91Ó°ÊÓ

Linear equations are mathematical expressions of the form \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants. They represent straight lines when plotted on a graph. The solution to a linear equation is any set of values \(x\) and \(y\) that make the equation true. For our exercise, the linear equation is \(x - 3y = -5\).

To work with linear equations, you often need to isolate one of the variables. Once one variable is expressed in terms of the others, it can be substituted into another equation. This is exactly what we did by rearranging \(x - 3y = -5\) to \(x = 3y - 5\). Once isolated, \(x\) can be substituted in other equations, helping solve systems involving multiple equations.

Quadratic equations take the general form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). The equation often represents a parabola when graphed. In our exercise, after substitution, we arrived at a quadratic equation in terms of \(y\): \((3y - 5)^2 + y^2 - 25 = 0\).

Solving quadratic equations involves methods such as factoring, completing the square, or using the quadratic formula, \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\] In this problem, simplifying the quadratic equation after substitution and expansion gives us: \(10y^2 - 30y = 0\). Factor by taking out the greatest common factor, \(10y(y - 3) = 0\), which results in solutions \(y = 0\) and \(y = 3\). These solutions correspond to potential solutions for \(x\) when plugged back into the linear equation.

The substitution method is a technique for solving systems of equations. It involves solving one equation for one variable and then substituting that expression into the other equation. This method is particularly effective when dealing with one linear and one nonlinear equation, as seen in our exercise.

In our example, we first solved the linear equation \(x - 3y = -5\) for \(x\), resulting in \(x = 3y - 5\). This expression for \(x\) was then substituted into the quadratic equation, creating a single equation with one unknown variable, \(y\). The substitution method simplifies the process, allowing you to solve for variables sequentially rather than simultaneously. Once we have the values for \(y\), they can be plugged back into the linear equation to find the corresponding \(x\) values.

Finding solutions to systems of equations means identifying the set of values for the variables that satisfies all equations simultaneously. In systems involving one linear and one quadratic equation, each solution represents a point of intersection between a line and a parabola.

In our problem, we found two solutions: \((-5, 0)\) and \((4, 3)\).

• For \(y = 0\), substituting back into \(x = 3y - 5\) yields \(x = -5\). This gives us the solution \((-5, 0)\).
• For \(y = 3\), substituting back leads to \(x = 4\), providing the solution \((4, 3)\).

These solutions indicate where both the line described by the linear equation and the curve described by the quadratic equation intersect. Understanding such solutions is crucial for skills in graphing, analyzing, and interpreting shapes formed by different types of equations.