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# Let x and y be real numbers whose absolute values are different and that satisfy

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Let x and y be real numbers whose absolute values are different and that satisfy

x^3 = 20x + 7y

y^3 = 7x + 20y

Find xy.

Thanks so much!

Jan 21, 2018

#1
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I will assume you want to find  x,y   where x and y have different absolute values...(NOT   x*y as written in your question)

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:

Jan 21, 2018
#3
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Thanks!  I actually was looking for x*y, however this was helpful nonetheless!

AnonymousConfusedGuy  Jan 21, 2018
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I believe this problem has appeared several times before....but, anyway

x^3 = 20x + 7y

y^3 = 7x + 20y     first, add these

x^3 + y^3  =  27x  + 27y

(x + y)( x^2  - xy  + y^2)  = 27(x + y)   divide by x + y

x^2 - xy  +  y^2   = 27   (1)

Next, subtract the first two equations

x^3 - y^3   = 13y - 13y

(x - y) (x^2 + xy+ y^2) =  13(x - y)   divide by x - y

x^2 + xy + y^2  =   13   (2)

Subtract  (2) from (1)

-2xy  =  14        divide both sides by  -2

xy  = -7

Jan 21, 2018
#4
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Thank you!  Sorry for the repeat question :/

AnonymousConfusedGuy  Jan 21, 2018
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No problem ACG....I wouldn't expect anyone to scroll through the thousands of questions on here to see if theirs has been asked before.....I just happened to remember this one....to give full credit, I believe that heureka  or alan may have solved it originally....

Jan 21, 2018
#6
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Oh...Ok    sorry....    If you multiply all of my x.y 's together (rounded)  they come out to -7

Jan 21, 2018
#7
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👍  :D

AnonymousConfusedGuy  Jan 21, 2018