Problem Solving Strategies
VI. Convert to algebra
Algebra is a powerful language that can help to organize information and
guide you to a solution. In many mathematics textbooks, it is featured as
part of a five-step problem solving process:
1.
2.
3.
4.
5.
Read the problem.
Choose a variable.
Write the equation.
Solve the equation.
Check your solution.
This can give the mistaken impression that writing down the correct
equation (#3 above) is an easy transitional step. It’s not. Representing the
problem with an equation is the hardest part of this process. However, artful
use of a guess-and-check method or the drawing of a diagram can be used to
help steer you toward the proper equation. Once this is done, the tools of
algebra often make solving the equation a straightforward process.
Example: The perimeter of a rectangular playing field measures 504 yards.
It is almost twice as long as it is wide. Specifically, its length is 6 yards
shorter than twice its width. What is the area?
Solution: To find the area, we’ll need both dimensions to multiply out. But
we know neither the length nor the width as of yet. Since these two
dimensions are related by the given information, let’s label both as variables
to be found and turn this information into an equation to solve. So we make
L = length, and W = width (both in yards) as our variables. The most direct
relationship tells us that L is 6 yards shorter than twice W, or in equation
form, L = 2W − 6 . Unfortunately, this is not enough information to solve for a
value of either variable. However, we have additional information: the
perimeter of the field is 504 yards. This is also related to the dimensions of
€ the field. The perimeter measures twice the length plus twice the width, so
we have that 2L + 2W = 504 . We can use the first equation to substitute in
for L in the second equation:
2( 2W − 6) + 2W = 504
4W −12 + 2W = 504
6W −12 = 504
6W = 516
W = 86
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Problem Solving Strategies
So the width of the field is 86 yards, whence its length is L = 2( 86) − 6 = 166
yards. (We can verify that the perimeter is indeed 2(166) + 2( 86) = 504 to
check our work so far.) It follows that the area of the field must be
166 yards × 86 yards = 14276 square yards. ☐
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€
Example:
A group of Japanese exchange students visited a convalescent
€
home to spend time with the residents and demonstrate how to fold origami
models to the seniors. Groups of students and seniors sat together at tables.
At each table there was either 1 student with 3 seniors or 2 students with 4
seniors. If there were 23 students and 61 seniors there all told, how many
tables were they using?
Solution: Since the tables are of two types, one with 1 student and 3 seniors
and one with 2 students and 4 seniors, the number of tables the groups were
using is the sum of the numbers of either type. So it makes sense to use
variables to represent the number of tables of either type:
x = number of tables at which 1 student sits with 3 seniors;
y = number of tables at which 2 students sit with 4 seniors.
Clearly we need to find the value of x + y , but how? The only other
information at hand is that there were 23 students and 61 seniors, so perhaps
we can use this to squeeze out another equation relating x and y. Since each
table counted by x has one student whereas each table counted by y has 2
€
students, it follows that the number of students sitting at tables of the first
type is x itself while the number students sitting at tables of the second type
is 2y. Thus, x + 2 y = 23 . Similar logic allows us to write a second equation
that uses information about how many seniors were sitting at tables of either
type. In fact, since 3 seniors sit at tables of the first type, there are 3x seniors
accounted for in this way; since 4 seniors sit at tables of the second type,
€
there are 4x seniors accounted for in the other way. Thus, 3x + 4 y = 61 is a
second equation involving the two variables. This allows us to solve for the
variables:
x + 2 y = 23
3x + 4 y = 61
−2x −€4 y = −46
3x + 4 y = 61
x = 15
⇒
€ x into the first of our equations,
Substituting this value of
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Problem Solving Strategies
15 + 2 y = 23
2y = 8
y=4
So there were 15 tables of the first type and 4 of the second type, for a total of
x + y = 15 + 4 = 19 tables in use by the students and seniors. . ☐
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€
Example: A pet store sells salt water for fish tanks. Unfortunately, the new
worker mixed a solution that is too weak for the fish. He’s made 150 pounds
of a 4% salt solution when he needed to make a 7% solution. What to do? He
can either add salt to the solution or evaporate some of the water; both
options will increase the concentration of salt. Determine how much salt he
would need to add, and alternatively, how much water he would need to have
evaporate, in order to get a 7% salt solution.
Solution: Let’s begin by trying to add some salt. If we add x pounds of salt to
the 4% solution he has already, he will have 150 + x pounds of salt water with
more salt in it than he had before. How much more? If the original 150pound solution was 4% salt, that represents 4% ×150 = 0.04 ×150 = 6 pounds
of salt. So the new solution has 6 + x€pounds of salt in it. Since this should
represent 7% of the weight of the new solution, we now know that
€
€
7% × (150 + x ) = 6 + x
0.07(150 + x ) = 6 + x
10.5 + 0.07x = 6 + x
4.5 = 0.93x
x = 4.84
That is, he must add 4.84 pounds of salt to his 4% solution to obtain a 7%
solution.
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To deal with the evaporation problem, suppose he lets y pounds of water
evaporate from the 150 pounds of the 4% solution he has already. Then he’ll
have a solution weighing only 150 − y pounds containing the same amount of
salt. The new solution will then contain 7% salt, but we recall that there was
6 pounds of salt in the 4% solution to begin with, so
€
7% × (150 − y ) = 6
0.07(150 − y ) = 6
10.5 − 0.07 y = 6
10.5 − 6 = 0.07 y
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Problem Solving Strategies
0.07 y = 4.5
y = 64.29
This means that he should allow the solution to lose 64.29 pounds of water to
evaporation. The resulting 150 − 64.29 = 85.71 pound solution will have the
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$4.40 4.5 hr
proper 7% concentration of salt.
⋅
= $18.70. ☐
1 hr
1
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