We have learned how to graph a line given its equation. In this section, we will learn what a system of two linear equations is, and how to use graphing to solve such a system.
Figure5.1.1Alternative Video Lesson
Subsection5.1.1Solving Systems of Equations by Graphing
Example5.1.2
Amy and d'Marie are running at constant speeds in parallel lanes on a track. Amy starts out ahead of d'Marie, but d'Marie is running faster. We want to determine when d'Marie will catch up with Amy. Let's start by looking at the graph of each runner's distance over time:
Figure5.1.3Amy and d'Marie's distances.
Each of the two lines in Figure 5.1.3 has an equation, as discussed in Chapter 4. The line representing Amy appears to have \(y\)-intercept \((0,4)\) and slope \(\frac{4}{3}\text{,}\) so its equation is \(y=\frac{4}{3}t+4\text{.}\) The line representing d'Marie appears to have \(y\)-intercept \((0,0)\) and slope \(2\text{,}\) so its equation is \(y=2t\text{.}\) When these two equations are together as a package, we have what is called a system of linear equations:
\begin{equation*}
\left\{
\begin{alignedat}{4}
y \amp {}={} \frac{4}{3}t \amp {}+{} \amp 4 \\
y \amp {}={} 2t \\
\end{alignedat}
\right.\text{.}
\end{equation*}
The large left brace indicates that this is a collection of two distinct equations, not one equation that was somehow algebraically manipulated into an equivalent equation.
As we can see in Figure 5.1.3, the graphs of the two equations cross at the point \((6,12)\text{.}\) We refer to the point \((6,12)\) as the solution to this system of linear equations. To denote the solution set, we write \(\{(6,12)\}\text{.}\) But it's much more valuable to interpret these numbers in context whenever possible: it took \(6\) seconds for the two runners to meet up, and when they met they were \(12\) meters up the track.
Remark5.1.4
In Example 5.1.2, we stated that the solution was the point \((6,12)\text{.}\) It makes sense to write this as an ordered pair when we're given a graph. In some cases when we have no graph, particularly when our variables are not \(x\) and \(y\text{,}\) it might not be clear which variable “comes first” and we won't be able to write an ordered pair. Nevertheless, given the context we can write meaningful summary statements.
Example5.1.5
Determine the solution to the system of equations graphed in Figure 5.1.6.
The two lines intersect where \(x=-3\) and \(y=-1\text{,}\) so the solution is the point \((-3,-1)\text{.}\) We write the solution set as \(\{(-3,-1)\}\text{.}\)
Checkpoint5.1.7
Now let's look at an example where we need to make a graph to find the solution.
Example5.1.8
Solve the following system of equations by graphing:
Notice that each of these equations is written in slope-intercept form. The first equation, \(y=\frac{1}{2}x+4\text{,}\) is a linear equation with a slope of \(\frac{1}{2}\) and a \(y\)-intercept of \((0,4)\text{.}\) The second equation, \(y=-x-5\text{,}\) is a linear equation with a slope of \(-1\) and a \(y\)-intercept of \((0,-5)\text{.}\) We'll use this information to graph both lines:
Figure5.1.9Graphs of \(y=\frac{1}{2}x+4\) and \(y=-x-5\text{.}\)
The two lines intersect where \(x=-6\) and \(y=1\text{,}\) so the solution of the system of equations is the point \((-6,1)\text{.}\) We write the solution set as \(\{(-6,1)\}\text{.}\)
Example5.1.10
Solve the following system of equations by graphing:
Since both line equations are given in standard form, we'll graph each one by finding the intercepts. Recall that to find the \(x\)-intercept of each equation, replace \(y\) with \(0\) and solve for \(x\text{.}\) Similarly, to find the \(y\)-intercept of each equation, replace \(x\) with \(0\) and solve for \(y\text{.}\)
So the \(x\)-intercept is \(\left(\frac{3}{2},0\right)\) and the \(y\)-intercept is \((0,1)\text{.}\)
Now we can graph each line by plotting the intercepts and connecting these points:
Figure5.1.11Graphs of \(x-3y=-12\) and \(2x+3y=3\)
It appears that the solution of the system of equations is the point of intersection of those two lines, which is \((-3,3)\text{.}\) It's important to check this is correct, because when making a hand-drawn graph, it would be easy to be off by a little bit. To check, we can substitute the values of \(x\) and \(y\) from the point \((-3,3)\) into each equation:
So we have checked that \((-3,3)\) is indeed the solution for the system. We write the solution set as \(\{(-3,3)\}\text{.}\)
Example5.1.12
A college has a north campus and a south campus. The north campus has \(18{,}000\) students, and the south campus has \(4{,}000\) students. In the past five years, the north campus lost \(4{,}000\) students, and the south campus gained \(3{,}000\) students. If these trends continue, in how many years would the two campuses have the same number of students? Write and solve a system of equations modeling this problem.
Since all the given student counts are in the thousands, we make the decision to measure student population in thousands. So for instance, the north campus starts with a student population of \(18\) (thousand students).
The north campus lost \(4\) thousand students in \(5\) years. So it is losing students at a rate of \(\frac{4\text{ thousand}}{5\text{ year}}\text{,}\) or \(\frac{4}{5}\,\frac{\text{thousand}}{\text{year}}\text{.}\) This rate of change should be interpreted as a negative number, because the north campus is losing students over time. So we have a linear model with starting value \(18\) thousand students, and a slope of \(-\frac{4}{5}\) thousand students per year. In other words,
We will graph both lines using their slopes and \(y\)-intercepts.
Figure5.1.13Number of Students at the South Campus and North Campus
According to the graph, the lines intersect at \((10,10)\text{.}\) So if the trends continue, both campuses will have \(10{,}000\) students \(10\) years from now.
Subsection5.1.2Special Systems of Equations
Recall that when we solved linear equations in one variable, we had two special cases. In one special case there was no solution and in the other case, there were infinitely many solutions. When solving systems of equations in two variables, we have two similar special cases.
Example5.1.14Parallel Lines
Let's look at the graphs of two lines with the same slope, \(y=2x-4\) and \(y=2x+1\text{:}\)
Figure5.1.15Graphs of \(y=2x-4\) and \(y=2x+1\)
For this system of equations, what is the solution? Since the two lines have the same slope they are parallel lines and will never intersect. This means that there is no solution to this system of equations. We write the solution set as \(\emptyset\text{.}\) (This is a special symbol to represent a set with nothing in it, not the number zero.)
Example5.1.16Coinciding Lines
Next we'll look at the other special case. Let's start with this system of equations:
To solve this system of equations, we want to graph each line. The first equation is in slope-intercept form and can be graphed easily using its slope of \(2\) and its \(y\)-intercept of \((0,-4)\text{.}\)
The second equation, \(6x-3y=12\text{,}\) can either be graphed by solving for \(y\) and using the slope-intercept form or by finding the intercepts. If we use the intercept method, we'll find that this line has an \(x\)-intercept of \((2,0)\) and a \(y\)-intercept of \((0,-4)\text{.}\) When we graph both lines we have:
Figure5.1.17Graphs of \(y=2x-4\) and \(6x-3y=12\)
Now we can see these are actually the same line, or coinciding lines. To determine the solution to this system, we'll note that they overlap everywhere. This means that we have an infinite number of solutions: all points that fall on the line. It may be enough to report that there are infinitely many solutions. In order to be more specific, all we can do is say that any ordered pair \((x,y)\) satisfying the line equation is a solution. In set-builder notation, we would write \(\{(x,y)\mid y=2x-4\}\text{.}\)
Remark5.1.18
In Example 5.1.16, what would have happened if we had decided to convert the second line equation into slope-intercept form?
This is the literally the same as the first equation in our system. This is a different way to show that these two equations are equivalent and represent the same line. Any time we try to solve a system where the equations are equivalent, we'll have an infinite number of solutions.
Warning5.1.19
Notice that for a system of equations with infinite solutions like Example 5.1.16, we didn't say that every point was a solution. Rather, every point that falls on that line is a solution. It would be incorrect to state this solution set as “all real numbers” or as “all ordered pairs.”
Let's summarize the three types of systems of equations and their solution sets:
Intersecting Lines:
If two linear equations have different slopes, the system has one solution.
Parallel Lines:
If the linear equations have the same slope with different \(y\)-intercepts, the system has no solution.
Coinciding Lines:
If two linear equations have the same slope and the same \(y\)-intercept (in other words, they are equivalent equations), the system has infinitely many solutions. This solution set consists of all ordered pairs on that line.
SubsectionExercises
Checking Solutions for System of Equations
1
Decide whether \((4,-4)\) is a solution to the system of equations:
