Solving Systems of Equations by Graphing

Tearrik is throwing a party and bought some sodas and chips.

Solution

To start, each equation in the system will be written in slope-intercept form. This means that the variable $y$ will be isolated in both equations.

$\{\begin{array}{ll}x+y=21& \text{(I)}\\ 2x+3y=51& \text{(II)}\end{array}$

▼

$\text{(I), (II): }$ Write in slope-intercept form

SubEqn

$\text{(I): }$ $\text{LHS}-x=\text{RHS}-x$

$\{\begin{array}{l}y=\text{-}x+21\\ 2x+3y=51\end{array}$

SubEqn

$\text{(II): }$ $\text{LHS}-2x=\text{RHS}-2x$

$\{\begin{array}{l}y=\text{-}x+21\\ 3y=\text{-}2x+51\end{array}$

DivEqn

$\text{(II): }$ $\text{LHS}/3=\text{RHS}/3$

$\{\begin{array}{l}y=\text{-}x+21\\ y=\frac{\text{-}2x+51}{3}\end{array}$

WriteSumFrac

$\text{(II): }$ Write as a sum of fractions

$\{\begin{array}{l}y=\text{-}x+21\\ y=\frac{\text{-}2x}{3}+\frac{51}{3}\end{array}$

MovePartNumRight

$\text{(II): }$ $\frac{a\cdot b}{c}=\frac{a}{c}\cdot b$

$\{\begin{array}{l}y=\text{-}x+21\\ y=\frac{\text{-}2}{3}x+\frac{51}{3}\end{array}$

MoveNegNumToFrac

$\text{(II): }$ Put minus sign in front of fraction

$\{\begin{array}{l}y=\text{-}x+21\\ y=\text{-}\frac{2}{3}x+\frac{51}{3}\end{array}$

CalcQuot

$\text{(II): }$ Calculate quotient

$\{\begin{array}{l}y=\text{-}x+21\\ y=\text{-}\frac{2}{3}x+17\end{array}$

Now, the slope and the $y\text{-}$intercept of each line will be used to draw the graphs on the same coordinate plane. Since the number of items cannot be negative, only the first quadrant will be considered for the graph.

Finally, the point of intersection $P$ can be identified.

The point of intersection of the lines is $P(12,9).$ In the context of the situation, this means that Tearrik bought $12$ sodas and $9$ bags of chips.

At Tearrik's party, Ignacio found one of Tearrik's homework assignments. He does not want to do Tearrik's homework for him, but Ignacio decides to quiz himself using the assignment.

Help Ignacio determine whether the given systems of equations are consistent systems and whether they are independent systems.

Solution

a The system will be solved graphically. To do so, both linear equations must to be expressed in slope-intercept form. In the first equation, the variable $y$ is already isolated. Therefore, only the second equation needs to be rewritten.

$\{\begin{array}{ll}y=\text{-}2x+6& \text{(I)}\\ 4x-2y=\text{-}20& \text{(II)}\end{array}$

▼

$\text{(II): }$ Write in slope-intercept form

SubEqn

$\text{(II): }$ $\text{LHS}-4x=\text{RHS}-4x$

$\{\begin{array}{l}y=\text{-}2x+6\\ \text{-}2y=\text{-}4x-20\end{array}$

DivEqn

$\text{(II): }$ $\text{LHS}/(\text{-}2)=\text{RHS}/(\text{-}2)$

$\{\begin{array}{l}y=\text{-}2x+6\\ y=\frac{\text{-}4x-20}{\text{-}2}\end{array}$

WriteDiffFrac

$\text{(II): }$ Write as a difference of fractions

$\{\begin{array}{l}y=\text{-}2x+6\\ y=\frac{\text{-}4x}{\text{-}2}-\frac{20}{\text{-}2}\end{array}$

DivNegNeg

$\text{(II): }$ $\frac{\text{-}a}{\text{-}b}=\frac{a}{b}$

$\{\begin{array}{l}y=\text{-}2x+6\\ y=\frac{4x}{2}-\frac{20}{\text{-}2}\end{array}$

MovePartNumRight

$\text{(II): }$ $\frac{a\cdot b}{c}=\frac{a}{c}\cdot b$

$\{\begin{array}{l}y=\text{-}2x+6\\ y=\frac{4}{2}x-\frac{20}{\text{-}2}\end{array}$

MoveNegDenomToFrac

$\text{(II): }$ Put minus sign in front of fraction

$\{\begin{array}{l}y=\text{-}2x+6\\ y=\frac{4}{2}x-\left(\text{-}\frac{20}{2}\right)\end{array}$

SubNeg

$\text{(II): }$ $a-(\text{-}b)=a+b$

$\{\begin{array}{l}y=\text{-}2x+6\\ y=\frac{4}{2}x+\frac{20}{2}\end{array}$

CalcQuot

$\text{(II): }$ Calculate quotient

$\{\begin{array}{l}y=\text{-}2x+6\\ y=2x+10\end{array}$

Now that both linear equations are expressed in slope-intercept form, their slopes and $y\text{-}$intercepts can be used to draw the lines on the same coordinate plane.

The lines intersect at exactly one point.

Since the system has one or more solutions, it is a consistent system. Furthermore, since it has exactly one solution, it is also an independent system.

b Again, the system of equations will be solved graphically. To do so, both equations will be rewritten in slope-intercept form.

$\{\begin{array}{ll}x+y=3& \text{(I)}\\ 4x+4y=12& \text{(II)}\end{array}$

▼

$\text{(I), (II): }$ Write in slope-intercept form

SubEqn

$\text{(I): }$ $\text{LHS}-x=\text{RHS}-x$

$\{\begin{array}{l}y=\text{-}x+3\\ 4x+4y=12\end{array}$

SubEqn

$\text{(II): }$ $\text{LHS}-4x=\text{RHS}-4x$

$\{\begin{array}{l}y=\text{-}x+3\\ 4y=\text{-}4x+12\end{array}$

DivEqn

$\text{(II): }$ $\text{LHS}/4=\text{RHS}/4$

$\{\begin{array}{l}y=\text{-}x+3\\ y=\frac{\text{-}4x+12}{4}\end{array}$

WriteSumFrac

$\text{(II): }$ Write as a sum of fractions

$\{\begin{array}{l}y=\text{-}x+3\\ y=\frac{\text{-}4x}{4}+\frac{12}{4}\end{array}$

MoveNegNumToFrac

$\text{(II): }$ Put minus sign in front of fraction

$\{\begin{array}{l}y=\text{-}x+3\\ y=\text{-}\frac{4x}{4}+\frac{12}{4}\end{array}$

MovePartNumRight

$\text{(II): }$ $\frac{a\cdot b}{c}=\frac{a}{c}\cdot b$

$\{\begin{array}{l}y=\text{-}x+3\\ y=\text{-}\frac{4}{4}x+\frac{12}{4}\end{array}$

CalcQuot

$\text{(II): }$ Calculate quotient

$\{\begin{array}{l}y=\text{-}x+3\\ y=\text{-}1x+3\end{array}$

IdPropMult

$\text{(II): }$ Identity Property of Multiplication

$\{\begin{array}{l}y=\text{-}x+3\\ y=\text{-}x+3\end{array}$

Note that both equations are simplified to the same equation. It can be graphed by using its slope and its $y\text{-}$intercept.

The lines overlap each other. Therefore, they intersect at infinitely many points. Since the system has one or more solutions, it is a consistent system. However, since it does not have exactly one solution, it is not an independent system.

Mark cannot go to Tearrik's party because he has recently started working with his father at a car dealership. At the dealership, Mark's father sells sedans and trucks.

Solution

To interpret the system in terms of consistency and independence, both linear equations will be drawn on the same coordinate plane. To do so, the slope-intercept form will be used. Since the first equation is already written in this form, the $y\text{-}$variable will be isolated in the second equation.

$\{\begin{array}{ll}y=2x+2& \text{(I)}\\ 2y-4x=10& \text{(II)}\end{array}$

▼

$\text{(II): }$ Write in slope-intercept form

AddEqn

$\text{(II): }$ $\text{LHS}+4x=\text{RHS}+4x$

$\{\begin{array}{l}y=2x+2\\ 2y=4x+10\end{array}$

DivEqn

$\text{(II): }$ $\text{LHS}/2=\text{RHS}/2$

$\{\begin{array}{l}y=2x+2\\ y=\frac{4x+10}{2}\end{array}$

WriteSumFrac

$\text{(II): }$ Write as a sum of fractions

$\{\begin{array}{l}y=2x+2\\ y=\frac{4x}{2}+\frac{10}{2}\end{array}$

MovePartNumRight

$\text{(II): }$ $\frac{a\cdot b}{c}=\frac{a}{c}\cdot b$

$\{\begin{array}{l}y=2x+2\\ y=\frac{4}{2}x+\frac{10}{2}\end{array}$

CalcQuot

$\text{(II): }$ Calculate quotient

$\{\begin{array}{l}y=2x+2\\ y=2x+5\end{array}$

Now that both equations are written in slope-intercept form, their slopes and $y\text{-}$intercepts can be used to draw the lines.

Next, recall the definitions of consistent, inconsistent, dependent, and independent systems.

Consistent System	A system of equations that has one or more solutions.
Independent System	A system of equations with exactly one solution.
Dependent System	A system of equations with infinitely many solutions.
Inconsistent System	A system of equations that has no solution.

With this information in mind, the lines will be considered one more time.

Since the lines do not intersect each other, the system has no solution. Therefore, it is an inconsistent system.

With the content learned in this lesson, the challenge presented at the beginning can be solved. It has been said that the number of students in the US participating in high school basketball and soccer has steadily increased over the past few years.

The following table shows some information about this.

Consider the above table to answer the following questions.

Solution

a To start, the equation for the number of students participating in high school basketball will be written. The variables can be defined as follows.

Variable	Meaning of the Variable
$x$	Number of years since $2015$
$y$	Number of students that participate in high school basketball (thousands)

From the given table, it is known that in the year $2015,$ $700$ thousands students participated in high school basketball. Also, the yearly average rate of increase is $20$ thousand students. With this information, an equation can be written.

$$\begin{array}{c}y=20x+700\end{array}$$

Next, the equation for the number of students participating in high school soccer will be written. The variables can be defined in a similar way as before.

Variable	Meaning of the Variable
$x$	Number of years since $2015$
$y$	Number of students that participate in high school soccer (thousands)

Again, it is seen in the table that in the year $2015,$ $400$ thousands students participated in high school soccer. Also, the yearly average rate of increase is $40$ thousand students. With this information, the second equation can be written.

$$\begin{array}{c}y=40x+400\end{array}$$

b The equations found in Part A form a system of equations.

$$\{\begin{array}{l}y=20x+700\\ y=40x+400\end{array}$$

Since both lines are already written in slope-intercept form, they can be graphed on the same coordinate plane.

The lines intersect at exactly one point. Therefore, this is a consistent and independent system. The solution is $x=15$ and $y=1000.$ This means that $15$ years after $2015$ — the year $2030$ — there will be one million high school students participating in both sports.