<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# Graphs of Linear Systems

## Graph lines to identify intersection points

Estimated12 minsto complete
%
Progress
Practice Graphs of Linear Systems

MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated12 minsto complete
%
Graphs of Linear Systems

### Graphs of Linear Systems

#### Determine Whether an Ordered Pair is a Solution to a System of Equations

A linear system of equations is a set of equations that must be solved together to find the one solution that fits them both.

Consider this system of equations:

{y=x+2y=2x+1\begin{align*}\begin{cases}y = x + 2\\ y = -2x + 1 \end{cases}\end{align*}

Since the two lines are in a system, we deal with them together by graphing them on the same coordinate axes. We can use any method to graph them; let’s do it by making a table of values for each line.

Line 1: y=x+2\begin{align*}y=x+2\end{align*}

x01y23\begin{align*}\begin{array}{|c|c|} \hline x & y \\\hline 0 & 2 \\\hline 1 & 3 \\\hline \end{array}\end{align*}

Line 2: y=-2x=1\begin{align*}y=\text{-}2x=1\end{align*}

x01y1-1\begin{align*}\begin{array}{|c|c|} \hline x & y \\\hline 0 & 1 \\\hline 1 & \text{-}1 \\\hline \end{array}\end{align*}

License: CC BY-NC 3.0

We already know that any point that lies on a line is a solution to the equation for that line. That means that any point that lies on both lines in a system is a solution to the whole system.

So in this system:

• Point A\begin{align*}A\end{align*} is not a solution to the system because it does not lie on either of the lines.
• Point B\begin{align*}B\end{align*} is not a solution to the system because it lies only on the blue line but not on the red line.
• Point C\begin{align*}C\end{align*} is a solution to the system because it lies on both lines at the same time.

In fact, point C\begin{align*}C\end{align*} is the only solution to the system, because it is the only point that lies on both lines. For a system of equations, the geometrical solution is the intersection of the two lines in the system. The algebraic solution is the ordered pair that solves both equations - in other words, it is the coordinates of the intersection point.

You can confirm the solution by plugging it into the system of equations, then checking that the solution works in each equation.

#### Finding the Solution to a System of Equations

Determine which of the points (1, 3), (0, 2), or (2, 7) is a solution to the following system of equations:

{y=4x1y=2x+3\begin{align*}\begin{cases}y = 4x - 1\\ y = 2x + 3 \end{cases}\end{align*}

To check if a coordinate point is a solution to the system of equations, we plug each of the x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} values into the equations to see if they work.

Point (1, 3):

y33y33=4x1=?4(1)1=3 solution checks=2x+3=?2(1)+35 solution does not check\begin{align*}y &= 4x - 1\\ 3 & \overset{?}{=} 4(1) - 1\\ 3 &= 3 \ \text{solution checks}\\ \\ y &= 2x + 3\\ 3 &\overset{?}{=} 2(1) + 3\\ 3 &\neq 5 \ \text{solution does not check}\\\end{align*}

Point (1, 3) is on the line y=4x1\begin{align*}y = 4x - 1\end{align*}, but it is not on the line y=2x+3\begin{align*}y = 2x + 3\end{align*}, so it is not a solution to the system.

Point (0, 2):

y22=4x1=?4(0)11 solution does not check\begin{align*}y &= 4x - 1\\ 2 &\overset{?}{=} 4(0) - 1\\ 2 &\neq -1 \ \text{solution does not check}\end{align*}

Point (0, 2) is not on the line y=4x1\begin{align*}y = 4x - 1\end{align*}, so it is not a solution to the system. Note that it is not necessary to check the second equation because the point needs to be on both lines for it to be a solution to the system.

Point (2, 7):

y77y77=4x1=?4(2)1=7 solution checks=2x+3=?2(2)+3=7 solution checks\begin{align*}y &= 4x - 1\\ 7 &\overset{?}{=} 4(2) - 1\\ 7 &= 7 \ \text{solution checks}\\ \\ y &= 2x + 3\\ 7 &\overset{?}{=} 2(2) + 3\\ 7 &= 7 \ \text{solution checks}\end{align*}

Point (2, 7) is a solution to the system since it lies on both lines.

The solution to the system is the point (2, 7).

#### Determine the Solution to a Linear System by Graphing

The solution to a linear system of equations is the point, (if there is one) that lies on both lines. In other words, the solution is the point where the two lines intersect.

We can solve a system of equations by graphing the lines on the same coordinate plane and reading the intersection point from the graph.

This method most often offers only approximate solutions, so it’s not sufficient when you need an exact answer. However, graphing the system of equations can be a good way to get a sense of what’s really going on in the problem you’re trying to solve, especially when it’s a real-world problem.

#### Solving a System of Equations by Graphing

Solve the system of equations by graphing:

{y=3x5y=-2x+5\begin{align*}\begin{cases} y = 3x - 5\\ y = \text{-}2x + 5 \end{cases}\end{align*}

Graph both lines on the same coordinate axis using any method you like.

In this case, let’s make a table of values for each line.

Line 1: y=3x5\begin{align*}y = 3x - 5\end{align*}

x12y-21\begin{align*}\begin{array}{|c|c|} \hline x & y \\\hline 1 & \text{-}2 \\\hline 2 & 1 \\\hline \end{array}\end{align*}

Line 2: y=2x+5\begin{align*}y = -2x + 5\end{align*}

x12y31\begin{align*}\begin{array}{|c|c|} \hline x & y \\\hline 1 & 3 \\\hline 2 & 1 \\\hline \end{array}\end{align*}

License: CC BY-NC 3.0

The solution to the system is given by the intersection point of the two lines. The graph shows that the lines intersect at point (2, 1). So the solution is \begin{align*}\mathbf{x = 2, \ y = 1}\end{align*} or (2, 1).

#### Solving a System of Equations Using a Graphing Calculator

As an alternative to graphing by hand, you can use a graphing calculator to find or check solutions to a system of equations.

Solve the system of equations using a graphing calculator.

\begin{align*}\begin{cases} x - 3y = 4\\ 2x + 5y = 8 \end{cases}\end{align*}

To input the equations into the calculator, you need to rewrite them in slope-intercept form (that is, \begin{align*}y = mx + b\end{align*} form).

\begin{align*}x - 3y = 4 \quad &\Rightarrow \quad y = \frac{1}{3}x - \frac{4}{3}\!\\ \\ 2x + 5y = 8 \quad &\Rightarrow \quad y = - \frac{2}{5}x + \frac{8}{5}\end{align*}

Press the [y=] button on the graphing calculator and enter the two functions as:

\begin{align*}Y_1 & = \frac{x}{3} - \frac{4}{3}\\ Y_2 & = \frac{-2x}{5} + \frac{8}{5}\end{align*}

Now press [GRAPH]. Here’s what the graph should look like on a TI-83 family graphing calculator with the window set to \begin{align*}-5 \le x \le 10\end{align*} and \begin{align*}-5 \le y \le 5\end{align*}.

License: CC BY-NC 3.0

There are a few different ways to find the intersection point.

Option 1: Use [TRACE] and move the cursor with the arrows until it is on top of the intersection point. The values of the coordinate point will be shown on the bottom of the screen. The second screen above shows the values to be \begin{align*}X = 4.0957447\end{align*} and \begin{align*}Y = 0.03191489\end{align*}.

Use the [ZOOM] function to zoom into the intersection point and find a more accurate result. The third screen above shows the system of equations after zooming in several times. A more accurate solution appears to be \begin{align*}X = 4\end{align*} and \begin{align*}Y = 0\end{align*}.

Option 2: Look at the table of values by pressing [2nd] [GRAPH]. The first screen below shows a table of values for this system of equations. Scroll down until the \begin{align*}Y-\end{align*}values for the two functions are the same. In this case this occurs at \begin{align*}X = 4\end{align*} and \begin{align*}Y = 0\end{align*}.

(Use the [TBLSET] function to change the starting value for your table of values so that it is close to the intersection point and you don’t have to scroll too long. You can also improve the accuracy of the solution by setting the value of the \begin{align*}\Delta\end{align*} Table smaller.)

License: CC BY-NC 3.0

Option 3: Using the [2nd] [TRACE] function gives the second screen shown above.

Scroll down and select “intersect.”

The calculator will display the graph with the question [FIRSTCURVE]? Move the cursor along the first curve until it is close to the intersection and press [ENTER].

The calculator now shows [SECONDCURVE]?

Move the cursor to the second line (if necessary) and press [ENTER].

The calculator displays [GUESS]?

Press [ENTER] and the calculator displays the solution at the bottom of the screen (see the third screen above).

The point of intersection is \begin{align*}X = 4\end{align*} and \begin{align*}Y = 0\end{align*}. Note that with this method, the calculator works out the intersection point for you, which is generally more accurate than your own visual estimate.

#### Solve Real-World Problems Using Graphs of Linear Systems

Consider the following problem:

Peter and Nadia like to race each other. Peter can run at a speed of 5 feet per second and Nadia can run at a speed of 6 feet per second. To be a good sport, Nadia likes to give Peter a head start of 20 feet. How long does Nadia take to catch up with Peter? At what distance from the start does Nadia catch up with Peter?

Start by drawing a sketch. Here’s what the race looks like when Nadia starts running. Call this time \begin{align*}t = 0\end{align*}.

License: CC BY-NC 3.0

Now, define two variables in this problem:

\begin{align*}t =\end{align*} the time from when Nadia starts running

\begin{align*}d =\end{align*} the distance of the runners from the starting point.

Since there are two runners, there need to be equations for each of them. That will be the system of equations for this problem.

For each equation, use the formula: \begin{align*}\text{distance} = \text{speed}\times \text{time}\end{align*}

Nadia’s equation: \begin{align*}d = 6t\end{align*}

Peter’s equation: \begin{align*}d = 5t + 20\end{align*}

(Remember that Peter was already 20 feet from the starting point when Nadia started running.)

Let’s graph these two equations on the same coordinate axes.

Time should be on the horizontal axis since it is the independent variable. Distance should be on the vertical axis since it is the dependent variable.

License: CC BY-NC 3.0

We can use any method for graphing the lines, but in this case we’ll use the slope–intercept method since it makes more sense physically.

To graph the line that describes Nadia’s run, start by graphing her equation's \begin{align*}y-\end{align*}intercept: (0, 0). (If you don’t understand why this is the \begin{align*}y-\end{align*}intercept, try plugging in the test-value of \begin{align*}x = 0\end{align*}.)

The slope tells us that Nadia runs 6 feet every one second, so another point on the line is (1, 6). Connecting these points gives us Nadia’s line:

License: CC BY-NC 3.0

To graph the line that describes Peter’s run, start with his \begin{align*}y-\end{align*}intercept. In this case, this is the point (0, 20).

The slope tells us that Peter runs 5 feet every one second, so another point on the line is (1, 25). Connecting these points gives Peter’s line:

License: CC BY-NC 3.0

In order to find when and where Nadia and Peter meet, graph both lines on the same graph and extend the lines until they cross. The crossing point is the solution to this problem.

License: CC BY-NC 3.0

The graph shows that Nadia and Peter meet 20 seconds after Nadia starts running, and 120 feet from the starting point.

These examples are great at demonstrating that the solution to a system of linear equations is the point at which the lines intersect. This is, in fact, the greatest strength of the graphing method - it offers a very visual representation of a system of equations and its solution. You can also see that finding the solution from a graph requires very careful graphing of the lines, and is really only practical when you’re sure that the solution gives integer values for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}. Often, this method can only offer approximate solutions to systems of equations, so you need to use other methods to get an exact solution.

### Example

#### Example 1

Solve the following system of equations by graphing:

\begin{align*}\begin{cases}2x + 3y = 6\\ \ \ 4x - y = \text{-}2 \end{cases}\end{align*}

Since the equations are in standard form, graph them by finding the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts of each of the lines.

Line 1: \begin{align*}2x + 3y = 6\end{align*}

\begin{align*}x-\end{align*}intercept: set \begin{align*}y = 0 \text{, giving } 2x = 6 \Rightarrow x = 3\end{align*} so the intercept is (3, 0)

\begin{align*}y-\end{align*}intercept: set \begin{align*}x = 0 \text{, giving } 3y = 6 \Rightarrow y = 2\end{align*} so the intercept is (0, 2)

Line 2: \begin{align*}\text{-}4x + y = 2\end{align*}

\begin{align*}x-\end{align*}intercept: set \begin{align*}y = 0 \text{, giving } \text{-}4x = 2 \Rightarrow x = \text{-} \frac{1}{2}\end{align*} so the intercept is \begin{align*}\left ( \text{-} \frac{1}{2}, 0 \right )\end{align*}

\begin{align*}y-\end{align*}intercept: set \begin{align*}x = 0 \text{, giving } y = 2\end{align*} so the intercept is (0, 2)

License: CC BY-NC 3.0

The graph shows that the lines intersect at (0, 2). Therefore, the solution to the system of equations is \begin{align*}\mathbf{x = 0, \ y = 2.}\end{align*}

### Review

Determine which ordered pair satisfies each system of linear equations.

1. \begin{align*}\begin{cases}y = 3x - 2\!\\ y = \text{-}x\end{cases}\end{align*}
1. (1, 4)
2. (2, 9)
3. \begin{align*}\left (\frac{1}{2}, \ \text{-}\frac{1}{2} \right )\\ \ \\\end{align*}
2. \begin{align*}\begin{cases}y = 2x - 3\\ y = x + 5 \end{cases}\end{align*}
1. (8, 13)
2. (-7, 6)
3. (0, 4)
3. \begin{align*}\\ \begin{cases}\ \ 2x + y = 8\\ 5x + 2y = 10\end{cases}\end{align*}
1. (-9, 1)
2. (-6, 20)
3. (14, 2)
4. \begin{align*}\\ \begin{cases}3x + 2y = 6\\ y = \frac{1}{2}x - 3 \end{cases}\end{align*}
1. \begin{align*}\left (3, \frac{-3}{2} \right )\end{align*}
2. (-4, 3)
3. \begin{align*}\left (\frac{1}{2}, \ 4 \right )\\ \ \\\end{align*}
5. \begin{align*}\begin{cases}2x - y = 10\\ 3x + y = \text{-}5 \end{cases}\end{align*}
1. (4, -2)
2. (1, -8)
3. (-2, 5)

Solve the following systems using the graphing method.

1. \begin{align*}\begin{cases}y = x + 3\\ y = \text{-}x + 3 \end{cases}\\ \ \\\end{align*}
2. \begin{align*}\begin{cases}y = 3x - 6\\ y = \text{-}x + 6 \end{cases}\\ \ \\\end{align*}
3. \begin{align*}\begin{cases}2x = 4\\ \ \ y = \text{-}3 \end{cases}\\ \ \\\end{align*}
4. \begin{align*}\begin{cases}y = \text{-}x + 5\\ \text{-}x + y = 1 \end{cases}\\ \ \\\end{align*}
5. \begin{align*}\begin{cases}\ \ x + 2y = 8\\ 5x + 2y = 0 \end{cases}\\ \ \\\end{align*}
6. \begin{align*}\begin{cases} 3x + 2y = 12\\ \ \ 4x - y = 5 \end{cases}\\ \ \\\end{align*}
7. \begin{align*}\begin{cases}5x + 2y = \text{-}4\\ x - y = 2 \end{cases}\\ \ \\\end{align*}
8. \begin{align*}\begin{cases} 2x + 4 = 3y\\ x - 2y + 4 = 0 \end{cases}\\ \ \\\end{align*}
9. \begin{align*}\begin{cases}y = \frac{1}{2}x - 3\\ 2x - 5y = 5 \end{cases}\\ \ \\\end{align*}
10. \begin{align*}\begin{cases}y = 4\\ x = 8 - 3y \end{cases}\\ \ \\\end{align*}
11. Try to solve the following system using the graphing method: \begin{align*}\begin{cases}y = \frac{3}{5}x + 5\\ y = \text{-}2x - \frac{1}{2} \end{cases}\end{align*}
1. What does it look like the \begin{align*}x-\end{align*}coordinate of the solution should be?
2. Does that coordinate really give the same \begin{align*}y-\end{align*}value when you plug it into both equations?
3. Why is it difficult to find the real solution to this system?
12. Try to solve the following system using the graphing method. Use a grid with \begin{align*}x-\end{align*}values and \begin{align*}y-\end{align*}values ranging from -10 to 10.\begin{align*}\begin{cases}y = 4x + 8\\ y = 5x + 1 \end{cases}\end{align*}
1. Do these lines appear to intersect?
2. Based on their equations, are they parallel?
3. What needs to be done to find their intersection point?
13. Try to solve the following system using the graphing method. Use a grid with \begin{align*}x-\end{align*}values and \begin{align*}y-\end{align*}values ranging from -10 to 10.\begin{align*}\begin{cases}y = \frac{1}{2}x + 4\\ y = \frac{4}{9}x + \frac{9}{2}\end{cases}\end{align*}
1. Can you tell exactly where the lines cross?
2. What would we have to do to make it clearer?

Solve the following problems by using the graphing method.

1. Mary’s car has broken down and it will cost her $1200 to get it fixed, or, for$4500, she can buy a new and more efficient car instead. Her present car uses about $2000 worth of gas per year, while gas for the new car would cost about$1500 per year. After how many years would the total cost of fixing the car equal the total cost of replacing it?
2. A tortoise and hare decide to race 30 feet. The hare, being much faster, decides to give the tortoise a 20 foot head start. The tortoise runs at 0.5 feet/sec and the hare runs at 5.5 feet per second. How long until the hare catches the tortoise?

### Review (Answers)

To view the Review answers, open this PDF file and look for section 7.1.

Solution

R

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Please to create your own Highlights / Notes
Show More

### Vocabulary Language: English

Consistent

A system of equations is consistent if it has at least one solution.

Dependent

A system of equations is dependent if every solution for one equation is a solution for the other(s).

Independent

A system of equations is independent if it has exactly one solution.

linear equation

A linear equation is an equation between two variables that produces a straight line when graphed.

Linear Function

A linear function is a relation between two variables that produces a straight line when graphed.

### Image Attributions

1. [1]^ License: CC BY-NC 3.0
2. [2]^ License: CC BY-NC 3.0
3. [3]^ License: CC BY-NC 3.0
4. [4]^ License: CC BY-NC 3.0
5. [5]^ License: CC BY-NC 3.0
6. [6]^ License: CC BY-NC 3.0
7. [7]^ License: CC BY-NC 3.0
8. [8]^ License: CC BY-NC 3.0
9. [9]^ License: CC BY-NC 3.0
10. [10]^ License: CC BY-NC 3.0

### Explore More

Sign in to explore more, including practice questions and solutions for Graphs of Linear Systems.
Please wait...
Please wait...