Solving System of Equations: Guidance for Australian Math Olympiad Problem

[murshid_islam]

can anyone tell me how to solve the system of equations:

(x-1)(y^2+6) = y(x^2+1)

(y-1)(x^2+6) = x(y^2+1)

note: this is not a homework problem. as far as i know, this is a problem from australia's national math olympiad.

 

[benorin]

I used the http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=solve&s3=advanced and put

(x-1)(y^2+6)=y(x^2+1)
(y-1)(x^2+6)=x(y^2+1)

into the field marked "EQUATION(S)" to get the solutions

(x,y)=(2,2),\, (2,3),\, (3,2),\, (3,3),\, \left( \frac{1}{2}\left(1\pm i\sqrt{15}\right) , \frac{1}{2}\left(1\mp i\sqrt{15}\right)\right).

 

[Integral]

I do not think that resorting to a software solution would be acceptable on a test.

Expand both equations, then add them together. You will now have quadratic equations in x any y. Complete the squares and you will easily get the real solutions.

 

[murshid_islam]

benorin, i also got those solutions using Maple. but what i want to know is how to solve them without using softwares.

Integral, i tried doing what you suggested. i expanded and subtracted one from the other and got an equation of a circle. but i don't know what to do next. can you give some more hints?

 

[cristo]

Try completing the square

 

[murshid_islam]

by adding the 2 equations i got, (x-y)(x+y+2xy+7) = 0

by subtracting one from the other i got, \left(x-\frac{5}{2}\right)^2 + \left(y-\frac{5}{2}\right)^2 = \frac{1}{2}

but i can't figure out what to do next.

 

[dextercioby]

That's the eqn of a circle in the Oxy plane with radius 1/2 and located with the center at (5/2,5/2). You'll have to intersect this circle with the 2 line y=x and with the curve x+y+2xy+7=0 to get the possible solutions.

Daniel.

 

[murshid_islam]

That's the eqn of a circle in the Oxy plane with radius 1/2 and located with the center at (5/2,5/2).

the radius is actually \frac{1}{\sqrt{2}}

You'll have to intersect this circle with the 2 line y=x and with the curve x+y+2xy+7=0 to get the possible solutions.

but how do i do that? if i take the intersection between the circle and the line y = x, i get the solutions (3,3) and (2,2). but how do i get the intersection of the circle and the curve x+y+2xy+7=0?

 

[murshid_islam]

well, i tried the following:
from x+y+2xy+7=0, we get y = \frac{-7-x}{1+2x}

substituting this into the equation of the circle and simplifying, i got the quartic equation:

x^4 - 4x^3 + 10x^2 + 33x + 24 = 0

now how do i solve this?

 

[arildno]

First of all, you got the first equation wrong, it should read:
(x-y)(x+y-2xy+7)=0

Secondly, eliminating the (x-y) factor, ADD the two equations you now have, yielding:
x^{2}-5x+y^{2}-5y+\frac{25}{2}+(x+y-2xy+7)=\frac{1}{2}
that is:
(x-y)^{2}-4(x+y)+19=0

Similarly, subtract the first one from your second one to get:
(x+y)^{2}-6(x+y)+5=0

Now, introduce the new variables u=x+y, v=x-y to rewrite your system of equations as:
u^{2}-6u+5=0
v^{2}-4u+19=0
This system is readily solvable.

 

[Gib Z]

Is it helpful to notice here that these equations are inverses of each other? Then would mean that the intersect on the like y=x correct? And a few other places of course...but thought it might help since if you have one solution, you immediately have another :)