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Problem Involving a System of Equations

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,682
Solve the system:

$x^3+3xy^2+49=0$

$x^2-8xy+y^2=8y-17x$

Hi all, I found this problem interesting and I think you may find it interesting too. I have solved it and am of course interested in seeing other approaches.

I will post my solution in a few days, so that everyone interested can have a chance to demonstrate how they would solve it.

Thanks.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Hey anemone!

I could find the solution with Wolfram|Alpha.
But that's cheating.
I haven't been able to come up with a nice solution.
So I'm looking forward to your solution! (Lipssealed)
 

Albert

Well-known member
Jan 25, 2013
1,225
from first equation we get -4<x<0

by observation if x=-1 then y=4 or y=-4

put this to equation 2 , it also satisfies

so we get the answer (x,y)=(-1,4) or (-1,-4)
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,682
Just so you know, I used an entirely different method to solve it. To encourage others to post their solutions, I'll wait until tomorrow to post my solution.

Thanks.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,682
My solution:

$x^3+3xy^2+49=0$---(1)

$x^2-8xy+y^2=8y-17x$---(2)

Multiply both left and right side of (2) by $3x$, we get:

$3x^3-24x^2y+(3xy^2)=24xy-51x^2$---(3)

Rearrange (1) to make $3xy^2$ the subject, we have:

$3xy^2=-49-x^3$

$\therefore 3x^3-24x^2y+(-49-x^3)=24xy-51x^2$

$ 2x^3-24x^2y-49=24xy-51x^2$

$ 2x^3-24x^2y-24xy-49+51x^2=0$

$ 2x^3-24x^2y-24xy-49+49x^2+2x^2=0$

$ (2x^3+2x^2)-24xy(x+1)+49(x^2-1)=0$

$ 2x^2(x+1)-24xy(x+1)+49(x+1)(x-1)=0$

$ (x+1)(2x^2-24xy+49(x-1))=0$

Hence, it must be $x+1=0$ or $2x^2-24xy+49(x-1)=0$

For the case with $x+1=0$, i.e. $x=-1$, we get

$-1-3y^2+49=0$

$48=3y^2$

$y^2=16$

$y=\pm4$

For the case with $2x^2-24xy+49(x-1)=0$, i.e. $24xy=2x^2+49(x-1)$, we'll do the following:

$x^2-8xy+y^2=8y-17x$---(2)

$3x^2-24xy+3y^2=24y-51x$

$3x^2-(2x^2+49(x-1))+3y^2=24y-51x$

$3x^2-(2x^2+49x-49)+3y^2-24y+51x=0$

$x^2+2x+1+3y^2-24y+48=0$

$(x+1)^2+3(y^2-6y+16)=0$

$(x+1)^2+3(y-4)^2=0$

This is true iff $x=-1$ and $y=4$.

We can conclude that the solutions for the original system are $x=-1$, $y=-4$ and $x=-1$, $y=4$.

 

Albert

Well-known member
Jan 25, 2013
1,225
Albert's solution:
$x^3+3xy^2+49=0$---(1)
$x^2-8xy+y^2=8y-17x$----(2)
(1)+(2):
$x^2(x+1)-8y(x+1)+3y^2(x+1)+17(x+1)-2(y^2-16)=0$
$(x+1)(x^2-8y+3y^2+17)-2(y^2-16)=0-----(3)$
$\therefore (x+1)=0,\,\, and, \,\, y^2-16=0----(4)$
$or \,\, x^2-8y+3y^2+17=0 \,\, and,\,\, y^2-16=0----(5)$
it is clear the solution of (4) are :
(x,y)=(-1,4) and (x,y)=(-1,-4)
equation (5) has no solution
in all we get the solutions of original equations (1) and (2) are :
(x,y)=(-1,4) and (x,y)=(-1,-4)
 
Last edited:

Klaas van Aarsen

MHB Seeker
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Mar 5, 2012
8,780
$(x+1)(x^2−8y+3y^2+17)−2(y^2−16)=0−−−−−(3)$

in all we get the solutions of original equations (1) and (2) are :
(x,y)=(-1,4) and (x,y)=(-1,-4)
How about the possible solution of (3) where neither $(x+1)$ nor $(y^2-16)$ are zero?
 

Albert

Well-known member
Jan 25, 2013
1,225
the plot of equation (1) is "Symmetrical" to the line y=0
and the plot of equation (2) is a "Hyperbola"
from (1) we get :
y=[FONT=MathJax_Main]±[/FONT]$ \sqrt{\dfrac{-49-x^3}{3x}}-----------(3)$
from (3) the range of x : $\sqrt[3]{-49}\leq x<0----(*)$
from (2) we rearrange it and obtain :
$y^2-y(8x+8)+x^2+17x=0$
$\therefore y=4(x+1)$[FONT=MathJax_Main]± [/FONT]$\sqrt {15x^2+15x+16}---------(4)$
if (3)=(4) :
y=[FONT=MathJax_Main]±[/FONT] $\sqrt{\dfrac{-49-x^3}{3x}}=4(x+1)$[FONT=MathJax_Main]±[/FONT] $\sqrt {15x^2+15x+16}----(5)$
the only possible solution is :x= - 1 and y=[FONT=MathJax_Main]±[/FONT]4
you may square both sides of (5) and find the solution of x
note: the range of x must meet the restriction of -----(*)
in all we conclude the solution of this equation system is x= -1 and [FONT=MathJax_Math]y=[/FONT][FONT=MathJax_Main]±4[/FONT]
 
Last edited:

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
I had some fun with this one.

Let's rewrite the top equation using mod 3:

[tex]x^3 + 1 = 0 \implies x = -1,~2,~5,~...[/tex] (all mod 3). We know that x = -1 since I cheated and graphed it. :)

So plugging in x = -1 into the bottom equation gives a simple quadratic equation in y. So we get:
[tex]y = \pm 4[/tex].

-Dan
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,682
Hi all,

Thank you so much for participating in this particular problem and I truly appreciate the time and effort you have put into it.:D

Best Regards,

anemone