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# Explain thougroughly... Again

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Let $$x$$ and $$y$$ be real numbers whose absolute values are different and that satisfy \begin{align*} x^3 &= 20x + 7y \\ y^3 &= 7x + 20y. \end{align*}

Find $$xy.$$

Dec 16, 2019

#1
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I used a graphical approach (desmos) to graph the two equations you will see the following sets of coordinates satisfy the equations :

x,y  = (-1.7 , 4.1)     (-4.1, 1.7)   and   (4.1, -1.7)  and (1.7, - 4.1)

0,0    -3.6, 3.6   and  3.6, -3.6     5.2 , 5.2     -5.2, -5.2 also work, but the absolute values of x and y are equal at these points, so they do not satisfy the question.

Here is the graph:

Dec 16, 2019
#2
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Let xy=k, so that y = k/x.

Substitute this into the first equation to get  $$x^2 - 20x^2 - 7k = 0$$

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 is not equal to 0,  it follows that xy = -7.

Dec 16, 2019
#3
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You can solve for y in the first equation, to get y = (x^3 - 20x)/7.

You can then substitute into the second equations, which gives you

$$\left( \frac{x^3 - 20x}{7} \right)^3 = 7x + 20 \left( \frac{x^3 - 20x}{7} \right)$$

This factors as x(x - 5)(x + 5)(x^2 + 12)(x^4 - 20x^2 + 49) = 0.  You can then work out the rest.

Dec 16, 2019