How to Solve a System of Three Non-linear Equations?

[solakis1]

Solve the fpllowing system:

[sp]1) $x^2(y-w)=a$

2) $y^2(w-x)=b$

2) $w^2(x-y) =c$

a,b,c non zero[/sp]

 

[jvicens]

To solve this system of equations, we can use the substitution method. We will start by solving for one variable in terms of the other two in each equation.

1) $x^2(y-w)=a$
We can rearrange this equation to get:
$y-w = \frac{a}{x^2}$

2) $y^2(w-x)=b$
Similarly, we can rearrange this equation to get:
$w-x = \frac{b}{y^2}$

3) $w^2(x-y) =c$
Again, we can rearrange this equation to get:
$x-y = \frac{c}{w^2}$

Now, we can substitute these expressions into each other to eliminate one variable at a time. Let's start by substituting $y-w$ into equation 2.

$w-x = \frac{b}{(\frac{a}{x^2})^2}$

Simplifying, we get:
$w-x = \frac{b}{\frac{a^2}{x^4}}$
$w-x = \frac{bx^4}{a^2}$

Next, we can substitute $w-x$ into equation 3.

$x-y = \frac{c}{(\frac{bx^4}{a^2})^2}$

Simplifying, we get:
$x-y = \frac{c}{\frac{b^2x^8}{a^4}}$
$x-y = \frac{ca^4}{b^2x^8}$

Now, we have two equations with only two variables, $x$ and $y$. We can solve for one variable in terms of the other and then substitute it into the other equation to solve for the remaining variable.

From equation 1, we know that:
$y = \frac{a}{x^2} + w$

Substituting this into the second equation, we get:
$x- (\frac{a}{x^2} + w) = \frac{b}{y^2}$
$x - \frac{a}{x^2} - w = \frac{b}{(\frac{a}{x^2} + w)^2}$
$x - \frac{a}{x^2} - w = \frac{bx^4}{a^2 + 2awx^2 + w^2x^