Topics
4. Systems of Linear Equations and Inequalities
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United States of America
Grades 9-12

4.01 Writing and graphing linear systems

Lesson
Practice

Introduction

In 8th grade, solving equations with variables on both sides led to equations with one solution, no solutions, or infinitely many solutions. We saw a connection to solutions of systems of equations that we graphed. We continue the concept of writing and graphing systems of linear equations here to solve problems in context and understand viable and non-viable solutions based on problem constraints.

Writing and graphing linear systems

System of equations

A set of equations that have the same variables

Solution to a system of equations

Any set of values of all variables in that system which is a solution to each equation in the system.

A solution can also be thought of graphically as the point(s) of intersection of the graphs of the equations (the points in common to all graphs).

A system of linear equations can have no solutions, infinitely many solutions, or one solution.

A four quadrant coordinate plane is drawn with two parallel lines leaning to the left.

When two lines are parallel and distinct, they have no points of intersection. The corresponding system of equations has no solutions.

A four quadrant coordinate plane is drawn with a dashed line of positive slope passing through the origin. The line is composed of two alternating colors of dashes and dots representing two lines of the same slope, and solutions.

When two lines are identical, they intersect at every point. The corresponding system of equations has infinitely many solutions.

A four quadrant coordinate plane with two lines drawn intersecting each other at a common point. The lines intersect at a point located at (2, negative 1)

When two lines are not parallel, they have exactly one point of intersection. The corresponding system of equations has one solution.

A solution to a system of equations in a given context is said to be viable if the solution makes sense in the context and non-viable if it does not make sense within the context, even if it would otherwise be algebraically valid.

Examples

Example 1

Graph each system of equations and state the solution.

a

\begin{cases} y = \dfrac{3}{4}x - 6 \\ 4y = 3x - 24 \end{cases}

Worked Solution
Create a strategy

We can convert the second equation in the system to slope-intercept form and graph both equations using the y-intercept and slope.

Apply the idea
\displaystyle 4y\displaystyle =\displaystyle 3x-24Second equation
\displaystyle \dfrac{4y}{4}\displaystyle =\displaystyle \dfrac{3x}{4}-\dfrac{24}{4}Divide equation by 4
\displaystyle y\displaystyle =\displaystyle \dfrac{3}{4}x-6Evaluate the division

Since the second equation is equivalent to the first equation in the system, the lines will be the same and therefore, the graph will show that the solution to the system of equations is infinite solutions.

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b

\begin{cases} 6x + 3y = 18 \\ 12x + 6y = 24 \end{cases}

Worked Solution
Create a strategy

We can calculate the intercepts of each equation in the system and graph them.

Apply the idea

Find the x-intercept of the first equation:

\displaystyle 6x+3y\displaystyle =\displaystyle 18First equation
\displaystyle 6x+3(0)\displaystyle =\displaystyle 18Substitute y=0
\displaystyle 6x\displaystyle =\displaystyle 18Evaluate the multiplication
\displaystyle x\displaystyle =\displaystyle 3Divide both sides by 6

Find the y-intercept of the first equation:

\displaystyle 6x+3y\displaystyle =\displaystyle 18First equation
\displaystyle 6(0)+3y\displaystyle =\displaystyle 18Substitute x=0
\displaystyle 3y\displaystyle =\displaystyle 18Evaluate the multiplication
\displaystyle y\displaystyle =\displaystyle 6Divide both sides by 3

The intercepts of the first equation are (3, 0) and (0, 6).

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Find the x-intercept of the second equation:

\displaystyle 12x+6y\displaystyle =\displaystyle 24Second equation
\displaystyle 12x+6(0)\displaystyle =\displaystyle 24Substitute y=0
\displaystyle 12x\displaystyle =\displaystyle 24Evaluate the multiplication
\displaystyle x\displaystyle =\displaystyle 2Divide both sides by 12

Find the y-intercept of the second equation:

\displaystyle 12x+6y\displaystyle =\displaystyle 24Second equation
\displaystyle 12(0)+6y\displaystyle =\displaystyle 24Substitute x=0
\displaystyle 6y\displaystyle =\displaystyle 24Evaluate the multiplication
\displaystyle y\displaystyle =\displaystyle 4Divide both sides by 6

The intercepts of the second equation are (2, 0) and (0, 4).

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The graphs of the equations never intersect, so there is no solution to the system of equations.

Reflect and check

Converting the system of equations to slope-intercept form before graphing would show that the slopes of the lines are the same and their y-intercepts are different, meaning that we would know the system has no solutions prior to graphing.

Example 2

Rodica and Yuwei purchased used cars at the same time. Rodica buys a car with 9000 miles on it and drives an average of 200 miles each week. Yuwei buys a car with 6000 miles on it and drives an average of 500 miles each week. Consider the equation that represents the number of weeks when Rodica's and Yuwei's cars will have the same mileage:

200x+9000=500x+6000

a

Rewrite the equation as a system of equations.

Worked Solution
Create a strategy

We can see that the expressions on each side of the equal sign are in slope-intercept form and use that to write two equations.

Apply the idea

\begin{cases}y=200x+9000\\y=500x+6000 \end{cases}

b

Construct a table of values to show when the cars will have the same mileage.

Worked Solution
Create a strategy

We can create a table of values beginning with the time at 0 weeks and going up by one week at a time. We will substitute the number of weeks into each equation to determine the mileage for that week.

Apply the idea
Time (weeks)Car Mileage (Rodica)Car Mileage (Yuwei)
0200\left(0\right)+9000 =9000500\left(0\right)+6000 =6000
1200\left(1\right)+9000 =9200500\left(1\right)+6000 =6500
2200\left(2\right)+9000 =9400500\left(2\right)+6000 =7000
3200\left(3\right)+9000 =9600500\left(3\right)+6000 =7500
4200\left(4\right)+9000 =9800500\left(4\right)+6000 =8000
5200\left(5\right)+9000 =10\,000500\left(5\right)+6000 =8500
6200\left(6\right)+9000 =10\,200500\left(6\right)+6000 =9000
7200\left(7\right)+9000 =10\,400500\left(7\right)+6000 =9500
8200\left(8\right)+9000 =10\,600500\left(8\right)+6000 =10\,000
9200\left(9\right)+9000 =10\,800500\left(9\right)+6000 =10\,500
10200\left(10\right)+9000 =11\,000500\left(10\right)+6000 =11\,000

When we get to 10 weeks in the table we can see that the mileage for both cars is 11\,000. So we know the mileage will be the same after 10 weeks.

c

Graph the system of equations. Choose an appropriate scale for the axes.

Worked Solution
Create a strategy

Based on the solution in part (b), we know that the cars will have the same mileage of 11\,000 miles after 10 weeks. This will help us decide on an appropriate scale.

Apply the idea

Since the number of weeks is the x-value, we can count by 1 along the x-axis to slightly beyond the point of intersection at 10 weeks. Since the mileage is the y-value, we can count by 500 along the y-axis to slightly beyond the point of intersection at 11\,000 miles.

Used car mileage
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\text{Time (weeks)}
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\text{Mileage}

Example 3

Bixia is saving up her quarters and dimes in a jar. She has a total of \$24.50 in 125 coins.

a

Write a system of equations that models this situation.

Worked Solution
Create a strategy

We'll need to write a system of equations with the information provided in the problem, and we're only given two numbers.

The two totals give us information about each equation: one of the equations utilizes the units of a dollar amount throughout, and the other equation must include the number of coins as its units.

We should define the unknown variables. Since we don't know how many of each type of coin Bixia has, we can say that x= the number of quarters and y= the number of dimes.

Apply the idea

\begin{cases}x+y=125\\0.25x+0.10y=24.50 \end{cases}

Reflect and check

We can confirm that the system of equations makes sense by checking the units of each equation.

Since the total of x+y=125 is the total number of coins, it should make sense that the number of quarters added to the number of dimes is equal to the total number of coins, which is what is being represented in this equation.

The total of 0.25x+0.10y=24.50 represents the fact that Bixia has \$24.50. The expression 0.25x represents the value of 1 quarter multiplied by how many quarters are in the jar. This term is the dollar amount of all the quarters combined. The expression 0.10x represents the value of 1 dime multiplied by how many dimes are in the jar. This term is the dollar amount of all the dimes combined. When these two expressions are added togehter, the result is the total value of all the coins, or \$24.50.

b

Graph the system of equations. Use appropriate axes, labels, and scales.

Worked Solution
Create a strategy

Based on the equation, x represents the number of quarters because we know that a quarter is \$0.25, and y represents the number of dimes because we know that a dime is \$0.10. We can calculate the intercepts of both equations and use those to determine the maximum values on our x- and y-axes.

Apply the idea

First, we'll calculate the x-intercept of x+y=125:

\displaystyle x+y\displaystyle =\displaystyle 125First equation
\displaystyle x+(0)\displaystyle =\displaystyle 125Substitute y=0
\displaystyle x\displaystyle =\displaystyle 125Evaluate the addition

Now, we'll calculate the y-intercept of x+y=125:

\displaystyle x+y\displaystyle =\displaystyle 125First equation
\displaystyle (0)+y\displaystyle =\displaystyle 125Substitute x=0
\displaystyle y\displaystyle =\displaystyle 125Evaluate the addition

The x-intercept of x+y=125 is (125,0) and the y-intercept is (0, 125).

Next, we'll calculate the x-intercept of 0.25x+0.10y=24.50:

\displaystyle 0.25x+0.10y\displaystyle =\displaystyle 24.50Second equation
\displaystyle 0.25x+0.10(0)\displaystyle =\displaystyle 24.50Substitute y=0
\displaystyle 0.25x\displaystyle =\displaystyle 24.50Evaluate the multiplication
\displaystyle x\displaystyle =\displaystyle 98Divide both sides by 0.25

Finally, we'll calculate the y-intercept of 0.25x+0.10y=24.50:

\displaystyle 0.25x+0.10y\displaystyle =\displaystyle 24.50Second equation
\displaystyle 0.25(0)+0.10y\displaystyle =\displaystyle 24.50Substitute x=0
\displaystyle 0.10y\displaystyle =\displaystyle 24.50Evaluate the multiplication
\displaystyle y\displaystyle =\displaystyle 245Divide both sides by 0.10

The x-intercept of 0.25x+0.10y=24.50 is (98,0) and the y-intercept is (0, 245).

Since the x-intercepts have a maximum value of 125, we can draw an x-axis up to 130 and count by 10. Since the y-intercepts have a maximum value of 245, we can draw a y-axis up to 250 and count by 50.

Bixia's coin jar
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\text{Number of quarters}
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\text{Number of dimes}
c

Interpret the solution to the system of equations.

Worked Solution
Create a strategy

We can use the graph to determine the solution to the system and the axes labels to interpret the meaning of the point of intersection. Since the solution to the system is not clear on the graph, we can use technology to graph the system of equations.

Apply the idea

The solution to the system of equations is (80, 45). Since x is the number of quarters, we know that Bixia had 80 quarters, and since y is the number of dimes, we know that Bixia had 45 dimes in the jar.

Example 4

Tyson is saving money in order to purchase a new smartphone for \$800 when the latest model is released. He currently has \$350 saved up and is able to put away \$100 each month.

a

Write a system of equations to represent the situation.

Worked Solution
Create a strategy

To write a system of equations, we will need to define some variables. Let's choose y to represent an amount of money (in dollars), and x to represent the number of months that have passed.

Apply the idea

Using these variables, the amount of money Tyson has saved over time can be represented by y = 350 + 100x. The price of the smartphone can be represented by y = 800.

b

Sketch the two lines representing these equations on the coordinate plane.

Worked Solution
Create a strategy

All of the values involved in the question are multiples of \$50, so we can use this for the scale of the y-axis. Also, both x and y only make sense for positive values in this context, so we only need to think about the first quadrant.

Apply the idea
Tyson's cell phone savings
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\text{Time (months)}
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\text{Amount of money}
c

If the new phone is to be released in 5 months' time, determine if Tyson will be able to afford it on release.

Worked Solution
Apply the idea

The point of intersection on the graph occurs at (4.5, 800), meaning that in 4.5 months, Tyson will have saved \$800. Therefore Tyson will have saved enough money before the phone is released.

Reflect and check

While the model equation of Tyson's savings is linear, in reality, he probably puts money away once per month or once per week, depending on how often he gets paid.

So although the point of intersection is at x = 4.5 months, Tyson might not actually reach \$800 in savings until the end of the 5th month.

Example 5

Gordiano made two trips to a flower shop to purchase roses and sunflowers. On his first trip, he purchased 4 roses and 4 sunflowers and paid \$12. The following day, Gordiano went back to the flower shop and purchased 12 roses and 8 sunflowers for \$16. Graph the system of equations and interpret the solution to the system. Use appropriate labels and scales for the axes.

\begin{cases}4x + 4y=12\\ 12x+8y=16 \end{cases}

Worked Solution
Create a strategy

First, define the variables.

Based on the problem, we see that each trip to the flower shop is represented in its own equation. Gordiano buys 4 roses, then 12, so the expressions 4x and 12x indicate the cost for the roses. Gordiano buys 4 sunflowers, then 8 sunflowers, so the terms 4y and 8y indicate the cost of the sunflowers.

Let x= the cost per rose and let y= the cost per sunflower.

Now that we have defined the variables, we can convert the equations to slope-intercept form and graph them on the same coordinate plane.

Apply the idea

Convert the first equation to slope-intercept form:

\displaystyle 4x+4y\displaystyle =\displaystyle 12First equation
\displaystyle 4x+4y-4x\displaystyle =\displaystyle 12-4xSubtract 4x from both sides
\displaystyle 4y\displaystyle =\displaystyle 12-4xCombine like terms
\displaystyle \dfrac{4y}{4}\displaystyle =\displaystyle \dfrac{-4x}{4}+ \dfrac{12}{4}Divide both sides by 4
\displaystyle y\displaystyle =\displaystyle -x + 3Evaluate the division

Convert the second equation to slope-intercept form:

\displaystyle 12x+8y\displaystyle =\displaystyle 16Second equation
\displaystyle 12x+8y-12x\displaystyle =\displaystyle 16-12xSubtract 12x from both sides
\displaystyle 8y\displaystyle =\displaystyle 16-12xCombine like terms
\displaystyle \dfrac{8y}{8}\displaystyle =\displaystyle \dfrac{-12x}{8}+ \dfrac{16}{8}Divide both sides by 8
\displaystyle y\displaystyle =\displaystyle -\dfrac{4}{3}x + 2Evaluate the division
Flower shop purchases
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\text{Cost per rose (dollars)}
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\text{Cost per sunflower (dollars)}

Each function represents the per-item costs of each type of flower on two different days. The solution to the system of equations is (-3,6), meaning that the cost per rose is -\$3, and the cost per sunflower is \$6. Since the cost cannot be negative, this is a non-viable solution to the system of equations. This means the per-item cost of the roses, or the per-item cost of the sunflowers (or both) must be different from one day to the next.

Idea summary

The solution to a system of linear equations is the ordered pair of the point of intersection of the lines.

Systems of equations may have one solution, no solutions, or infinitely many solutions.

Graphing a system can help to determine the number of solutions a system will have:

  • One solution: the lines will intersect at a single point
  • No solution: the lines will be parallel
  • Infinitely many solutions: the lines will be the same line

Outcomes

A.CED.A.3

Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context.

A.REI.C.6

Solve systems of linear equations exactly and approximately (e.g. With graphs), focusing on pairs of linear equations in two variables.

A.REI.D.11

Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the solutions approximately.

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