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        riddles >> easy >> Weird system of equations
        
(Message started by: codpro880 on Oct 6^th, 2009, 8:07am)

Title: Weird system of equations
Post by codpro880 on Oct 6^th, 2009, 8:07am

x+y=1
(x^2+y^2)(x^3+y^3)=12

Show appropriate algebra.

Title: Re: Weird system of equations
Post by Wiles_euler on Oct 6^th, 2009, 8:22am

Since, (x+y)=1, then we can create an identity:

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

(x+y)^3=x^3+3x^2y+3xy^2+y^3=1
x^3+y^3+3xy(x+y)=1
x^3+y^3+3xy(1)=1
x^3+y^3=1-3xy

Now we can substitute the identities in the (x^2+y^2)(x^3+y^3)=12

(1-2xy)(1-3xy)=12

Then use quadratic formula to solve for xy and you should be able to do the rest

Title: Re: Weird system of equations
Post by codpro880 on Oct 6^th, 2009, 8:32am

Good work Matt :)

Title: Re: Weird system of equations
Post by Michael Dagg on Oct 12^th, 2009, 10:00pm

Clever work. One hopes to not lose sight of the solution
set by doing that.

So how many solutions are there ?

Can you characterize the solutions of that system
by looking at (1-2xy)(1-3xy) = 12 or perhaps more
specifically from

(1-2xy)(1-3xy) - 12 = 6x^2 y^2 - 5xy - 11 = (xy+1)(6xy-11) = 0 ?

Notice that the obvious relations between x and y in
the polynomial factors (zeros) above, regardless of
which one you choose, do not seem to coincide
with the first equation in the system, that is, y = 1 - x.

Title: Re: Weird system of equations
Post by jpk2009 on Oct 17^th, 2009, 1:47pm

Based on the various definitions of "characterize", the answer is no. To characterize the solutions like I suppose the question is asking we need a polynomial P(x,y)=0 such that each x and y pair that satisfies P(x,y)=0 must also satisfy the system. The polynomial P(x,y)=(xy+1)(6xy-11) does not do that. Besides, it has an infinite number of solutions.

The substitutions that Grendel5 made just "reduce" the second equation in the system to a simplier equation. You can't just use the quadratic formula on it without using the first equation in the system again. Also, you can see that the system equations are multiples of each other by factoring the left side of the second equation.

x+y=1
(x+y)(x²-xy+y²)(x²+y²)=(x ²+y²)(x³+y³)=12

Another thing that I see is that if the system does have solutions then x and y are both real or both complex but not a mixture. If not then x+y has no chance of being real because the imaginary part of x+y must be zero. The substitutions that Grendel5 made make a simpler problem.

x+y=1
(xy+1)(6xy-11)=0

This new problem makes two new problems

x+y=1
xy=-1

x+y=1
xy=11/6

Solving the first system by substituting x=-1/y in the first equation makes

1/y+y-1=0
1/y(y²-y-1)=0

y²-y-1 is zero when y=1/2+1/2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5 or y=1/2-1/2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5. But x=-1/y so there are pairs of solutions for x,y in this same form x=1/2-1/2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5,x=1/2+1/2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5.

Solving the second system by substituting x=11/(6y) in the first equation makes

11/(6y)+y-1=0
1/(6y)(6y²-6y+11)=0

6y²-6y+11 is zero when y=1/2+1/6http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif57 i or y=1/2-1/6http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif57 i. But x=11/(6y) so the solutions are conjugate pairs x=1/2-1/6http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif57 i,x=1/2+1/6http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif57 i.

There are four pairs of solutions to this system

(x1,y1)=(1/2-1/2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5,1/2+1/2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5)
(x2,y2)=(1/2+1/2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5,1/2-1/2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5)
(x3,y3)=(1/2+1/6http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif57 i,1/2-1/6http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif57 i)
(x4,y4)=(1/2-1/6http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif57 i,1/2+1/6http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif57 i)

I confirmed the solution by doing it another way. Using x+y=1, the system

x+y=1
(x+y)(x²-xy+y²)(x²+y²)=12

is now

x+y=1
(x²-xy+y²)(x²+y²)=12

x+y=1
x⁴+2x²y²-x³y-xy³+y⁴-12=0

Using the first equation again and substituting it in the second equation as x=1-y makes

6y⁴-12y³+11y²-5y-11=0

I now cheat an use my TI calculator and the solutions for y are y=1/2+1/6*sqrt(57)*i,y=1/2-1/6*sqrt(57)*i. You can use x=1-y to get x but if you do a substitution and you get the same polynomial except that the variable is x

6x⁴-12x³+11x²-5x-11=0

Maybe this explains why the solutions are conjugate pairs.

I guess the solutions to the system can be characterized one way with the set {(1-y,y)|6y⁴-12y³+11y²-5y-11=0}. I don't know any simpler way to make it.

:)

Title: Re: Weird system of equations
Post by nega1sqrt on Oct 19^th, 2009, 5:07pm

This is from Wolfram Alpha:
[hideb]x = 1/2 (1-sqrt(5)), y = 1/2 (1+sqrt(5))
or
x = 1/2 (1+sqrt(5)), y = 1/2 (1-sqrt(5)) [/hideb]

Title: Re: Weird system of equations
Post by jpk2009 on Oct 19^th, 2009, 5:24pm

Same real solution I produced. What Alpha don't do is show you how to compute the solutions. I showed how to do that.

Title: Re: Weird system of equations
Post by jpk2009 on Oct 24^th, 2009, 12:43pm

After I did that I thought there might be a multiplicity but that is not true. I thought that because the second equation being a multiple of the first might was hiding that additional solution kuz (x+y)=1 can easily removed from the second equation.

Title: Re: Weird system of equations
Post by Michael Dagg on Oct 25^th, 2009, 2:15pm

Multiplicities don't hide. They are quite obvious and
you can gather from the linear factorization of the degree
4 polynomial that there aren't any for at least one factor
would need to occur an even number of times.

You didn't mention it but notice that the product of the
two quadratics (that you produced) is the degree 4 polynomial:

(y^2-y-1)(6y^2-6y+11) = 6y^4-12y^3+11y^2-5y-11 .

This should make complete sense -- the roots of each
quadratic are roots of the degree 4 polynomial.

Title: Re: Weird system of equations
Post by jpk2009 on Oct 25^th, 2009, 5:01pm

Only after the fact did I see that the polynomial could be made as a product of the two quad polynomials. I might have seen that earlier if I had started with the full polynomial, probably maybe. But the real thing I see is that by just making the substitution y=1-x in the second equation "tames" an "otherwise strange looking problem". Also this makes me wonder if there is a method that you can follow so to easily factor polynomials by making systems that have solutions or
factors that are also factors of the underlying polynomial like I did with substitution.

Title: Re: Weird system of equations
Post by Michael Dagg on Oct 28^th, 2009, 7:51pm

Nice idea. But what you seem to be talking about
is not so much about working backwards to
discover a factorization but instead discovering
one by substitution from the start.

There is no such method that I know of that will
work for all polynomials.