STEM 698 Algebra Initiative
HW 5 Due Tuesday April 30, 2013
1. Read “Rethinking the Teaching of Systems of Equations: Highlighting multiple
representation of systems of equation deepens student understanding” by Jerome Proulx
et al. from the Mathematics Teacher March 2009.
a. Solve one of the problems (your choice) in Fig. 2. Remember to define your variable
(or variables) carefully.
Problem: The school cafeteria offered two difference meals at lunch. Three times more
hamburgers than pizzas were served. If, in total, 212 meals were served, how many
hamburgers and how many pizzas were served?
One possible solution.
Let p be the number of pizzas sold. Then 3p is the number of hamburgers sold. So
4p = 212. Dividing both sides by 4, we obtain p = 53. So there were 53 pizzas sold and
159 hamburgers were sold.
Problem. Two children have a stamp collection. Alex has 37 more stamps than Josie. If,
together, they have 181 stamps, how many stamps does each have in his or her
collection?
One possible answer.
Let J be the number of stamps that Alex has. Then Alex has J + 37 stamps. So
J + (J + 37) = 181. Solving for J, you get 2J = 144. So J = 72. So Josie has 72 stamps
and Alex as 109 stamps.
b. The authors feel that algebraic problems like those in figure 2 “should not be included
in a unit on system of equation, because such problems do not show the relevance of
using systems of equations.” Do you agree? Briefly explain why or why not.
While it would inappropriate to show only questions such as those in Figure 2, I think
that showing students a few problems that could be solved both ways gives an
opportunity to compare and contrast the methods.
c. Solve the problem in Fig. 3. Remember to define your variables carefully.
At the museum, three children accompanied by an adult have to pay $54, whereas two
children in the company of two adults have to pay $60. If the cost of admission is
different for adults and children, what is the price for one adult? What is the price for
one child?
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Let x be the cost of admission (in dollars) for a child and let y be the cost of admission for
an adult. Then
3x  y  54

2 x  2 y  60
I would solve this by substitution or elimination. I will use substitution this time. Solving
for y in terms of x in the first equation, you get y = 54 − 3x. Now substitute into the
second equation to get 2x + 2(54 − 3x) = 60. Solving this equation, you find that x = 12
and so y = 54 − 3(12) = 18. Therefore the price one child is $12 and the price one adult
is $18.
d. How does the problem in Fig. 3 differ from the problems in Fig. 2?
It is much more difficult to do the problem in Fig. 3 with a single variable. The problem
in Fig. 3 lends itself naturally to using two variables.
e. Solve the problem in Fig. 4.
 y  3x  8

4 x  6 y  12
Substituting, the first equation into the second, we get 4 x  6(3x  8)  12 . So
4x 18x  48  12
14 x  36
x  187
y  3( 187 )  8   72
 4( 187 )  6( 72 )  847  12
f. Solve the problem in Fig. 7. I would use a calculator, but you could do it by hand.
Show your work either way. You will need the Pythagorean theorem to find the distance
between points. If ( x1 , y1 ) and ( x2 , y2 ) are two points in the plane, the distance between
them by the Pythagorean Theorem is
( x2  x1 )2  ( y2  y1 )2 .
You first have to find the intersection points. I would probably find them using a
calculator. They are A = (3,6), B = (5,−2), and C = (−5,3).
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Dist(A,B) =
22  82  68  2 17
Dist(A,C) =
82  32  73
Dist(A,B) = 102  52  125  5 5
No, they do not form an isosceles triangle because no pair of sides have equal lengths.
g. Finally, in your words, in about a paragraph, describe the central points of this article.
2. Do p.353 5, 6, 7, 8 in CME Algebra I.
5. Derman has the best time.
6. Graph on next page
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Here is the graph in yards and seconds:
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7. Derman will overtake Tony. In units of yards and minutes, the equation for Derman’s
line is y  (440 /1.375) x . The equation for Tony’s line is y  20  (440 /1.5) x . To find
when Derman overtakes Tony, set these equations equal and solve for x. You will get x
= ¾ minutes (i.e. 45 seconds).
If you use units of yards and seconds, the formulas are y  (440 / 82.5) x and
y  20  (440 / 90) x respectively.
8. Derman takes 1.375 minutes; Tony takes 420/(440/1.5) ≈ 1.432 minutes. Note that
Tony is only running 420 yards in this race. (In terms of seconds, Derman takes 82.5
seconds, and Tony approximately 85.9 seconds.)
3. Do p. 363 Question 6 in CME Algebra I.
Tony made an error using the distributive law in going from line 1 to line 2.
4. Do p. 379 Question 5c and d (Note: do only c and d).
c. no solution
5d.
2 x  3 y  1

4 x  2 y  10
Multiply the first equation by 2.
4 x  6 y  2

4 x  2 y  10
Subtract the equations:
8 y  8
So y  1
Usually, we substitute back into one of the equations and solve for x.
I’ll choose the second equation.
4 x  2(1)  10
4x  8
x2
The solution is x = 2 and y = −1
I usually check my answer in the other equation 2(2)+3(−1) = 1 which checks.
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