5.1: Solving Systems of Linear Equations
Solving Systems of Linear Equations by Graphing
Objectives
The lesson objectives for Solving Systems of Linear Equations by Graphing are:
 Understanding what is meant by a system of linear equations
 Consistent systems of linear equations
 Independent and dependent systems of linear equations
 Equivalent systems of linear equations
 Inconsistent systems of linear equations
Introduction
In a previous chapter you learned how to solve linear equations with one variable. The solution was a number that belonged to the real number system. When the solution replaced the variable in the original equation and when this solution was correct, both sides of the equation were equal.
In this lesson you will learn to solve a linear equation that has two variables by using two linear equations that have two variables. This is known as a \begin{align*}2 \times 2\end{align*} system of linear equations. In general: \begin{align*}\begin{Bmatrix} a_1 {\color{red}x}+b_1{\color{blue}y}=c_1\\a_2{\color{red}x}+b_2{\color{blue}y}=c_2 \end{Bmatrix}\end{align*} where ‘\begin{align*}a\end{align*}’ is the coefficient of ‘\begin{align*}{\color{red}x}\end{align*}’; ‘\begin{align*}b\end{align*}’ is the coefficient of ‘\begin{align*}{\color{blue}y}\end{align*}’; and ‘\begin{align*}c\end{align*}’ is a constant. The subscripts of 1 and 2 simply refer to the first equation and the second equation.
The first method of solving a system of linear equations will be graphing.
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Khan Academy Solving Systems by Graphing 3
Guidance
When you graph two equations with two variables on the same Cartesian plane, the resulting lines may intersect. The following linear equations will be graphed by using the slopeintercept method.
\begin{align*}& \begin{Bmatrix} 2x+y = 5 \\ xy = 1 \end{Bmatrix}\\ & 2x+y = 5\\ & 2x {\color{red}2x}+y = 5 {\color{red}2x}\\ & y = 5 {\color{red}2x}\\ & \boxed{y = 2x+5}\\ & xy = 1\\ & x {\color{red}x}y = 1 {\color{red}x}\\ & y = 1x\\ & \frac{y}{{\color{red}1}} = \frac{1}{{\color{red}1}}\frac{x}{{\color{red}1}}\\ & y = 1+x\\ & \boxed{y = x1}\end{align*}
The two lines intersect at one point. The coordinates of the point of intersection are (2, 1).
Example A
Solve the following system of linear equations graphically:
\begin{align*}\begin{Bmatrix} x2y 2= 0 \\ 3x+4y = 16 \end{Bmatrix}\end{align*}
Both equations will be graphed on the same Cartesian plane using the slopeintercept method.
(NOTE): You could graph the equations using the \begin{align*}x\end{align*} and \begin{align*}y\end{align*}intercepts or the slopeintercept method. The same result will occur regardless of the method used to graph the lines.
Begin by writing each linear equation in slopeintercept form.
\begin{align*}& x2y2 = 0\\ & x {\color{red}x}2y2 = 0 {\color{red}x}\\ & 2y2 = x\\ & 2y2 {\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}x1} \qquad \text{Equation One}\end{align*}
\begin{align*}& 3x+4y = 16\\ & 3x {\color{red}3x}+4y = 16 {\color{red}3x}\\ & 4y = 163x\\ & \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 of the 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*}x2y2 &= 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.\\ 422 &= 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} 2y3x = 6 \\ 4y6x = 12 \end{Bmatrix}\end{align*}
Both equations will be graphed on the same Cartesian plane using the intercept method.
Let \begin{align*}x = 0\end{align*}. Solve for \begin{align*}y\end{align*}
\begin{align*}& 2y3x = 6\\ & 2y3 ({\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*}. Solve for \begin{align*}x\end{align*}.
\begin{align*}& 2y3x = 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} \ x.\\ & \boxed{x = 2} \qquad \qquad \text{The} \ x \text{intercept is} \ (2, 0)\end{align*}
\begin{align*}& 4y6x = 12\\ & 4y6 ({\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*}& 4y6x = 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} \ x.\\ & \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} 2y3x = 6 \\ 4y6x = 12 \end{Bmatrix}\end{align*}, equation two is simply a multiple of equation one.
Example C
Although example A had only one ordered pair as its solution and example B had an infinite number of ordered pairs as its solution, both systems are consistent systems of equations.
Two systems of equations are equivalent if they have the same solution set. If example A is considered \begin{align*}\begin{Bmatrix} x2y2=0 \\ 3x+4y = 16 \end{Bmatrix}\end{align*} the ordered pair (4, 1) is the solution for this system.
If the equations are both set equal to zero, then they are equal to each other. (Commutative Property)
\begin{align*}\begin{Bmatrix} x2y2=0 \\ 3x+4y16=0 \end{Bmatrix}\end{align*}
If the equations are then added, \begin{align*}(x2y2)+(3x+4y16)=0\end{align*} the result is \begin{align*}4x+2y18=0\end{align*}.
The graph of the three equations, on the same Cartesian grid, is shown below:
The three equations all intersect at the same point (4, 1). The point (4, 1) can be tested in the equation \begin{align*}4x+2y18=0\end{align*}.
\begin{align*}4x+2y18 &=0 && \text{Use the original equation}\\ 4({\color{red}4})+2({\color{red}1})18 &= 0 && \text{Replace} \ x \ \text{with} \ 4 \ \text{and replace} \ y \ \text{with} \ 1.\\ 16+218 &= 0 && \text{Perform the indicated operations and simplify the result.}\\ 18 {\color{red}18} &= 0\\ {\color{red}0} &= 0 && \text{Both sides of the equation are equal. The ordered pair} \ (4, 1) \ \text{satisfies the equation.}\end{align*}
The ordered pair (4, 1) satisfies the three equations and is a solution for all of the equations.
Therefore: \begin{align*}& \begin{Bmatrix} x2y2=0 \\ 3x+4y16=0 \end{Bmatrix}\end{align*} and \begin{align*}\begin{Bmatrix} x2y2=0 \\ 4x+2y18=0 \end{Bmatrix}\end{align*} and \begin{align*}\begin{Bmatrix} 3x+4y16=0 \\ 4x+2y18=0 \end{Bmatrix}\end{align*} are equivalent equations.
This same concept can be extended to include multiples of an equation.
Example D
Solve the following system of linear equations graphically:
\begin{align*}\begin{Bmatrix} 3x+4y=12 \\ 6x+8y=8 \end{Bmatrix}\end{align*}
Both equations will be graphed on the same Cartesian plane using the slopeintercept method.
Begin by writing each linear equation in slopeintercept 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}x1}\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 slopeintercept form \begin{align*}y=\frac{3}{4}x+3\end{align*} and \begin{align*}y=\frac{6}{8}x1\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 E
Before technology was introduced, 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, technology will be used to determine the solution for \begin{align*}\begin{Bmatrix} x+4y=14 \\ 2xy=4 \end{Bmatrix}\end{align*}.
To use technology, the equations must be written in slopeintercept 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*}& 2xy = 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 = 2x4}\end{align*}
The equations are both in slopeintercept form. Set the window on the calculator as shown below:
The intersection point of the linear equations is (0.22, –3.56).
The following represent the keys that were pressed on the calculator to obtain the above results:
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
 A 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, an infinite number of ordered pairs satisfy 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 inconsistent systems 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 \\ x2y=8 \end{Bmatrix}\end{align*}
Is the system consistent and dependent, consistent and independent, or inconsistent?
2. For the following system of consistent and independent linear equations, write equivalent systems of equations. \begin{align*}\begin{Bmatrix} 4x+2y=8 \\ 2x+3y=20 \end{Bmatrix}\end{align*}.
3. Use technology to determine whether the system is consistent and independent, consistent and dependent, or inconsistent.
i) \begin{align*}\begin{Bmatrix} 3x2y=8 \\ 6x4y=20 \end{Bmatrix}\end{align*}
ii) \begin{align*}\begin{Bmatrix} x+3y=4 \\ 5xy=4 \end{Bmatrix}\end{align*}
Answers
1. \begin{align*}\begin{Bmatrix} 3x+4y=20 \\ x2y=8 \end{Bmatrix}\end{align*} Begin by writing the equations in slopeintercept 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*}& x2y = 8\\ & x {\color{red}x}2y = 8 {\color{red}x}\\ & 2y = 8x\\ & \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} 4x+2y=8 \\ 2x+3y=20 \end{Bmatrix}\end{align*}
The system is consistent and independent. Set each equation equal to zero.
\begin{align*}& 4x+2y = 8 && 2x+3y=20\\ & 4x+2y {\color{red}8} = 8 {\color{red}8} && 2x+3y {\color{red}20}=20 {\color{red}20}\\ & \boxed{4x+2y8 = 0} && \boxed{2x+3y20=0}\end{align*}
Add the equations:
\begin{align*}& (4x+2y8)+(2x+3y20) = 0\\ & 4x+2y82x+3y20 = 0\\ & \boxed{2x+5y28 = 0}\end{align*}
The equivalent systems are:
\begin{align*}& \begin{Bmatrix} 4x+2y8=0 \\ 2x+3y20=0 \end{Bmatrix} && \begin{Bmatrix} 4x+2y8=0 \\ 2x+5y28=0 \end{Bmatrix} && \begin{Bmatrix} 2x+3y20=0 \\ 2x+5y28=0 \end{Bmatrix}\end{align*}
3. i) \begin{align*}\begin{Bmatrix} 3x2y=8 \\ 6x4y=20 \end{Bmatrix}\end{align*}
\begin{align*}& 3x2y = 8 && 6x4y=20\\ & 3x3x2y = 3x+8 && 6x6x4y=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}x4} \quad \text{Slopeintercept form} && \boxed{y = \frac{6}{4}x5}\end{align*}
The lines are parallel. The lines will never intersect so there is no solution. The system is inconsistent.
ii) \begin{align*}& \begin{Bmatrix} x+3y=4 \\ 5xy=4 \end{Bmatrix}\end{align*}
\begin{align*}& x+3y = 4 && 5xy=4\\ & xx+3y = x+4 && 5x5xy=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=5x4}\end{align*}
There is one point of intersection (1, 1). The system is consistent and independent.
Summary
In this lesson you have learned that a \begin{align*}2 \times 2\end{align*} system of linear equations consists of two equations with two variables. You learned to solve these systems by graphing and you learned that the resulting lines may a) intersect at one point, b) coincide or c) be parallel. You also learned that a system that resulted in one point of intersection is consistent and independent. A system that resulting in lines that coincided is consistent and dependent. A system that resulted in two parallel lines is inconsistent.
You also learned that a system of equations that was consistent had equivalent systems that could be determined by adding the original equations of the system. The final concept that you learned was how to solve the systems using technology.
Problem Set
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} 2xy=14 \\ 12x6y=11 \end{Bmatrix}\end{align*}
 .

 \begin{align*}\begin{Bmatrix} 3x+2y=14 \\ 5xy=6 \end{Bmatrix}\end{align*}
 .

 \begin{align*}\begin{Bmatrix} 2x+3y=12 \\ 3xy=3 \end{Bmatrix}\end{align*}
 .

 \begin{align*}\begin{Bmatrix} 20x+15y=30 \\ 4x+3y=18 \end{Bmatrix}\end{align*}
For each of the following linear systems that are consistent, write equivalent systems:
 .

 \begin{align*}\begin{Bmatrix} x+2y=8 \\ 3x+6y=24 \end{Bmatrix}\end{align*}
 .

 \begin{align*}\begin{Bmatrix} 4x+2y=2 \\ 2x3y=9 \end{Bmatrix}\end{align*}
 .

 \begin{align*}\begin{Bmatrix} 3x+5y=11 \\ 4x2y=20 \end{Bmatrix}\end{align*}
 .

 \begin{align*}\begin{Bmatrix} 2x+y=5 \\ 6x=153y \end{Bmatrix}\end{align*}
 .

 \begin{align*}\begin{Bmatrix} 2xy=2 \\ 4x3y=2 \end{Bmatrix}\end{align*}
Solve the following systems of linear equations by graphing. (Use technology if the solution is consistent and independent and the solution is not integers)
 .

 \begin{align*}\begin{Bmatrix} 2x3y=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} x3y=7 \\ 2x+5y=19 \end{Bmatrix}\end{align*}
 .

 \begin{align*}\begin{Bmatrix} x+3y=9 \\ xy=3 \end{Bmatrix}\end{align*}
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