Foundational Mathematics for Corequisite Courses: Succeeding at College Mathematics

Section 15.1 Replace This With That

In the previous sections, we were looking at lines in isolation of each other. In this section, we're going to look at the ways that two lines can interact with each other.

A solution to a linear equation is a pair of values \((x, y)\) that make the equation true. When we talk about a simultaneous system of linear equations, we are looking for a pair of values \((x, y)\) that satisfy all of the equations at the same time. This is probably best understood graphically.

The solution of a single linear equation can be represented as a line. Every point on this line can be plugged into that equation and yield a true result. If we add in a second linear equation, we get a different line. These are represented as the two separate graphs below.

When we talk about solutions of simultaneous equations, we want to find points that satisfy both equations at the same time. In other words, we're looking for points that are on both lines, and so we can merge the two images into one.

Activity 15.1.1. Verifying Solutions of Simultaneous Linear Systems.

We are going to focus on systems of two linear equations with two variables. Consider the following system:

\begin{equation*} \left\{ \begin{array}{rcrcr} x \amp + \amp y \amp = \amp 2 \\ x \amp - \amp y \amp = \amp 4 \end{array} \right. \end{equation*}

As noted in the definition, a solution must solve all the equations. This means that even though \((1, 1)\) is a solution of \(x + y = 2\) and \((2, -2)\) is a solution of \(x - y = 4\text{,}\) neither nor would be a solution to the system because when we plug in the values, we don't solve both.

Try it!

Show that \((3, -1)\) is a solution to the system of equations by direct substitution.

Solution.

When thinking about different possibilities for solutions, we need to think about the configurations we can get when we graph the two lines together. It turns out that there are three possibilities:

In the first case, we can see that there are no solutions because there do not exist any points that lie on both lines at the same time. In the second case, we can see that every point on one line is also a point on the other line, so there are infinitely many solutions. In the last case, we can see that there is exactly one point that lies on both lines.

Activity 15.1.2. Systems in Slope-Intercept Form.

If the equations are given in slope-intercept form, it's possible to immediately identify which situation you are in, and it's also possible to use substitution to solve the equations. Suppose that you have the following system of equations, where and are all constants.

\begin{equation*} \left\{ \begin{array}{rcrcr} y \amp = \amp m_1 x \amp + \amp b_1 \\ y \amp = \amp m_2 x \amp + \amp b_2 \end{array} \right. \end{equation*}

We can match these up with the images above.

• Two different parallel lines: Same slope, different intercept

• Two overlapping lines: Same slope, same intercept

• Two non-parallel lines: Different slopes

If the lines are not parallel, you can find the solution by substituting one equation into the other.

Try it!

Describe the configuration of the following system of equations. If they intersect at a single point, determine the coordinates of that point.

\begin{equation*} \left\{ \begin{array}{rcrcr} y \amp = \amp 4x \amp + \amp 3 \\ y \amp = \amp 2x \amp - \amp 1 \end{array} \right. \end{equation*}

Solution.

This is an example of solving an equation by substitution. The idea of a substitution here is the same as we used earlier, where we simply replace a symbol in an equation with some other collection of symbols, which then reduces the equation to a single variable equation that we can solve.

Activity 15.1.3. Solving a System of Linear Equations.

In general, the equations will not be given to you in slope-intercept form. This means that you will have to choose which variable to solve for and from which equation. There are no rules for this. You can sometimes avoid fractions, but you often can't. Here is the same system of equations solved two ways. Each column represents a different approach, but notice that they end up in the same place.

Try it!

Solve the system of equations above two more times by solving for \(x\) and \(y\) from the second equation.

Solution.

If the system of equations involves two lines with the same slope, but the equations are not written in slope-intercept form, you may not immediately recognize that they have the same slope. However, in the process of working through the algebra, you will end up with all of the variable terms canceling out. If that happens, the equation that you're left with will tell you whether the lines overlap or not. If you get a valid mathematical equation, such as \(0 = 0\text{,}\) then the two lines overlap. If you end up with an invalid mathematical equation, such as \(2 = 5\text{,}\) then the lines do not overlap.

Subsection 15.1.1 Brief Fraction Review

We will take a deeper dive into fractions in a later section. For now, we're just going to review the basic mechanics of fractions by looking at some examples.

For addition and subtraction of fractions, you need to use a common denominator. When rewriting a fraction with a different denominator, you must multiply both the numerator and denominator by the same value.

For multiplication of fractions, you need to multiply straight across. If you are multiplying by an integer, you can put that integer over a denominator of 1. If you remember how to reduce before multiplying, that's fine. If you don't, you can just reduce after multiplying. The problems in this section will not involve large enough numbers for that to be important. Make sure to reduce your fractions when you finish.