Explainer: Solving Systems of Linear Equations Graphically

In this explainer, we will learn how to solve a system of two linear equations by considering their graphs and identifying the point of intersection.

When we are asked to solve a system of linear equations (sometimes called simultaneous equations), we are trying to find any points at which the equations are equal, that is, any points where their graphs intersect.

If we are asked to solve a system of equations and are given their graphs, all we need to do is identify points of intersection. If the lines intersect at points with integer coordinates, these are generally easy to identify, but when they do not, it can be very difficult to accurately find the solutions of the system from their graphs. When this is the case, it is often only reasonable to state a range in which the solution must lie.

Additionally, if the lines are parallel, there are no points of intersection and, thus, no solutions to the system of equations. Equally, if the lines are coincident, that is, they are in the same position, then the system has infinitely many solutions.

Let us have a look at a couple of examples for this.

Example 1: Solving a System of Linear Equations from a Graph

Use the shown graph to solve the simultaneous equations 2π‘₯βˆ’3𝑦=6,7π‘₯+3𝑦=21.

Answer

We can see from the graph here that the two lines intersect at the point (3,0) which is the solution to our system of equations, that is, the point with coordinates π‘₯=3 and 𝑦=0.

It is always worth checking this, however, in case the graphs are drawn inaccurately or the intersection is not clear. Substituting into the left-hand side of each equation, we get 2(3)βˆ’3(0)=6,7(3)+3(0)=21, which are all consistent with the right-hand sides of the equations, meaning our solution is correct.

Let us now look at a couple of examples of solving a system of equations graphically when we have to plot the graph ourselves. The most common way to plot the graphs is to construct a table of values, but we can also use the slope–intercept method if preferred.

Example 2: Finding the Solution of a System of Equations by Graphing

By plotting the graphs of 𝑦=π‘₯βˆ’1 and 𝑦=5π‘₯+7, find the point that satisfies both equations simultaneously.

Answer

Our first step is to plot the graphs of the two equations. We will do this by constructing a table of values for each equation and then plotting the points to form the lines. The following table of values is for the equation 𝑦=π‘₯βˆ’1.

π‘₯βˆ’3βˆ’2βˆ’10123
π‘¦βˆ’4βˆ’3βˆ’2βˆ’1012

The next table of values is for the equation 𝑦=5π‘₯+7.

π‘₯βˆ’3βˆ’2βˆ’10123
π‘¦βˆ’8βˆ’327121722

We can then use these two tables to plot the shown lines.

Notice, for the second equation, that we have only plotted five of the points due to the scale of the graph (but you only actually need two points to define a line). Now that we have plotted the lines, we can see that the solution to our system is, in fact, the point (βˆ’2,βˆ’3). We can actually see this directly from the tables of values as both contain the point (βˆ’2,3).

Example 3: Using Graphs to Solve a System of Linear Equations

Plot the graphs of the following simultaneous equations: 𝑦=2π‘₯+7,𝑦=2π‘₯βˆ’4, and then solve the system.

Answer

Our first step is to plot the graphs of the two equations. We will do this by constructing a table of values for each equation and then plotting the points to form the lines. We get the following table of values for the first equation.

π‘₯βˆ’3βˆ’2βˆ’10123
𝑦135791113

And we get the next table for the second equation.

π‘₯βˆ’3βˆ’2βˆ’10123
π‘¦βˆ’10βˆ’8βˆ’6βˆ’4βˆ’202

We can then use these two tables to plot the shown lines.

Notice, for the first equation, that we have only plotted five of the points due to the scale of the graph (but you only actually need two points to define a line). Now that we have plotted the lines, we can clearly see that the two lines are parallel and, therefore, the system has no solutions. In fact, we could have saved ourselves the effort of drawing the graphs: if we look at the equations of the two lines, they both have the same slope and different 𝑦-intercepts which means the system has no solutions.

A further skill that is useful to master is the ability to identify which system of equations can be solved using which graphs. The easiest way to do this is to use the slopes and intercepts to identify the different equations. Let us have a look at an example.

Example 4: Matching a System of Equations to Their Graphs

Which of the following graphs could be used to help solve the given set of simultaneous equations: 𝑦=3π‘₯βˆ’4,𝑦=βˆ’12π‘₯+3?

Answer

Here, we are given graphs of five different systems of two equations. In order to identify which of the graphs is the system that we are looking for, we should check which lines have the correct 𝑦-intercepts. Our system has one line with a 𝑦-intercept of βˆ’4 and another line with a 𝑦-intercept of 3. If we look at the five graphs of the systems, only C has lines with these two intercepts. We then need to check that the slopes of these two lines are correct. If we start by looking at the green line (with 𝑦-intercept βˆ’4), we can see that it contains the points (0,βˆ’4) and (1,βˆ’1). Using the formula for the slope, π‘š=𝑦π‘₯,differenceindifferencein or more formally π‘š=π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯, we find that π‘š=βˆ’1βˆ’βˆ’41βˆ’0=31=3.

Similarly, for the blue line, if we use the points (2,2) and (6,0), we find that π‘š=2βˆ’02βˆ’6=2βˆ’4=βˆ’12.

This confirms that the lines in C are indeed the same as in our system of equations as they have the same intercepts and slopes.

We mentioned in the beginning that it is sometimes difficult to accurately identify an intersection point from two graphs and it is only reasonable to provide an estimate for the solution or a range in which the solution must lie. Let us look at an example where this is the case.

Example 5: Estimating the Solutions to a System of Equations by Graphing

Use the shown graph to find appropriate ranges for the solution to the given simultaneous equations: 𝑦=3π‘₯βˆ’3,5π‘₯+7𝑦=βˆ’2.

Answer

If we look at the graph, we can see that the intersection point is not at a point with integer coordinates. We can see that the π‘₯-coordinate is between 0 and 1 and the 𝑦-coordinate is between 0 and βˆ’1. We can, however, significantly improve these intervals using the grid lines. There are four increments between each two integers which means each line represents a step of 0.25. We can see that the intersection happens between the second and third grid lines on the π‘₯-axis and the third and fourth grid lines on the 𝑦-axis. This means that the range for the value of π‘₯ is 0.5<π‘₯<0.75, and the interval for the 𝑦-value is βˆ’1<𝑦<βˆ’0.75.

Using the grid lines, these are, as accurately as we can state, the ranges in which the coordinates lie. This being said, we can see that the intersection is closer to 0.75 for π‘₯ and closer to βˆ’0.75 for 𝑦 but we cannot clearly state the range as we do not have further grid lines to reference.

Key Points

To solve a system of equations by graphing, remember the following steps:

  1. If you are given the graphs of the system of equations, to solve the system you need to identify any points of intersection.
  2. If you are not given the graphs, you can plot them by creating a table of values for each equation and then identify any points of intersection.
  3. Sometimes, you will not be able to identify the exact coordinates of intersection, but you will be able to give a range in which the solutions must lie.

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