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# Graphs of Linear Systems

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Practice Graphs of Linear Systems
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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?

$& \begin{Bmatrix}2x+y = 5 \\x-y = 1\end{Bmatrix}$

### Guidance

A $2 \times 2$ system of linear equations consists of two equations with two variables, such as the one below:

$& \begin{Bmatrix}2x+y = 5 \\x-y = 1\end{Bmatrix}$

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:

1. The lines might never intersect (they are parallel lines)
2. 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:

$\begin{Bmatrix}x-2y -2= 0 \\3x+4y = 16\end{Bmatrix}$

Solution: Begin by writing each linear equation in slope-intercept form.

$& x-2y-2 = 0\\& x {\color{red}-x}-2y-2 = 0 {\color{red}-x}\\& -2y-2 = -x\\& -2y-2 {\color{red}+2} = -x {\color{red}+2}\\& -2y = -x+2\\& \frac{-2y}{{\color{red}-2}} = \frac{-x}{{\color{red}-2}}+\frac{2}{{\color{red}-2}}\\& \boxed{y = \frac{1}{2}x-1} \qquad \text{Equation One}$

$& 3x+4y = 16\\& 3x {\color{red}-3x}+4y = 16 {\color{red}-3x}\\& 4y = 16-3x\\& \frac{4y}{{\color{red}4}} = \frac{16}{{\color{red}4}}-\frac{3x}{{\color{red}4}}\\& y = 4-\frac{3}{4}x\\& \boxed{y = -\frac{3}{4}x+4} \qquad \text{Equation Two}$

Graph both equations on the same Cartesian plane.

The lines intersect at the point (4, 1). The solution is an ordered pair that should satisfy both of the equations in the system.

Test (4, 1) in equation one:

$x-2y-2 &= 0 && \text{Use the original equation}\\({\color{red}4})-2({\color{red}1})-2 &= 0 && \text{Replace} \ x \ \text{with} \ 4 \ \text{and replace} \ y \ \text{with} \ 1.\\4-2-2 &= 0 && \text{Perform the indicated operations and simplify the result.}\\4-{\color{red}4} &= 0\\{\color{red}0} &= 0 && \text{Both sides of the equation are equal. The ordered pair} \ (4, 1) \ \text{satisfies the equation.}$

Test (4, 1) in equation two:

$3x+4y &= 16 && \text{Use the original equation}\\3({\color{red}4})+4({\color{red}1}) &= 16 && \text{Replace} \ x \ \text{with} \ 4 \ \text{and replace} \ y \ \text{with} \ 1.\\12+4 &= 16 && \text{Perform the indicated operations and simplify the result.}\\{\color{red}16} &= 16 && \text{Both sides of the equation are equal. The ordered pair} \ (4, 1) \ \text{satisfies the equation.}$

This system of 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:

$\begin{Bmatrix}2y-3x = 6 \\4y-6x = 12\end{Bmatrix}$

Solution: Graph both equations on the same Cartesian plane using the intercept method. Let $x = 0$ . Solve for $y$

$& 2y-3x = 6\\& 2y-3 ({\color{red}0}) = 6 \quad \text{Replace} \ x \ \text{with zero.}\\& 2y = 6 \qquad \qquad \text{Simplify}\\& \frac{2y}{{\color{red}2}} = \frac{6}{{\color{red}2}} \qquad \quad \ \ \text{Solve for} \ y.\\& \boxed{y = 3} \qquad \qquad \text{The} \ y \text{-intercept is} \ (0, 3)$

Let $y = 0$ . Solve for $x$ .

$& 2y-3x = 6\\& 2({\color{red}0})-3x = 6 \qquad \text{Replace} \ y \ \text{with zero.}\\& -3x = 6 \qquad \quad \ \ \text{Simplify}\\& \frac{-3x}{{\color{red}-3}} = \frac{6}{{\color{red}-3}} \qquad \ \ \ \text{Solve for} \ y.\\& \boxed{x = -2} \qquad \qquad \text{The} \ x \text{-intercept is} \ (-2, 0)$

$& 4y-6x = 12\\& 4y-6 ({\color{red}0}) = 12 \qquad \text{Replace} \ x \ \text{with zero.}\\& 4y = 12 \qquad \qquad \quad \text{Simplify}\\& \frac{4y}{{\color{red}4}} = \frac{12}{{\color{red}4}} \qquad \qquad \ \ \text{Solve for} \ y.\\& \boxed{y = 3} \qquad \qquad \quad \ \text{The} \ y \text{-intercept is} \ (0, 3)$

Let $y = 0$ . Solve for $x$ .

$& 4y-6x = 12\\& 4 ({\color{red}0})-6x = 12 \qquad \text{Replace} \ y \ \text{with zero.}\\& -6x = 12 \qquad \quad \ \ \text{Simplify}\\& \frac{-6x}{{\color{red}-6}} = \frac{12}{{\color{red}-6}} \qquad \quad \ \text{Solve for} \ y.\\& \boxed{x = -2} \qquad \qquad \ \text{The} \ x \text{-intercept is} \ (-2, 0)$

When the $x$ and $y$ -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 $\begin{Bmatrix}2y-3x = 6 \\4y-6x = 12\end{Bmatrix}$ , equation two is simply a multiple of equation one.

#### Example C

Solve the following system of linear equations graphically:

$\begin{Bmatrix}3x+4y=12 \\6x+8y=-8\end{Bmatrix}$

Solution: Graph both equations on the same Cartesian plane using the slope-intercept method. Begin by writing each linear equation in slope-intercept form.

$& 3x+4y = 12\\& 3x {\color{red}-3x}+4y = 12 {\color{red}-3x}\\& 4y = 12 {\color{red}-3x}\\& \frac{4y}{{\color{red}4}} = \frac{12}{{\color{red}4}}-\frac{3x}{{\color{red}4}}\\& y = 3-\frac{3}{4}x\\& \boxed{y = -\frac{3}{4}x+3}$

$& 6x+8y = -8\\& 6x {\color{red}-6x}+8y = -8 {\color{red}-6x}\\& 8y = -8 {\color{red}-6x}\\& \frac{8y}{{\color{red}8}} = \frac{-8}{{\color{red}8}}-\frac{6x}{{\color{red}8}}\\& y = -1-\frac{6}{{\color{red}8}}x\\& \boxed{y = -\frac{6}{8}x-1}$

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 $y=-\frac{3}{4}x+3$ and $y=-\frac{6}{8}x-1$ , the slopes are the same $\left(-\frac{6}{8}=-\frac{3}{4}\right)$ . 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 $\begin{Bmatrix}x+4y=-14 \\2x-y=4\end{Bmatrix}$ .

Solution: To use a graphing calculator, the equations must be written in slope-intercept form:

$& x+4y = -14\\& x {\color{red}-x}+4y = {\color{red}-x}-14\\& 4y = {\color{red}-x}-14\\& \frac{4y}{{\color{red}4}} = -\frac{x}{{\color{red}4}}-\frac{14}{{\color{red}4}}\\& y = -\frac{1}{4}x-\frac{14}{4}\\& \boxed{y = -\frac{1}{4}x-\frac{7}{2}}$

$& 2x-y = 4\\& 2x {\color{red}-2x}-y = {\color{red}-2x}+4\\& -y = -2x+4\\& \frac{-y}{{\color{red}-1}} = \frac{-2x}{{\color{red}-1}}+\frac{4}{{\color{red}-1}}\\& \boxed{y = 2x-4}$

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.

$& \begin{Bmatrix}2x+y = 5 \\x-y = 1\end{Bmatrix}\\& 2x+y = 5\\& 2x {\color{red}-2x}+y = 5 {\color{red}-2x}\\& y = 5 {\color{red}-2x}\\& \boxed{y = -2x+5}\\& x-y = 1\\& x {\color{red}-x}-y = 1 {\color{red}-x}\\& -y = 1-x\\& \frac{-y}{{\color{red}-1}} = \frac{1}{{\color{red}-1}}-\frac{x}{{\color{red}-1}}\\& y = -1+x\\& \boxed{y = x-1}$

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
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, 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 more than one linear equation. Two equations with two unknowns is called a $2 \times 2$ system of linear equations.

### Guided Practice

1. Solve the following system of linear equations by graphing: $\begin{Bmatrix}-3x+4y=20 \\x-2y=-8\end{Bmatrix}$

Is the system consistent and dependent, consistent and independent, or inconsistent?

For #2 and #3, use technology to determine whether the system is consistent and independent, consistent and dependent, or inconsistent.

2. $\begin{Bmatrix}3x-2y=8 \\6x-4y=20\end{Bmatrix}$

3. $\begin{Bmatrix}x+3y=4 \\5x-y=4\end{Bmatrix}$

1. $\begin{Bmatrix}-3x+4y=20 \\x-2y=-8\end{Bmatrix}$ Begin by writing the equations in slope-intercept form.

$& -3x+4y = 20\\& -3x {\color{red}+3x}+4y = 20 {\color{red}+3x}\\& 4y = 20+3x\\& \frac{4y}{{\color{red}4}} = \frac{20}{{\color{red}4}}+\frac{3x}{{\color{red}4}}\\& y = {\color{red}5}+\frac{3}{4}x\\& \boxed{y = \frac{3}{4}x+5}$
$& x-2y = -8\\& x {\color{red}-x}-2y = -8 {\color{red}-x}\\& -2y = -8-x\\& \frac{-2y}{{\color{red}-2}} = \frac{-8}{{\color{red}-2}}-\frac{x}{{\color{red}-2}}\\& y = {\color{red}4}+\frac{1}{2}x\\& \boxed{y = \frac{1}{2}x+4}$
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.

$\begin{Bmatrix}3x-2y=8 \\6x-4y=20\end{Bmatrix}$
$& 3x-2y = 8 && 6x-4y=20\\& 3x-3x-2y = -3x+8 && 6x-6x-4y=-6x+20\\& -2y=-3x+8 && -4y=-6x+20\\& \frac{-2y}{-2} = \frac{-3x}{-2}+\frac{8}{-2} && \frac{-4y}{-4}=\frac{-6x}{-4}+\frac{20}{-4}\\& \boxed{y = \frac{3}{2}x-4} \quad \text{Slope-intercept form} && \boxed{y = \frac{6}{4}x-5}$
The lines are parallel. The lines will never intersect so there is no solution. The system is inconsistent.

3.

$& \begin{Bmatrix}x+3y=4 \\5x-y=4\end{Bmatrix}$
$& x+3y = 4 && 5x-y=4\\& x-x+3y = -x+4 && 5x-5x-y=-5x+4\\& 3y=-x+4 && -y=-5x+4\\& \frac{3y}{3} = \frac{-x}{3}+\frac{4}{3} && \frac{-y}{-1}=\frac{-5x}{-1}+\frac{4}{-1}\\& \boxed{y = \frac{-1}{3}x+\frac{4}{3}} && \boxed{y=5x-4}$
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.

1. .

$\begin{Bmatrix}2x+3y=6 \\6x+9y=18\end{Bmatrix}$

1. .

$\begin{Bmatrix}2x-y=-14 \\12x-6y=-11\end{Bmatrix}$

1. .

$\begin{Bmatrix}3x+2y=14 \\5x-y=6\end{Bmatrix}$

1. .

$\begin{Bmatrix}2x+3y=-12 \\3x-y=3\end{Bmatrix}$

1. .

$\begin{Bmatrix}20x+15y=-30 \\4x+3y=18\end{Bmatrix}$

Solve the following systems of linear equations by graphing.

1. .

$\begin{Bmatrix}x+2y=8 \\3x+6y=24\end{Bmatrix}$

1. .

$\begin{Bmatrix}4x+2y=-2 \\2x-3y=9\end{Bmatrix}$

1. .

$\begin{Bmatrix}3x+5y=11 \\4x-2y=-20\end{Bmatrix}$

1. .

$\begin{Bmatrix}2x+y=5 \\6x=15-3y\end{Bmatrix}$

1. .

$\begin{Bmatrix}2x-y=2 \\4x-3y=-2\end{Bmatrix}$

1. .

$\begin{Bmatrix}2x-3y=15 \\4x+y=2\end{Bmatrix}$

1. .

$\begin{Bmatrix}2x+3y=-6 \\9y+6x+18=0\end{Bmatrix}$

1. .

$\begin{Bmatrix}6x+12y=-24 \\5x+10y=30\end{Bmatrix}$

1. .

$\begin{Bmatrix}x-3y=7 \\2x+5y=-19\end{Bmatrix}$

1. .

$\begin{Bmatrix}x+3y=9 \\x-y=-3\end{Bmatrix}$