Let 
and 
be real numbers whose absolute values are different and that satisfy
Find 
Please explain every part thoroughly. Thank you!
+893 Let xy=k, so that y = k/x.
Substitute this into the first equation to get $$x^{4}-20x^{2}-7k=0.$$
Solve this (using the usual formula) as a quadratic in x squared to get $$x^{2}=10\pm\sqrt{100+7k}$$.
The original equations are symmetric in x and y so repeating the procedure will produce an identical result for y.
Since x and y are different, one will have the positive sign in the middle, the other the negative sign.
Multiplying the two results produces (difference between two squares)
$$x^{2}y^{2}=100-(100+7k)=-7xy.$$
Since $$xy\ne 0,$$ it follows that xy = -7.
+129852 x^3 = 20x + 7y (1)
y^3 = 7x + 20y (2) subtract (2) from(1)
x^3 - y^3 = 13(x-y) factor the left side
(x-y)(x^2 + xy + y^2) = 13(x-y) divide through by (x-y)
x^2 + xy + y^2) = 13 subtract 13 from both sides
x^2 + xy +y^2 - 13 = 0 (3) .... this is a rotated ellipse....
We need to find the intersection points of (1),(2) and (3)
Here's a graph
The intersection points we're after are where the absolute values of x and y are different
These occur at (-4.14, 1.691), (4.14, -1.691),(1.691, -4.14) and (-1.691, 4.14)
And xy will be -(1.691)(4.14) which will be ≈ -7
BTW... here are the exact values generated by WolframAlpha
I have a feeling that the math was pretty sticky getting these.....!!!
+893 Let xy=k, so that y = k/x.
Substitute this into the first equation to get $$x^{4}-20x^{2}-7k=0.$$
Solve this (using the usual formula) as a quadratic in x squared to get $$x^{2}=10\pm\sqrt{100+7k}$$.
The original equations are symmetric in x and y so repeating the procedure will produce an identical result for y.
Since x and y are different, one will have the positive sign in the middle, the other the negative sign.
Multiplying the two results produces (difference between two squares)
$$x^{2}y^{2}=100-(100+7k)=-7xy.$$
Since $$xy\ne 0,$$ it follows that xy = -7.
