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For Teachers!
Systems of Linear Equations Modeling
Notes on teaching the lesson:
Lesson 1
Focus and Review:
Starting the lesson with an introductory problem that the students can relate to is a good idea. Try one like this:
The junior class prom committee is selling boxes of candy to raise money for a dance hall. There are two box sizes.
One box cost $2.00 and the other cost $5.00. The committee sold 500 boxes. They raised $1450. How many boxes
were sold for $2.00 and how many were sold for $5.00?
Letting the students explore this problem on paper as well as with a calculator is a good way to enhance their curiosity.
Statement of Objectives:
Let the students know that one way to solve this problem is by graphing.
Teacher Input:
Ask the students what is the problem asking. The problem is asking: How many boxes were sold for $2.00 and how
many were sold for $5.00? We have two unknowns, x and y. Let x represent the number of boxes sold for $2.00, and y
represent the number of boxes sold for $5.00.
What do we know about x and y? We know that x + y = 500, and that $2.00x+$5.00y=$1450. If we graph these two
equations on the same coordinate plane, we will find our answer to how many boxes were sold for $2.00 and how
many were sold for $5.00?
Graph the two lines on the same coordinate plane, finding the point of intersection. The point where the two lines
intersect is your answer. What is the point of intersection? (350, 150) If x equals 350 and y equals 150, how many
boxes were sold for $2.00 and how many boxes were sold for $5.00? There were 350 boxes of candy sold for $2.00
and 150 boxes of candy sold for $5.00.
Put your equations into the systems of equations calculator.
Now is a good time to tell the students that they have just solved a system of equations. Let the students know that
these two equations had the same variables and that
this is what made them a system. The first variable is the # of boxes for
$2 and the second variable is the # of boxes for $5.
Guided Practice:
Now let us try another example:
You are starting a bumper sticker business. The start up cost is $150.00. Each bumper sticker cost $3.00 to make.
You are selling each bumper sticker for $5.00. How many bumper stickers must you sell in order to break even?
To break even means to reach the point where the income is equal to the cost. How many would you need to sell to make
a $200 profit? Let us work out this problem.
The cost of making x bumper stickers is represented
by y = 150 + 3x
The income from x bumper stickers is represented
by y = 5x
What are the variables in this system? x represents how many bumper stickers you need in order to break even. y
represents the amount of money you need to break even.
By putting these two equations in the systems of equations calculator, we will find the point where you will “break
even.” These equations are written in slope- intercept form. You should use the calculator
that fits this form of equation. The point of intersection is (75,375). That means that you need to sell 75 bumper stickers and make $375 in order to
“break even”.
In order to make a $200.00 profit, you have to change your equations. The first equation becomes y = 350 + 3x. Ask
the students to tell you why the 150 changed to 350. The second equation remains the same unless you change the price
of the bumper stickers. By putting these two equations into the calculator, we will find how many bumper stickers you
must sell in order to make a $200.00 profit.
y = 350 + 3x
y = 5x
(175,875) You will need to sell 175 bumper stickers and make $875 in order to make a $200 profit.
Independent Practice:
You are renting a 15-passenger van for the day after graduation trip to the beach. One company
charges $55.49 per day plus $.25 for each mile and another company charges $71.30 a day plus $.07 a mile. You need to determine which
company would best serve your needs in each of these different situations.
Explain why?
A. You are going to a beach 80 miles away.
B. You are going to a beach 50 miles away and are staying 2 days.
C. You are going to tour a major city 75 miles away and the tour is 15 miles.
Encourage the students to use the System of Equations Calculator.
Answers to A, B, and C
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by the National Science FoundationOpinions expressed are those of the authorsand not necessarily those of the National Science Foundation. |