91Ó°ÊÓ

To solve a system of equations means finding the values of the variables that satisfy all the equations simultaneously. In our problem, we are given two equations involving the variables x and y:
\[2x^2 - y^2 = 6\]
\[y = x^2 - 3\]
We will use the substitution method to solve this system. The substitution method involves substituting one equation into the other. This is possible because one of the equations is already solved for y. By substituting \(y = x^2 - 3\) into the first equation, we will end up with a single equation containing only one variable, x. This makes it easier to solve.
After substitution, we need to simplify and solve the resultant equation to find the values of x. Finally, once we find the values of x, we will substitute them back into the second equation to find the corresponding values of y.

A quadratic equation is an equation of the form \(ax^2 + bx + c = 0\). In our problem, after substituting \(y = x^2 - 3\) into the first equation, we end up with a quadratic equation in x. To find the values of x that satisfy this equation, we need to simplify and re-arrange the terms. After combining like terms, our equation might look something like:
\[ax^2 + bx + c = 0\]
To solve this, we can use the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This formula gives us the possible values of x. Depending on the discriminant \(b^2 - 4ac\), we might have:

• Two real solutions
• One real solution
• No real solutions (if the discriminant is negative)

Once we have the solutions for x, these solutions are used in the next step called back-substitution to find y.

Back-substitution is the process of substituting the solutions for one variable back into the original equations to find the corresponding values of the other variables. In our problem, after solving the quadratic equation for x, we might get two solutions:
\(x_1\) and \(x_2\).
To find the corresponding values of y, we substitute these values of x back into the second equation:
\(y = x^2 - 3\)
For each value of x, we perform the substitution and calculate the value of y. This gives us the full set of solutions (x, y) that satisfy both original equations.