Non linear system of 4 equations, how to solve it?

[fluidistic]

Homework Statement

I'm stuck in a problem where I deduced a set of non linear equations. I must solve for A_0, A_1[/itex], x_0 and x_1. I just don't know how to tackle this.
The system is:
A_0+A_1=\frac{2}{3}
A_0x_0+A_1x_1=0
A_0x_0^2+A_1x_1^2=\frac{2}{5}
A_0x_0^3+A_1x_1^3=0.

Homework Equations

No clue.

The Attempt at a Solution

Dead ends. Too many variables, no restriction on these variables except that they must be real. I can't divide any equation by any variable since they can be worth 0, I'm really stuck here.
Thanks for any tip on how to solve this system.

 

[I like Serena]

Hi fluidistic!

The method is: substitution, substitution, substitution.

Use the first equation to express A1 using A0.
Substitute in the other 3 expressions.

Use the new second equation to express A0 using x0 and x1.
Substitute in the new 3rd and 4th expression.

Repeat once more and you'll have solved x0 (or x1).

Use the previous expressions to calculate the other variables.

 

[fluidistic]

Hi fluidistic!

The method is: substitution, substitution, substitution.

Use the first equation to express A1 using A0.
Substitute in the other 3 expressions.

Use the new second equation to express A0 using x0 and x1.
Substitute in the new 3rd and 4th expression.

Repeat once more and you'll have solved x0 (or x1).

Use the previous expressions to calculate the other variables.

Thanks a lot for this tip. Actually this is very nasty :D
I reached \frac{2x_0 ^3}{3(x_0-1)}+ \left [ \frac{2}{3}- \frac{2}{3(x_0-1)} \right ] \left ( \pm \sqrt {\frac{\frac{2}{5}-\frac{2x_0 ^2}{3(x_0-1)}}{\frac{2}{3}-\frac{2}{3(x_0-1)}}} \right ) ^3=0. I'll continue with this :/

 

[I like Serena]

Yes, I knew it would be nasty. Didn't want to spoil the surprise. ;)

 

[hunt_mat]

Newton's method perhaps?

 

[I like Serena]

If you only want the solution, of course you can always use wolfram:
http://www.wolframalpha.com/input/?i={a%2Bb%3D2%2F3%2C+a+x+%2B+b+y+%3D0%2C+a+x^2+%2B+b+y^2+%3D+2%2F5%2C+a+x^3+%2B+b+y^3%3D0}

It turns out that the solution is surprisingly simple. :)

 

[hunt_mat]

There is a trick...
Divide the second equation by A_{0}x_{0} to find that:
<br /> \frac{a_{1}x_{1}}{A_{0}x_{0}}=-1<br />
Now divide the last equation through by A_{0}x_{0} to find that:
<br /> x_{0}^{2}+\frac{A_{1}x_{1}^{3}}{A_{0}x_{0}}=0=&gt;x_{0}^{2}+\frac{A_{1}x_{1}}{A_{0}x_{0}}x_{1}^{2}=0<br />
From here it is pretty much straight forward.

 

[fluidistic]

Thanks guys. In fact I need(ed) the exact values and using the values given in Wolfram seems to solve my problem so that the set of equation I've fell over seems right.
By the way I had no time to continue the algebra to solve for the 4 unknowns. At least I know how to do and I know what I did was ok.
So we can consider the problem as solved :)

Edit.: I just read your last post hunt_mat. Wow, nice trick. I instantly get x_0= \pm x_1 which indeed simplifies things a lot.

 

[hunt_mat]

I told you how to solve the system. I don't think the idea was to use wolframalpha to do your homework...

 

[fluidistic]

I told you how to solve the system. I don't think the idea was to use wolframalpha to do your homework...

Yes sorry, just read your post. (I posted almost in same time as you, as a result I missed the trick). I edited my previous post.

 

[hunt_mat]

So can you say how x_{0} is related to x_{1}?

 

[hunt_mat]

Edit.: I just read your last post hunt_mat. Wow, nice trick. I instantly get x_0= \pm x_1 which indeed simplifies things a lot.

What can I say to such praise.