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

• fluidistic
In summary, the conversation discusses a problem involving a set of non-linear equations with four unknown variables. The participants suggest using substitution to solve the system, and one participant reveals a helpful trick that simplifies the solution. The conversation concludes with the problem being solved and the participants discussing the solution.
fluidistic
Gold Member

## 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:
$A_0+A_1=\frac{2}{3}$
$A_0x_0+A_1x_1=0$
$A_0x_0^2+A_1x_1^2=\frac{2}{5}$
$A_0x_0^3+A_1x_1^3=0$.

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.

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.

I like Serena said:
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 :/

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

Newton's method perhaps?

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

Last edited:
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.

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

hunt_mat said:
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.

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

fluidistic said:
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.

## 1. How do I identify a non-linear system of 4 equations?

A non-linear system of 4 equations is a set of four equations where the variables are raised to a power higher than 1, or where the variables are multiplied together. This is different from a linear system where the variables are only raised to the power of 1 and are not multiplied together.

## 2. What is the difference between solving a linear and non-linear system of 4 equations?

The main difference is that a linear system can be solved using standard algebraic methods such as substitution or elimination, while a non-linear system requires more advanced techniques such as graphing or using a computer program to find approximate solutions.

## 3. Can a non-linear system of 4 equations have more than one solution?

Yes, it is possible for a non-linear system to have multiple solutions. In fact, most non-linear systems have multiple solutions, making them more complex to solve compared to linear systems.

## 4. What are some common methods for solving a non-linear system of 4 equations?

Some common methods for solving non-linear systems include graphing, substitution, elimination, and using a computer program or calculator to find approximate solutions. It is also helpful to rewrite the equations in terms of one variable and then solve for that variable.

## 5. Are there any special cases or exceptions when solving a non-linear system of 4 equations?

Yes, there are some special cases that may arise when solving a non-linear system. For example, there may be no solutions, infinitely many solutions, or only one solution. It is important to carefully analyze the equations and the solutions obtained to ensure they make sense in the given context.

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