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The substitution method is a powerful tool for solving systems of equations. This approach involves isolating one variable in one equation and then substituting that expression into the other equation.

To use this method effectively, you start by identifying a linear equation within the system that can be easily manipulated to express one variable in terms of the other. For instance, given a system with equations like the ones in our example, it’s straightforward to isolate the variable x in the linear equation x + 2y = 6. Once rearranged, you can express x as 6 - 2y.

The next step involves replacing the isolated variable, x, in the quadratic equation with the expression 6 - 2y. This substitution gives you a single-variable quadratic equation, which is often easier to solve than the original system.

Quadratic equations are mathematical expressions of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a \eq 0\). They are called 'quadratic' because 'quadra' refers to the square in Latin, indicating the variable is squared (raised to the power of two).

The characteristic 'U' shape on a graph of these equations is known as a parabola. The solutions to a quadratic equation are the points at which this parabola crosses the x-axis. These can be found using different methods, such as factoring, completing the square, or applying the quadratic formula. Recognizing the forms of these equations is crucial for factoring and finding the solutions.

Factoring is the process of breaking down the quadratic equation into a product of binomials. The goal is to find two numbers that multiply to the constant term c and add up to the coefficient of the linear term b.

For instance, to factor \(2y^2 - 6y + 2 = 0\), you can look for two numbers that multiply together to make the constant term, \(2\), and also add up to the coefficient of \(y\), which is \( -6\) when factored by 2. After simplification, we get the factored form \(2(y - 1)^2 = 0\), where the solution becomes evident as \(y = 1\).

Factoring can be straightforward, as in this case, but it's trickier when the quadratic doesn't easily factor into neat binomials. In such instances, other methods, like the quadratic formula, might be more efficient.

When solving systems that involve both linear and quadratic equations, such as the one presented in our exercise, it's essential to balance methodical approach with strategic thinking. These systems consist of at least one linear equation, which graphs as a straight line, and one quadratic equation, resulting in a parabola.

The intersection points between the line and the parabola represent the solutions to the system. By using methods like the substitution method, we transform a system of a linear and a quadratic equation into a more straightforward problem, usually reducing it to a single quadratic equation that can be solved by factoring or other means.