations are solved like this. p = (s2 &minus; 25)/2 x2 &minus; sx + p = 0 x2 &minus; sx + (s2 &minus; 25)/2 = 0 x = (s + (50 &minus; s2)1/2)/2 y = s &minus; x (s &minus; 7)(s2 + 7 s &minus; 26) = 0

There are six solutions: the following three and their symmetricals. s1 = 7 ; x1 = (7 + (50 &minus; 72)1/2)/2 = 4 ; y1 = 7 &minus; 4 = 3 s2 = (3&times;171/2 &minus; 7)/2 ; x2 = (s2  + (50 &minus; s2 2)1/2)/2 ; y2 = s2 &minus; x2 s3 = (&minus; 3&times;171/2 &minus; 7)/2 ; x3 = (s3  + (50 &minus; s3 2)1/2)/2 ; y3 = s3 &minus; x3 Bo Jacoby (talk) 21:53, 28 July 2013 (UTC).
 * Really, six solutions? The graph of the two only suggests the existence of four, which makes two of those solutions extraneous.--Jasper Deng (talk) 23:20, 28 July 2013 (UTC)
 * Only four of the solutions are real.  Sławomir Biały  (talk) 00:07, 29 July 2013 (UTC)
 * Bézout's theorem guarantees 6 solutions (including multiples and complex).RDBury (talk) 00:46, 29 July 2013 (UTC)

Solutions: (3, 4), (4, 3), (4.61313, &minus;1.92847), (&minus;1.92847, 4.61313), (&minus;4.84233+3.3088i, &minus;4.84233&minus;3.3088i), (&minus;4.84233&minus;3.3088i, &minus;4.84233+3.3088i) Bo Jacoby (talk) 05:08, 29 July 2013 (UTC).
 * I'm sure it was one of the last two that the original poster was looking for. :) Naraht (talk) 15:37, 29 July 2013 (UTC)


 * i imagine so. :-) StuRat (talk) 06:11, 1 August 2013 (UTC)

The OP was looking for tricks. It is luck that these equations are solvable by square roots. Algebraic equations 0=f(x,y)=g(x,y) are solved by eliminating y getting 0=h(x) where h is a polynomial. In our case h(x) has degree 6, as pointed out by Jasper Deng, and so there are six solutions. Bo Jacoby (talk) 05:14, 2 August 2013 (UTC).