Teaching Algebra
i
In 1953, I was recruited for my high
school’s math team. That event
changed my life. No longer simply
repeating techniques shown previously, I was taught to focus intensively
on problem solving. As a result, I
majored in mathematics in college.
For much of my teaching career,
I coached math teams and was often
amazed at the number of well-known
mathematics teachers I met whose
backgrounds were similar to mine. My
teaching career, like many of theirs, was
built on using rich, engaging problems
to teach content. One key to this approach is to allow students to create
their own ways to solve the problems,
which helps them actively own the
concepts they use. I acted as a moderator, encouraging different individuals
to explain aloud different solutions to
each problem, adding my comments
only as needed. Material was automatically reviewed intensively by eliciting
several concepts for every problem. In
addition, many students participated
voluntarily and often made surprising connections. See “The Value of
Multiple Solutions” (Kalman 2004) for
an amplification of this point.
In 1963, I was asked to teach a prealgebra course for which there was no
text. One of the strategies I used with
Richard Kalman, rkalman@
moems.org, is the
executive director of the
Mathematical Olympiads
for Elementary and Middle
Schools. He has loved problem solving
since 1954 on all levels from grade 4
through grade 12.
334
the students was to explore algebra 1
problems nonalgebraically. The hope
was that by helping students visualize
the problems conversationally, they
might be better prepared to understand
the algebraic techniques required.
Algebra is, among other things, a
shorthand way to express quantitative
reasoning. To me, elementary algebra
actually is common sense written in
symbols. For a student, its concise
language can obscure the actual thinking that occurs when solving a problem. A series of equations that forms
a solution tells us how to proceed but
is silent about why we perform these
particular steps. A clear conversational
approach that examines the rationale
behind each step can set a good foundation for the technical work to follow.
In 1994, I retired from teaching
to join the Mathematical Olympiads
for Elementary and Middle Schools
(MOEMS, or Math Olympiads) but
continued to work frequently with
both students and teachers. Since 1979,
the Math Olympiads has made a point
of eliciting nonalgebraic solutions to
algebra 1 problems from students in
grades 4–6 (Lenchner 1997; MOEMS
1995–2005). As educators, one of our
most compelling obligations is to build
a strong foundation for future studies.
In keeping with that obligation, this
article will illustrate ways for the classroom teacher to convert algebraic solutions to verbal problems into conversational solutions that can be understood
by students in the lower grades. Three
reasonably typical verbal problems that
either appeared as or resemble Math
Olympiad tasks are presented here.
Mathematics Teaching in the Middle School
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Vol. 13, No. 6, February 2008
Copyright © 2008 The National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved.
This material may not be copied or distributed electronically or in any other format without written permission from NCTM.
NICKELS AND QUARTERS
Mrs. Bailey has the same number
of nickels as quarters. The value
of the quarters is $1.80 more than
the value of the nickels. What is
the total value of her nickels and
quarters? (Lenchner 1997, p. 70)
Algebraic solution: Suppose she has
n nickels. Then she has n quarters.
Their values are 5n cents and 25n
cents, respectively. Expressing the second sentence of the problem in cents
produces equation (1):
(1)
25n = 5n + 180
Subtracting 5n from each side, we get
(2)
20n = 180.
Dividing both sides by 20, we get
(3)
n = 9.
Therefore, the total value is
(4) 25n + 5n = 25(9) + 5(9)
= 225 + 45 = 270.
Mrs. Bailey has a total value of $2.70.
How can we help students employ
the algebraic thinking explained above
before they have learned algebraic techniques? We need to translate each step
above into conversational terms to plant
the seeds of appropriate procedural understanding in their minds. I believe it is
valuable for students in the earlier grades
to experience the pathways of thought
that they will need subsequently.
without Algebra
Richard S. Kalman
This same conversational approach also teaches students to solve
problems, a major goal of mathematics education according to Principles
and Standards for School Mathematics
(NCTM 2000) and many educators.
Learning to think mathematically
now will help them form a stronger
foundation for high school mathematics and college entrance tests later
(Bay-Williams 2004).
Verbal (nonalgebraic) solution: First,
let us examine the actions behind each
equation. The first two sentences of
the problem state that the two types
of coins are equal in quantity and that
they differ by $1.80. Equation (1) is a
combination of those facts. Equation
(2) is the result of subtracting the 5n
from 25n to leave 20n. Equation (3) is
the result of dividing by 20. In equation
(4), we merely use the solution to equation (1) to generate the requested value.
To parallel the algebraic steps from
above, we want to translate each step
from algebra to verbal descriptions.
Since the numbers of quarters and
nickels are the same, we can pair
them. In each pair, the value of the
quarter is 20 cents more than the
value of the nickel. Consider all pairs:
The given $1.80 is the product of the
number of pairs and 20 cents. Thus,
Mrs. Bailey has 9 pairs. The value of
the nine quarters is $2.25, the value of
the nine nickels is $0.45, and the total
value of her quarters and nickels is
$2.70. Checking, the difference
between $2.25 and $0.45 is $1.80.
Some additional comments can be
helpful to the teacher.
Vol. 13, No. 6, February 2008
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Mathematics Teaching in the Middle School
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• Sometimes students’ short answers
make little sense. For example, if
asked for, say, a number of buses,
they may reply, “3 1/2.” Routinely
requiring an answer to be a full
sentence forces students to consider the context. In this case, they
are unlikely to respond, “There are
3 1/2 buses.”
• Sometimes students answer questions they like instead of the ones
you ask. Training them to answer
using the wording given in the
question encourages them to check
for appropriateness.
• Some students may notice that considering a single pair of coins allows
us to use the distributive law twice.
The first instance occurs when $1.80
is divided by the difference in values,
which is 20 cents. The second can
occur if we choose to multiply the
number of pairs, 9, by the value of
1 pair, 30 cents, instead of translating equation (4). Such students are
using mathematical thinking instead
of just performing mechanical operations and should be encouraged
to explain it to everyone.
• Traditionally, students are asked
only to find the value of the variable. This could lead a teacher
to suggest erroneously that the
variable should always represent
the quantity to be found. A nice
feature of problem 1 is that the
variable and the answer are two
different quantities, which forces
the student to analyze the problem
carefully before assigning a meaning to their variable.
• Once students solve for a variable, they may stop automatically.
Requesting an answer that is an
outgrowth of the variable can help
train students to keep working and
perhaps to look back at the question before answering.
• A great feature of coin problems
is that they ask students to supply
values not stated explicitly. In this
336
case, students had to furnish the
values of both coins and had to use
them properly.
• Many students at first will employ
some version of guess and check,
perhaps using a table. Over the
years, I found that this is typical for
beginning problem solvers as they
slowly develop the ability to think
mathematically. Even experienced problem solvers tackling an
unfamiliar problem may just “try
numbers” at first to get the feel for
the problem.
BENJI’S BOXES OF MARBLES
Benji lines up several boxes. The
first box contains 1 marble, the
second box contains 3 marbles, the
third box contains 5 marbles, and
so on, with each box containing 2
more marbles than the previous box.
The sum of the marbles in the last
five boxes is 85. What is the sum of
the marbles in the last two boxes?
This problem is really a dressed-up
version of “Find each of 5 consecutive
odd integers if their sum is 85,” which
is a standard algebra 1 problem. Using
marbles in a box allows students to
visualize the situation. Drawing a diagram or using manipulatives are other
solution strategies.
Algebraic solution 1: If we assume
that the first of the five boxes contains
x marbles, then the contents of the
other boxes can be represented by x +
2, x + 4, x + 6, and x + 8.
(1)x + (x + 2) + (x + 4) + (x + 6) +
(x + 8) = 85
(2)5x + 20 = 85
(3)5x = 65
(4)x = 13
(5)x + 6 = 19 and x + 8 = 21
The sum of the marbles in the last two
boxes is 40. To check, the five boxes
contain a total of 13 + 15 + 17 + 19 +
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Vol. 13, No. 6, February 2008
21 = 85 marbles, and each box contains 2 more marbles than the one box
before it. This solution could have been
shortened slightly by letting the variable
represent the contents of the last box.
In my workshops with teachers, I
have noticed that many of them have
difficulty envisioning a nonalgebraic
solution to such a problem. The difficulty seems to increase substantially
when they are confronted with a standard coin or motion problem. Perhaps
we have been taught too well! At
this point, you, the reader, may want
to test your ability to create a purely
verbal solution to problem 1 before
reading further. Perhaps when you
reach the other problems, you might
wish to try your hand with them, too.
Verbal solution 1: Consider only the
last five boxes. Suppose we skip the
first box, remove 2 marbles from the
second box, 4 marbles from the third,
6 from the fourth, and 8 from the
last. Then all five boxes would have
the same number of marbles, and the
total number would be 65. Thus, each
box would have 13 marbles. Restore
the marbles we removed, and the last
two boxes would contain 19 and 21
marbles. The sum of the marbles in
the last two boxes is 40.
As in the algebraic solution, we
added the 2, 4, 6, and 8; subtracted
last five boxes. Since there are an odd
number of boxes, the average of their
contents is the same as the number
of marbles in the middle box. To
understand this, picture a transfer of
2 marbles from the fourth box to the
second and 4 marbles from the last
box to the first. As a result, the total
number of marbles is unchanged, but
every box now has the same number
of marbles. Thus, the middle box contains one-fifth of 85 marbles, or 17,
marbles. The last two boxes originally
contained a total of 40 marbles.
The two additional problems below can extend Benji’s box problem.
the resulting 20 from 85; divided the
65 by 5 to get the contents of the first
box, 13 marbles; and then added back
to the 13 both the 6 and the 8 to get
the contents of the last two boxes. It
mirrors the algebraic solution perfectly.
Interestingly enough, a different conversational approach also mirrors the
algebraic one. The following solution
appears to be similar but is conceptually opposite. In the first solution, we
“leveled” the numbers by removing
marbles. In the second solution, we
will choose the simplest related case
and then add marbles.
Verbal solution 2: Each box contains
2 more marbles than its predecessor.
Choose the simplest case: Suppose the
boxes contain 0, 2, 4, 6, and 8 marbles.
Our total of 20 marbles, then, is 65
short of the given total of 85. To maintain that common difference of 2, distribute the 65 marbles equally to each
of the 5 boxes, adding 13 to each. Then
the last two boxes contain 19 and 21
marbles, and their sum is 40 marbles.
Our perception, and therefore our
strategy, may have changed radically,
but our procedure did not. In all three
solutions above, the first step was 2 + 4 +
6 + 8 = 20, our second step was 85 – 20 =
65, our third step was 65 ÷ 5 = 13, our
fourth step was 13 + 6 = 19 and 13 + 8 =
21, and our fifth step was 19 + 21 = 40.
Algebraic solution 2: Suppose the
third box contains y marbles. Then,
in order, the five boxes contain y – 4,
y – 2, y , y + 2, and y + 4 marbles,
respectively.
(1)(y – 4) + (y – 2) + (y) + (y + 2) +
(y + 4) = 85
(2)5y = 85
(3)y = 17
(4)y + 2 = 19 and y + 4 = 21
The sum of the marbles in the last
two boxes is 40. Subtly hidden in this
solution is an interesting principle.
Given an odd number of equally
spaced terms (“arithmetic sequence”),
the middle term is also the average of
all the terms. For example, the average
of 2, 5, and 8 is 5; the average of 35,
40, 45, 50, and 55 is 45; and the average of 90, 91, 92, 93, 94, 95, 96, 97,
and 98 is 94. In this case, y represents
both the number of marbles in the
middle box and the average number
of marbles in the five boxes. For an
even number of terms, the principle is
slightly different.
Verbal solution 2: Solution 2 uses
an approach that I have seen both
students and teachers use increasingly
in recent years. Although it is easy to
show algebraically, I have only seen
it used verbally. Consider only the
Vol. 13, No. 6, February 2008
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1. Suppose Benji’s boxes contained, in
order, 1, 4, 7, . . . marbles, and he
has 20 boxes, each with three more
than the box before it.
a. How many marbles are in the
last box?
b. How many marbles are in all the
boxes, in all?
2. In a triangular array, each row has
two more Xs than the row above it.
How many Xs are in the first ten
rows?
X
XXX
XXXXX
XXXXXXX
.
.
.
CD AND DVD SALE
Two-variable problems are also
important.
CDs and DVDs are on sale. Amanda buys 4 CDs and 1 DVD for $38,
and Carlos buys 1 CD and 4 DVDs
for $47. Each of the CDs is one
price, and each of the DVDs is another price. If Kobe buys 1 CD and
1 DVD, how much does he pay?
Mathematics Teaching in the Middle School
337
her one DVD. Therefore, each DVD
costs $10, and Kobe pays $17 for 1 CD
and 1 DVD. No difference in ease of
solution, whether algebraic or verbal,
would exist if we were to solve first for
the DVD instead of the CD.
Algebraic solution 2: If we subtract
the two original equations, we get
3d – 3c = 9, which simplifies to d – c =
3. Thus, d = c + 3. Substituting into
4c + d = 38 produces
4c + (c + 3) = 38
5c + 3 = 38
5c = 35
c=7
Continue as in algebraic solution 1 to
derive d = 10 and the total of $17 for
Kobe’s purchase.
Algebraic solution 1: Suppose that
each CD costs c dollars, and each
DVD costs d dollars. Then:
(1) Given
4c + d = 38
(2) Given c + 4d = 47
(3) Multiply (1) by 4: 16c + 4d = 152
(4) Multiply (2) by 1: c + 4d = 47
(5) Subtract (4)
from (3):
15c = 105
(6) Divide (5) by 15: c = 7
Each CD
costs $7.
(7) Repeat (1) or (2): 4c + d = 38
(8) Replace c by 7: 4(7) + d = 38
(9) Evaluate:
28 + d = 38
(10)Subtract 28 d = 10
from each side: Each DVD
costs $10.
Kobe pays $17 for one CD and one
DVD.
You can check the answer by
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reconstructing Amanda’s and Carlos’
purchases from the prices. Again, the
reader is urged to try to develop a
nontechnical explanation before reading the one provided below. The mass
of equations above may make it appear
that verbalizing the solution will prove
quite difficult, but actually the opposite
seems to be true. This verbal solution
looks to be much simpler, clearer, and
swifter. Algebra is often the best way to
handle a problem—but not always.
Verbal solution 1: If Amanda makes
her purchase four times, then both she
and Carlos would buy 4 DVDs. In
this case, she would pay $152 for 16
CDs and 4 DVDs, while Carlos pays
$47 for 1 CD and 4 DVDs. Thus, she
would pay an additional $105 for 15
additional CDs. Each CD costs $7.
Because Amanda paid $28 for her 4
CDs, she paid the remaining $10 for
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Verbal solution 2: Compare Carlos’
purchase to Amanda’s. He pays $9
more than she, but in effect he swaps
3 of her CDs for 3 DVDs. Thus, each
DVD must cost $3 more than each
CD. If Amanda had bought a fifth
CD instead of her 1 DVD, she would
have paid $3 less than the $38 she
really pays. That is, her 5 CDs would
have cost $35, making the price $7 per
CD. Since Amanda pays $28 for her 4
CDs, she pays the remaining $10 for
her 1 DVD. Kobe therefore pays $17
for 1 CD and 1 DVD.
Algebraic solution 3: If we add the
equations 4c + d = 38 and c + 4d = 47,
we get 5c + 5d = 85. Dividing both
sides by 5 yields c + d = 17. Thus, Kobe
pays $17 for 1 CD and 1 DVD. This
solution, somewhat less routine for algebra students, is wonderfully elegant.
It drives directly toward the required
answer much more efficiently than the
more usual approaches used in algebraic solutions 1 and 2. Notice that it
tells exactly what Kobe paid, without
ever revealing the price of either item,
a fact we did not have to know.
Verbal solution 3: This method
is very simple to translate. Suppose
that their mother gives them enough
money for all the purchases. She gives
Amanda and Carlos a total of $85 to
buy 5 CDs and 5 DVDs. But Kobe
only wants 1 CD and 1 DVD. Then
the mother gives Kobe one-fifth as
much, $17. Kobe pays $17 for 1 CD
and 1 DVD.
SUMMARY
As mathematics teachers, we have
many responsibilities. Among the
most important is that of providing
the firmest possible foundation for future studies and for adulthood. In this
article, three facets were addressed.
Foreshadowing
By using problems similar to those
that students will see in a later grade
and helping them find their own way
to a solution (Lenchner 2005, p. 11;
Pólya 1957), we begin to embed the
thinking processes in their minds.
This can allow students to acquire
a greater clarity and more intuitive
comprehension when studying the appropriate procedures in algebra 1.
As an aside, I also recommend that
algebra 1 teachers themselves have
students solve verbal problems conversationally as well as algebraically
so as to impart the reasoning behind
the technical steps (Lenchner 2005,
pp. 103–9). The more ways a student
looks at something, the deeper the
understanding (Kalman 2004).
Problem Solving
I believe that problem solving, not
memorization, is the cornerstone
of all mathematics. It seems to me
that mathematics grew as the result
of people looking for better ways to
solve challenging problems. What
we learned in school was usually the
polished final result of their efforts,
stripped of the very things that fascinated them so. I believe very strongly
that we should do all in our power to
put back the mystery. Students respond much better to the journey than
they do to the arrival. You need not
take my word for it; NCTM named
Problem Solving as its first Process
Standard in Principles and Standards
(NCTM 2000).
There are other benefits. Over the
years, I found several things to hold
true. Students that were assigned a
steady diet of demanding nonroutine
problems tended to see standardized
tests as nonthreatening and performed up to their abilities. Students
who reasoned their own way through
a problem retained principles and
relationships more readily and fully.
Students who were fed a steady diet
of nonroutine problems knew they
were shown respect. With experience,
students improved in their problemsolving abilities, which resulted in
greater self-confidence and greater
mathematical thinking.
Mathematical Thinking
Often, young students are satisfied
just to get “the answer.” To them,
guess and check works well enough.
But later on when they are in high
school courses and taking college entrance exams, success is possible only
by thinking mathematically. Questions often involve multiple principles,
long chains of reasoning, and sometimes unexpected manipulations. Rote
memorization of techniques can move
the student toward success but only
part of the way. The student must
be able to reason his or her own way
through to a solution. By our accepting their approaches to problems like
those above, we help them become
active learners and own the concepts,
and we allow them to embed the algebraic reasoning they will need later. By
encouraging several students to share
with classmates several approaches to
a problem, we expose every student to
many concepts within each problem,
Vol. 13, No. 6, February 2008
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which intensifies the review of each
concept over time, and we accustom
students to speak publicly.
An interesting by-product of
building my teaching on appropriate
nonroutine problems and encouraging
the students to provide all the solutions was the timing, especially for
average and below-average classes. For
many problems, they volunteered two
or three different methods. This took
time. For the first third of the year,
my classes usually fell far behind the
other classes. As they began to “catch
the feel” of thinking mathematically,
their work sped up. By the end of the
second third of the year, they always
caught up to the other classes. By the
end of the school year, they had covered all the required material in four
to six weeks less time than those other
classes, and they felt well prepared for
the formal end-of-year exams.
REFERENCES
Bay-Williams, Jennifer, and Sherri
Martinie. “Families Ask: What Does
Algebraic Thinking Look Like?”
Mathematics Teaching in the Middle
School 9 (November 2004): 198–99.
Kalman, Richard. “The Value of Multiple
Solutions.” Mathematics Teaching in
the Middle School 9 (November 2004):
174–79.
Lenchner, George. Math Olympiad Contest
Problems for Elementary and Middle
Schools. Bellmore, NY: MOEMS, 1997.
———. Creative Problem Solving in School
Mathematics Schools. Bellmore, NY:
MOEMS, 2005.
Mathematical Olympiads for Elementary
and Middle Schools (MOEMS). Various Contests. Bellmore, NY: MOEMS,
1995–2005.
National Council of Teachers of Mathematics (NCTM). Principles and Standards for School Mathematics. Reston,
VA: NCTM, 2000.
Pólya, George. How to Solve It. 2nd ed.
Garden City, NY: Doubleday and Co.,
1957. l
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