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The substitution method is a technique used to solve systems of equations by expressing one variable in terms of another, then substituting this expression into a second equation.
This creates a single equation in one variable, which can be solved normally.

• First, isolate one variable in one of the equations, like we did with the equation \( y = 3x + 3 \).
• Substitute this expression for the isolated variable into the other equation.

By repeating these steps, the problem reduces to solving a simpler equation. Let's look at an example with our system: given the equations:
\( 9x^2 + y^2 = 9 \) and \( y = 3x + 3 \).

We start by substituting \( y = 3x + 3 \) into the first equation, resulting in:
\[ 9x^2 + (3x + 3)^2 = 9 \]. This effectively turns our system of equations into one, simpler quadratic equation in terms of \(x\). This brings us to our next core concept.

A quadratic equation is a second-degree polynomial equation in a single variable, usually in the form \( ax^2 + bx + c = 0 \). In our example, after substituting, we get:
\[ 9x^2 + (3x + 3)^2 = 9 \].
Expanding and combining like terms, the equation becomes:
\[ 18x^2 + 18x + 9 = 9 \].
After simplifying, it results in:
\[ 18x^2 + 18x = 0 \].
Now, we have a standard quadratic equation where \(a = 18\), \(b = 18\), and \(c = 0\). Quadratic equations can be solved by various methods such as factoring, completing the square, or using the quadratic formula. In this instance, factoring is straightforward.

Factoring involves expressing a polynomial as a product of its factors. This is particularly useful for solving quadratic equations. In our problem, we have the equation:
\[ 18x^2 + 18x = 0 \].
We can factor out the greatest common factor (GCF), which in this case is 18x:
\[ 18x(x + 1) = 0 \].
This factorization reveals two potential solutions: \( x = 0 \), or \( x = -1 \).
To find the corresponding values of \( y \), we substitute each \( x \)-value back into the second original equation \( y = 3x + 3 \):

• For \( x = 0 \), \( y = 3(0) + 3 = 3 \).
• For \( x = -1 \), \( y = 3(-1) + 3 = 0 \).

This gives us the two solutions to the system of equations: \( (0, 3) \) and \( (-1, 0) \).
Understanding these methods and how they are used together allows you to approach and solve a wide range of problems involving systems of equations.