91Ó°ÊÓ

The substitution method is a powerful tool used for solving systems of equations, including nonlinear ones. This involves isolating one variable in one equation and substituting it into the other. Here’s how it works:

In our exercise, we start with:
1) \,2x^2 - 3y^2 = 12\
2) \,6x^2 + 5y^2 = 36
First, we solve the first equation for \( x^2 \):
\(2x^2 - 3y^2 = 12\)
We rearrange it to:
\(2x^2 = 12 + 3y^2\)
then:
\(x^2 = \frac{12 + 3y^2}{2}\)
You now have \(x^2\) in terms of \(y^2\) and can substitute this expression into the second equation. By replacing \(x^2\) in the second equation with \( \frac{12 + 3y^2}{2} \), you eliminate one variable, making the equation solvable. This reduces complexity and allows for step-by-step simplification.

The term 'nonreal complex solutions' refers to solutions that include imaginary numbers. Imaginary numbers are based on the imaginary unit \( i \), where \(i^2 = -1\). Such solutions often arise in higher-degree polynomials or when dealing with square roots of negative numbers.

In our exercise, we primarily focus on real solutions, but it's important to recognize that complex solutions can also be part of the solution set in other cases:

• If during solving, you encounter a negative number inside a square root, it indicates potential nonreal complex solutions.
• These would generally be represented as \( a + bi \), where \(a\) is the real part and \(b\) is the imaginary part.

For example, solving \(x^2 = -1\) yields \(x = \pm i\). While our given system didn't lead to nonreal complex solutions, knowing how they arise prepares you for various equations.

Quadratic equations are second-degree polynomial equations of the form \(ax^2 + bx + c = 0\). They may have two real solutions, one real solution, or two complex solutions depending on the discriminant \(b^2 - 4ac\):

• When \(b^2 - 4ac > 0\), there are two distinct real solutions.
• When \(b^2 - 4ac = 0\), there’s one real solution (a repeated root).
• When \(b^2 - 4ac < 0\), there are two complex solutions.

In our exercise, the equations we solved were quadratic in form. For example, solving \(2x^2 = 12 + 3y^2 \) reduces to:
\(x^2 = 6 + \frac{3y^2}{2}\)
Given that \(y^2 = 0\), we find:
\(x^2 = 6\)
This quadratic equation has real solutions \(x = \pm \sqrt{6}\). Quadratic equations frequently surface in various algebraic contexts, making them essential to master.