5.1: Graphical Solutions to Systems of Equations
When you graph two linear functions on the same Cartesian plane, the resulting lines may intersect. Do the following two lines intersect? If so, where?
\begin{align*}& \begin{Bmatrix}
2x+y = 5 \\
x-y = 1
\end{Bmatrix}\end{align*}
Watch This
Khan Academy Solving Systems by Graphing 3
Guidance
A \begin{align*}2 \times 2\end{align*}
\begin{align*}& \begin{Bmatrix}
2x+y = 5 \\
x-y = 1
\end{Bmatrix}\end{align*}
When graphed, a system of linear equations is two lines. To solve a system of linear equations, figure out if the two lines intersect and if so, at what point. One way to solve a system of equations is by graphing. Graph both lines and look for the point where they intersect.
Keep in mind that even though most of the time when you graph two lines they will intersect in just one point, there are two other possibilities:
- The lines might never intersect (they are parallel lines)
- The lines might coincide (be exactly the same line)
A system that results in one point of intersection is consistent and independent. A system that results in lines that coincide is consistent and dependent. A system that results in two parallel lines is inconsistent.
Example A
Solve the following system of linear equations graphically:
\begin{align*}\begin{Bmatrix}
x-2y -2= 0 \\
3x+4y = 16
\end{Bmatrix}\end{align*}
Solution:
Both equations will be graphed on the same Cartesian plane using the slope-intercept method. Begin by writing each linear equation in slope-intercept form.
\begin{align*}& x-2y-2 = 0\\
& x {\color{red}-x}-2y-2 = 0 {\color{red}-x}\\
& -2y-2 = -x\\
& -2y-2 {\color{red}+2} = -x {\color{red}+2}\\
& -2y = -x+2\\
& \frac{-2y}{{\color{red}-2}} = \frac{-x}{{\color{red}-2}}+\frac{2}{{\color{red}-2}}\\
& \boxed{y = \frac{1}{2}x-1} \qquad \text{Equation One}\end{align*}
\begin{align*}& 3x+4y = 16\\
& 3x {\color{red}-3x}+4y = 16 {\color{red}-3x}\\
& 4y = 16-3x\\
& \frac{4y}{{\color{red}4}} = \frac{16}{{\color{red}4}}-\frac{3x}{{\color{red}4}}\\
& y = 4-\frac{3}{4}x\\
& \boxed{y = -\frac{3}{4}x+4} \qquad \text{Equation Two}\end{align*}
Graph both equations on the same Cartesian plane.
The lines have intersected at the point (4, 1). When two equations place two conditions on two same variables at the same time, a system of simultaneous equations is formed. The solution is an ordered pair which satisfies both of the equations in the system. If an ordered pair satisfies an equation, replacement of the variables with the values will result in both sides of the equation being equal.
Test (4, 1) in equation one:
\begin{align*}x-2y-2 &= 0 && \text{Use the original equation}\\
({\color{red}4})-2({\color{red}1})-2 &= 0 && \text{Replace} \ x \ \text{with} \ 4 \ \text{and replace} \ y \ \text{with} \ 1.\\
4-2-2 &= 0 && \text{Perform the indicated operations and simplify the result.}\\
4-{\color{red}4} &= 0\\
{\color{red}0} &= 0 && \text{Both sides of the equation are equal. The ordered pair} \ (4, 1) \ \text{satisfies the equation.}\end{align*}
Test (4, 1) in equation two:
\begin{align*}3x+4y &= 16 && \text{Use the original equation}\\
3({\color{red}4})+4({\color{red}1}) &= 16 && \text{Replace} \ x \ \text{with} \ 4 \ \text{and replace} \ y \ \text{with} \ 1.\\
12+4 &= 16 && \text{Perform the indicated operations and simplify the result.}\\
{\color{red}16} &= 16 && \text{Both sides of the equation are equal. The ordered pair} \ (4, 1) \ \text{satisfies the equation.}\end{align*}
This system of simultaneous equations has a solution and is therefore called a consistent system. Because it has only one ordered pair as a solution, it is called an independent system.
Example B
Solve the following system of linear equations graphically:
\begin{align*}\begin{Bmatrix}
2y-3x = 6 \\
4y-6x = 12
\end{Bmatrix}\end{align*}
Solution:
Both equations will be graphed on the same Cartesian plane using the intercept method. Let \begin{align*}x = 0\end{align*}
\begin{align*}& 2y-3x = 6\\
& 2y-3 ({\color{red}0}) = 6 \quad \text{Replace} \ x \ \text{with zero.}\\
& 2y = 6 \qquad \qquad \text{Simplify}\\
& \frac{2y}{{\color{red}2}} = \frac{6}{{\color{red}2}} \qquad \quad \ \ \text{Solve for} \ y.\\
& \boxed{y = 3} \qquad \qquad \text{The} \ y \text{-intercept is} \ (0, 3)\end{align*}
Let \begin{align*}y = 0\end{align*}
\begin{align*}& 2y-3x = 6\\
& 2({\color{red}0})-3x = 6 \qquad \text{Replace} \ y \ \text{with zero.}\\
& -3x = 6 \qquad \quad \ \ \text{Simplify}\\
& \frac{-3x}{{\color{red}-3}} = \frac{6}{{\color{red}-3}} \qquad \ \ \ \text{Solve for} \ y.\\
& \boxed{x = -2} \qquad \qquad \text{The} \ x \text{-intercept is} \ (-2, 0)\end{align*}
\begin{align*}& 4y-6x = 12\\
& 4y-6 ({\color{red}0}) = 12 \qquad \text{Replace} \ x \ \text{with zero.}\\
& 4y = 12 \qquad \qquad \quad \text{Simplify}\\
& \frac{4y}{{\color{red}4}} = \frac{12}{{\color{red}4}} \qquad \qquad \ \ \text{Solve for} \ y.\\
& \boxed{y = 3} \qquad \qquad \quad \ \text{The} \ y \text{-intercept is} \ (0, 3)\end{align*}
Let \begin{align*}y = 0\end{align*}. Solve for \begin{align*}x\end{align*}.
\begin{align*}& 4y-6x = 12\\ & 4 ({\color{red}0})-6x = 12 \qquad \text{Replace} \ y \ \text{with zero.}\\ & -6x = 12 \qquad \quad \ \ \text{Simplify}\\ & \frac{-6x}{{\color{red}-6}} = \frac{12}{{\color{red}-6}} \qquad \quad \ \text{Solve for} \ y.\\ & \boxed{x = -2} \qquad \qquad \ \text{The} \ x \text{-intercept is} \ (-2, 0)\end{align*}
When the \begin{align*}x\end{align*} and \begin{align*}y\end{align*}-intercepts were calculated for each equation, they were the same for both lines. The graph resulted in the same line being graphed twice. The blue line is longer to show that the same line is graphed directly on top of the red line. The system does have solutions so it is also known as a consistent system. However, the system does not have one solution; it has an infinite number of solutions. This type of consistent system is called a dependent system. All the ordered pairs found on the line will satisfy both equations. If you look at the two given equations \begin{align*}\begin{Bmatrix} 2y-3x = 6 \\ 4y-6x = 12 \end{Bmatrix}\end{align*}, equation two is simply a multiple of equation.
Example C
Solve the following system of linear equations graphically:
\begin{align*}\begin{Bmatrix} 3x+4y=12 \\ 6x+8y=-8 \end{Bmatrix}\end{align*}
Solution:
Both equations will be graphed on the same Cartesian plane using the slope-intercept method. Begin by writing each linear equation in slope-intercept form.
\begin{align*}& 3x+4y = 12\\ & 3x {\color{red}-3x}+4y = 12 {\color{red}-3x}\\ & 4y = 12 {\color{red}-3x}\\ & \frac{4y}{{\color{red}4}} = \frac{12}{{\color{red}4}}-\frac{3x}{{\color{red}4}}\\ & y = 3-\frac{3}{4}x\\ & \boxed{y = -\frac{3}{4}x+3}\end{align*}
\begin{align*}& 6x+8y = -8\\ & 6x {\color{red}-6x}+8y = -8 {\color{red}-6x}\\ & 8y = -8 {\color{red}-6x}\\ & \frac{8y}{{\color{red}8}} = \frac{-8}{{\color{red}8}}-\frac{6x}{{\color{red}8}}\\ & y = -1-\frac{6}{{\color{red}8}}x\\ & \boxed{y = -\frac{6}{8}x-1}\end{align*}
The lines do not intersect. This means that the system of equations has no solution. The lines are parallel and will never intersect. If you look at the equations that were written in slope-intercept form \begin{align*}y=-\frac{3}{4}x+3\end{align*} and \begin{align*}y=-\frac{6}{8}x-1\end{align*}, the slopes are the same \begin{align*}\left(-\frac{6}{8}=-\frac{3}{4}\right)\end{align*}. A system of linear equations that has no solution is called an inconsistent system.
Example D
Before graphing calculators, graphing was not considered the best way to determine the solution for a system of linear equations, especially if the solutions were not integers. However, technology has changed this outlook. In this example, a graphing calculator will be used to determine the solution for \begin{align*}\begin{Bmatrix} x+4y=-14 \\ 2x-y=4 \end{Bmatrix}\end{align*}.
Solution:
To use a graphing calculator, the equations must be written in slope-intercept form:
\begin{align*}& x+4y = -14\\ & x {\color{red}-x}+4y = {\color{red}-x}-14\\ & 4y = {\color{red}-x}-14\\ & \frac{4y}{{\color{red}4}} = -\frac{x}{{\color{red}4}}-\frac{14}{{\color{red}4}}\\ & y = -\frac{1}{4}x-\frac{14}{4}\\ & \boxed{y = -\frac{1}{4}x-\frac{7}{2}}\end{align*}
\begin{align*}& 2x-y = 4\\ & 2x {\color{red}-2x}-y = {\color{red}-2x}+4\\ & -y = -2x+4\\ & \frac{-y}{{\color{red}-1}} = \frac{-2x}{{\color{red}-1}}+\frac{4}{{\color{red}-1}}\\ & \boxed{y = 2x-4}\end{align*}
The equations are both in slope-intercept form. Set the window on the calculator as shown below:
The intersection point of the linear equations is (0.22, -3.56). The following represents the keys that were pressed on the calculator to obtain the above results:
Concept Problem Revisited
The following linear equations will be graphed by using the slope-intercept method.
\begin{align*}& \begin{Bmatrix} 2x+y = 5 \\ x-y = 1 \end{Bmatrix}\\ & 2x+y = 5\\ & 2x {\color{red}-2x}+y = 5 {\color{red}-2x}\\ & y = 5 {\color{red}-2x}\\ & \boxed{y = -2x+5}\\ & x-y = 1\\ & x {\color{red}-x}-y = 1 {\color{red}-x}\\ & -y = 1-x\\ & \frac{-y}{{\color{red}-1}} = \frac{1}{{\color{red}-1}}-\frac{x}{{\color{red}-1}}\\ & y = -1+x\\ & \boxed{y = x-1}\end{align*}
The two lines intersect at one point. The coordinates of the point of intersection are (2, 1).
Vocabulary
- Consistent System of Linear Equations
- A consistent system of linear equations is a system of linear equations that has a solution. The solution may be one solution or an infinite number of solutions.
- Dependent System of Linear Equations
- An dependent system of linear equations is a system of linear equations that has an infinite number of solutions. The equations are multiples and the graphs of each equation are the same. Therefore, the infinite number of ordered pairs satisfies both equations.
- Equivalent Systems of Linear Equations
- Equivalent systems of linear equations are systems of linear equations that have the same solution set.
- Inconsistent System of Linear Equations
- An inconsistent system of linear equations is a system of linear equations that has no solution. The graphs of an inconsistent system of linear equations are parallel lines. The lines never intersect so there is no common point of intersection.
- Independent System of Linear Equations
- An independent system of linear equations is a system of linear equations that has one solution. There is only one ordered pair that satisfies both equations.
- System of Linear Equations
- A system of linear equations is two linear equations each having two variables. This type of system – two equations with two unknowns-is called a \begin{align*}2 \times 2\end{align*} system of linear equations.
Guided Practice
1. Solve the following system of linear equations by graphing: \begin{align*}\begin{Bmatrix} -3x+4y=20 \\ x-2y=-8 \end{Bmatrix}\end{align*}
- Is the system consistent and dependent, consistent and independent, or inconsistent?
Use technology to determine whether the system is consistent and independent, consistent and dependent, or inconsistent.
2. \begin{align*}\begin{Bmatrix} 3x-2y=8 \\ 6x-4y=20 \end{Bmatrix}\end{align*}
3. \begin{align*}\begin{Bmatrix} x+3y=4 \\ 5x-y=4 \end{Bmatrix}\end{align*}
Answers:
1. \begin{align*}\begin{Bmatrix} -3x+4y=20 \\ x-2y=-8 \end{Bmatrix}\end{align*} Begin by writing the equations in slope-intercept form.
- \begin{align*}& -3x+4y = 20\\ & -3x {\color{red}+3x}+4y = 20 {\color{red}+3x}\\ & 4y = 20+3x\\ & \frac{4y}{{\color{red}4}} = \frac{20}{{\color{red}4}}+\frac{3x}{{\color{red}4}}\\ & y = {\color{red}5}+\frac{3}{4}x\\ & \boxed{y = \frac{3}{4}x+5}\end{align*}
- \begin{align*}& x-2y = -8\\ & x {\color{red}-x}-2y = -8 {\color{red}-x}\\ & -2y = -8-x\\ & \frac{-2y}{{\color{red}-2}} = \frac{-8}{{\color{red}-2}}-\frac{x}{{\color{red}-2}}\\ & y = {\color{red}4}+\frac{1}{2}x\\ & \boxed{y = \frac{1}{2}x+4}\end{align*}
- The lines intersect at the point (-4, 2). This ordered pair is the one solution for the system of linear equations. The system is consistent and independent.
2.
- \begin{align*}\begin{Bmatrix} 3x-2y=8 \\ 6x-4y=20 \end{Bmatrix}\end{align*}
- \begin{align*}& 3x-2y = 8 && 6x-4y=20\\ & 3x-3x-2y = -3x+8 && 6x-6x-4y=-6x+20\\ & -2y=-3x+8 && -4y=-6x+20\\ & \frac{-2y}{-2} = \frac{-3x}{-2}+\frac{8}{-2} && \frac{-4y}{-4}=\frac{-6x}{-4}+\frac{20}{-4}\\ & \boxed{y = \frac{3}{2}x-4} \quad \text{Slope-intercept form} && \boxed{y = \frac{6}{4}x-5}\end{align*}
- The lines are parallel. The lines will never intersect so there is no solution. The system is inconsistent.
3.
- \begin{align*}& \begin{Bmatrix} x+3y=4 \\ 5x-y=4 \end{Bmatrix}\end{align*}
- \begin{align*}& x+3y = 4 && 5x-y=4\\ & x-x+3y = -x+4 && 5x-5x-y=-5x+4\\ & 3y=-x+4 && -y=-5x+4\\ & \frac{3y}{3} = \frac{-x}{3}+\frac{4}{3} && \frac{-y}{-1}=\frac{-5x}{-1}+\frac{4}{-1}\\ & \boxed{y = \frac{-1}{3}x+\frac{4}{3}} && \boxed{y=5x-4}\end{align*}
- There is one point of intersection (1, 1). The system is consistent and independent.
Practice
Without graphing, determine whether the system is consistent and independent, consistent and dependent, or inconsistent.
\begin{align*}\begin{Bmatrix} 2x+3y=6 \\ 6x+9y=18 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} 2x-y=-14 \\ 12x-6y=-11 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} 3x+2y=14 \\ 5x-y=6 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} 2x+3y=-12 \\ 3x-y=3 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} 20x+15y=-30 \\ 4x+3y=18 \end{Bmatrix}\end{align*}
Solve the following systems of linear equations by graphing.
\begin{align*}\begin{Bmatrix} x+2y=8 \\ 3x+6y=24 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} 4x+2y=-2 \\ 2x-3y=9 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} 3x+5y=11 \\ 4x-2y=-20 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} 2x+y=5 \\ 6x=15-3y \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} 2x-y=2 \\ 4x-3y=-2 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} 2x-3y=15 \\ 4x+y=2 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} 2x+3y=-6 \\ 9y+6x+18=0 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} 6x+12y=-24 \\ 5x+10y=30 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} x-3y=7 \\ 2x+5y=-19 \end{Bmatrix}\end{align*}
\begin{align*}\begin{Bmatrix} x+3y=9 \\ x-y=-3 \end{Bmatrix}\end{align*}
Notes/Highlights Having trouble? Report an issue.
Color | Highlighted Text | Notes | |
---|---|---|---|
Please Sign In to create your own Highlights / Notes | |||
Show More |
Image Attributions
Here you'll learn to solve a system of linear equations by graphing.
We need you!
At the moment, we do not have exercises for Graphical Solutions to Systems of Equations.
To add resources, you must be the owner of the Modality. Click Customize to make your own copy.