91Ó°ÊÓ

Nonlinear equations differ from linear equations because they contain variables raised to a power greater than one or variables multiplied together. As a result, their graphs are not straight lines but curves such as parabolas or circles. To solve nonlinear systems of equations, we look for values that satisfy all given equations simultaneously.

In our particular problem, we have the system of equations:
\[x^{2}-3xy+2y^{2}=0\]
\[x^{2}+xy=6\]
These are both quadratic in nature due to the term \(x^2\) and the product \(xy\). Our goal is to find pairs of \(x\) and \(y\) that make both equations true.

Solving such systems might involve different techniques such as substitution or elimination. In this problem, we use the substitution method to find solutions effectively.

The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This helps reduce the system to a single equation with one variable.

First, let's solve the second equation for \(y\):
\[x^2 + xy = 6\]
Subtract \(x^2\) from both sides:
\[xy = 6 - x^2\]
Then, divide by \(x\):
\[y = \frac{6 - x^2}{x}\]
Now, we substitute this expression into the first equation where \(y\) appears:
\[x^2 - 3x \left( \frac{6 - x^2}{x} \right) + 2 \left( \frac{6 - x^2}{x} \right)^2 = 0\]
By doing this, we transform the problem into a single equation with only one variable, \(x\). This equation can further be simplified and solved for \(x\).

Factoring is a key method used to solve quadratic equations. After substitution and simplification in our problem, we reached a quadratic equation in terms of \(x^2\):
\[x^4 - 7x^2 - 12 = 0\]
Here, we let \(z = x^2\), reformulating the equation as:
\[z^2 - 7z - 12 = 0\]
This is a standard quadratic form that can be factored as:
\[(z - 9)(z + 2) = 0\]
Solving for \(z\), we get:
\[z = 9\] or \[z = -2\]
Since \(z = x^2\) and \(x^2\) cannot be negative, only \(z = 9\) is valid. Thus;
\[x^2 = 9\]
Which leads to:
\[x = 3\] or \[x = -3\]
We then substitute these values of \(x\) back into \(y = \frac{6 - x^2}{x}\) to find the corresponding \(y\) values, eventually providing us with solutions to the original system:
\[(3, -1)\]
\[(-3, 1)\]