MHB Problem Involving a System of Equations

AI Thread Summary
The discussion centers around solving a system of equations involving cubic and quadratic expressions. Participants found that the solutions are (x, y) = (-1, 4) and (-1, -4). Various methods were employed, including graphical approaches and algebraic manipulation. The symmetry of the first equation and the hyperbolic nature of the second were noted as important characteristics. Overall, the community engaged in sharing their solutions and methods, highlighting the collaborative effort in problem-solving.
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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.
 
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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)
 
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)
 
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.
 
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'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:
Albert said:
$(x+1)(x^2−8y+3y^2+17)−2(y^2−16)=0−−−−−(3)$

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?
 
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]±$ \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]± $\sqrt {15x^2+15x+16}---------(4)$
if (3)=(4) :
y=[FONT=MathJax_Main]± $\sqrt{\dfrac{-49-x^3}{3x}}=4(x+1)$[FONT=MathJax_Main]± $\sqrt {15x^2+15x+16}----(5)$
the only possible solution is :x= - 1 and y=[FONT=MathJax_Main]±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=MathJax_Main]±4
 
Last edited:
I had some fun with this one.

Let's rewrite the top equation using mod 3:

x^3 + 1 = 0 \implies x = -1,~2,~5,~... (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:
y = \pm 4.

-Dan
 
  • #10
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
 

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