

We now need to discuss the section that most students
hate. We need to talk about applications
to linear equations. Or, put in other
words, we will now start looking at story problems or word problems. Throughout history students have hated
these. It is my belief however that the
main reason for this is that students really don’t know how to work them. Once you understand how to work them, you’ll
probably find that they aren’t as bad as they may seem on occasion. So, we’ll start this section off with a
process for working applications.
Process for Working
Story/Word Problems
 READ THE PROBLEM.
 READ THE PROBLEM AGAIN. Okay, this may be a little bit of
overkill here. However, the point
of these first two steps is that you must read the problem. This step is
the MOST important step, but it is also the step that most people don’t
do properly.
You need to read the problem very carefully and as many times as it
takes. You are only done with
this step when you have completely understood what the problem is asking
you to do. This includes
identifying all the given information and identifying what you being
asked to find.
Again, it can’t be stressed enough that you’ve got to carefully read the
problem. Sometimes a single word
can completely change how the problem is worked. If you just skim the problem you may
well miss that very important word.
 Represent
one of the unknown quantities with a variable and try to relate all the
other unknown quantities (if there are any of course) to this variable.
 If
applicable, sketch a figure illustrating the situation. This may seem like a silly step, but
it can be incredibly helpful with the next step on occasion.
 Form
an equation that will relate known quantities to the unknown quantities. To do this make use of known formulas
and often the figure sketched in the previous step can be used to
determine the equation.
 Solve
the equation formed in the previous step and write down the answer to
all the questions. It is
important to answer all the questions that you were asked. Often you will be asked for several
quantities in the answer and the equation will only give one of them.
 Check
your answer. Do this by plugging
into the equation, but also use intuition to make sure that the answer
makes sense. Mistakes can often
be identified by acknowledging that the answer just doesn’t make sense.

Let’s start things off with a couple of fairly basic
examples to illustrate the process. Note
as well that at this point it is assumed that you are capable of solving fairly
simple linear equations and so not a lot of detail will be given for the actual
solution stage. The point of this
section is more on the set up of the equation than the solving of the equation.
Example 1 In
a certain Algebra class there is a total of 350 possible points. These points come from 5 homework sets that
are worth 10 points each and 3 hour exams that are worth 100 points
each. A student has received homework
scores of 4, 8, 7, 7, and 9 and the first two exam scores are 78 and 83. Assuming that grades are assigned according
to the standard scale and there are no weights assigned to any of the grades
is it possible for the student to receive an A in the class and if so what is
the minimum score on the third exam that will give an A? What about a B?
Solution
Okay, let’s start off by defining p to be the minimum required score on the third exam.
Now, let’s recall how grades are set. Since there are no weights or anything on
the grades, the grade will be set by first computing the following
percentage.
Since we are using the standard scale if the grade
percentage is 0.9 or higher the student will get an A. Likewise if the grade percentage is between
0.8 and 0.9 the student will get a B.
We know that the total possible points is 350 and the
student has a total points (including the third exam) of,
The smallest possible percentage for an A is 0.9 and so
if p
is the minimum required score on the third exam for an A we will have the
following equation.
This is a linear equation that we will need to solve for p.
So, the minimum required score on the third exam is
119. This is a problem since the exam
is worth only 100 points. In other
words, the student will not be getting an A in the Algebra class.
Now let’s check if the student will get a B. In this case the minimum percentage is
0.8. So, to find the minimum required
score on the third exam for a B we will need to solve,
Solving this for p
gives,
So, it is possible for the student to get a B in the
class. All that the student will need
to do is get at least an 84 on the third exam.

Example 2 We
want to build a set of shelves. The
width of the set of shelves needs to be 4 times the height of the set of
shelves and the set of shelves must have three shelves in it. If there are 72 feet of wood to use to
build the set of shelves what should the dimensions of the set of shelves be?
Solution
We will first define x
to be the height of the set of shelves.
This means that 4x is width
of the set of shelves. In this case we
definitely need to sketch a figure so we can correctly set up the
equation. Here it is,
Now we know that there are 72 feet of wood to be used and
we will assume that all of it will be used.
So, we can set up the following word equation.
It is often a good idea to first put the equation in words
before actually writing down the equation as we did here. At this point, we can see from the figure
there are two vertical pieces; each one has a length of x. Also, there are 4
horizontal pieces, each with a length of 4x.
So, the equation is then,
So, it looks like the height of the set of shelves should
be 4 feet. Note however that we
haven’t actually answered the question however. The problem asked us to find the dimensions. This means that we also need the width of
the set of shelves. The width is
4(4)=16 feet. So the dimensions will
need to be 4x16 feet.

Pricing Problems
The next couple of problems deal with some basic principles
of pricing.
Example 3 A
calculator has been marked up 15% and is being sold for $78.50. How much did the store pay the manufacturer
of the calculator?
Solution
First, let’s define p
to be the cost that the store paid for the calculator. The stores markup on the calculator is
15%. This means that 0.15p has been added on to the original
price (p) to get the amount the
calculator is being sold for. In other
words, we have the following equation
that we need to solve for p. Doing this gives,
The store paid $68.26 for the calculator. Note that since we are dealing with money
we rounded the answer down to two decimal places.

Example 4 A
shirt is on sale for $15.00 and has been marked down 35%. How much was the shirt being sold for
before the sale?
Solution
This problem is pretty much the opposite of the previous
example. Let’s start with defining p to be the price of the shirt before
the sale. It has been marked down by
35%. This means that 0.35p has been subtracted off from the
original price. Therefore, the
equation (and solution) is,
So, with rounding it looks like the shirt was originally
sold for $23.08.

Distance/Rate
Problems
These are some of the standard problems that most people
think about when they think about Algebra word problems. The standard formula that we will be using
here is
All of the problems that we’ll be doing in this set of
examples will use this to one degree or another and often more than once as we
will see.
Example 5 Two
cars are 500 miles apart and moving directly towards each other. One car is moving at a speed of 100 mph and
the other is moving at 70 mph.
Assuming that the cars start moving at the same time how long does it
take for the two cars to meet?
Solution
Let’s let t
represent the amount of time that the cars are traveling before they
meet. Now, we need to sketch a figure
for this one. This figure will help us
to write down the equation that we’ll need to solve.
From this figure we can see that the Distance Car A
travels plus the Distance Car B travels must equal the total distance
separating the two cars, 500 miles.
Here is the word equation for this problem in two separate
forms.
We used the standard formula here twice, once for each
car. We know that the distance a car
travels is the rate of the car times the time traveled by the car. In this case we know that Car A travels at
100 mph for t hours and that Car B
travels at 70 mph for t hours as
well. Plugging these into the word
equation and solving gives us,
So, they will travel for approximately 2.94 hours before
meeting.

Example 6 Repeat
the previous example except this time assume that the faster car will start 1
hour after slower car starts.
Solution
For this problem we are going to need to be careful with
the time traveled by each car. Let’s
let t be the amount of time that
the slower travel car travels. Now,
since the faster car starts out 1 hour after the slower car it will only
travel for hours.
Now, since we are repeating the problem from above the
figure and word equation will remain identical and so we won’t bother
repeating them here. The only
difference is what we substitute for the time traveled for the faster
car. Instead of t as we used in the previous example we will use since it travels for one hour less that the
slower car.
Here is the equation and solution for this example.
In this case the slower car will travel for 3.53 hours before
meeting while the faster car will travel for 2.53 hrs (1 hour less than the
faster car…).

Example 7 Two
boats start out 100 miles apart and start moving to the right at the same
time. The boat on the left is moving
at twice the speed as the boat on the right.
Five hours after starting the boat on the left catches up with the
boat on the right. How fast was each
boat moving?
Solution
Let’s start off by letting r be the speed of the boat on the right (the slower boat). This means that the boat to the left (the
faster boat) is moving at a speed of 2r. Here is the figure for this situation.
From the figure it looks like we’ve got the following word
equation.
Upon plugging in the standard formula for the distance
gives,
For this problem we know that the time each is 5 hours and
we know that the rate of Boat A is 2r
and the rate of Boat B is r. Plugging these into the work equation and
solving gives,
So, the slower boat is moving at 20 mph and the faster
boat is moving at 40 mph (twice as fast).

Work/Rate Problems
These problems are actually variants of the Distance/Rate
problems that we just got done working.
The standard equation that will be needed for these problems is,
As you can see this formula is very similar to the formula
we used above.
Example 8 An
office has two envelope stuffing machines.
Machine A can stuff a batch of envelopes in 5 hours, while Machine B
can stuff a batch of envelopes in 3 hours.
How long would it take the two machines working together to stuff a
batch of envelopes?
Solution
Let t be the
time that it takes both machines, working together, to stuff a batch of
envelopes. The word equation for this
problem is,
We know that the time spent working is t however we don’t know the work rate
of each machine. To get these we’ll
need to use the initial information given about how long it takes each
machine to do the job individually. We
can use the following equation to get these rates.
Let’s start with Machine A.
Now, Machine B.
Plugging these quantities into the main equation above
gives the following equation that we need to solve.
So, it looks like it will take the two machines, working
together, 1.875 hours to stuff a batch of envelopes.

Example 9 Mary
can clean an office complex in 5 hours.
Working together John and Mary can clean the office complex in 3.5
hours. How long would it take John to
clean the office complex by himself?
Solution
Let t be the
amount of time it would take John to clean the office complex by
himself. The basic word equation for
this problem is,
This time we know that the time spent working together is
3.5 hours. We now need to find the
work rates for each person. We’ll
start with Mary.
Now we’ll find the work rate of John. Notice however, that since we don’t know
how long it will take him to do the job by himself we aren’t going to be able
to get a single number for this. That
is not a problem as we’ll see in a second.
Notice that we’ve managed to get the work rate of John in
terms of the time it would take him to do the job himself. This means that once we solve the equation
above we’ll have the answer that we want.
So, let’s plug into the work equation and solve for the time it would
take John to do the job by himself.
So, it looks like it would take John 11.67 hours to clean
the complex by himself.

Mixing Problems
This is the final type of problems that we’ll be looking at
in this section. We are going to be
looking at mixing solutions of different percentages to get a new
percentage. The solution will consist of
a secondary liquid mixed in with water.
The secondary liquid can be alcohol or acid for instance.
The standard equation that we’ll use here will be the
following.
Note as well that the percentage needs to be a decimal. So if we have an 80% solution we will need to
use 0.80.
Example 10 How
much of a 50% alcohol solution should we mix with 10 gallons of a 35%
solution to get a 40% solution?
Solution
Okay, let x be
the amount of 50% solution that we need.
This means that there will be gallons of the 40% solution once we’re done
mixing the two.
Here is the basic work equation for this problem.
Now, plug in the volumes and solve for x.
So, we need 5 gallons of the 50% solution to get a 40%
solution.

Example 11 We
have a 40% acid solution and we want 75 liters of a 15% acid solution. How much water should we put into the 40%
solution to do this?
Solution
Let x be the
amount of water we need to add to the 40% solution. Now, we also don’t how much of the 40% solution
we’ll need. However, since we know the
final volume (75 liters) we will know that we will need liters of the 40% solution.
Here is the word equation for this problem.
Notice that in the first term we used the “Amount of acid
in the water”. This might look a
little weird to you because there shouldn’t be any acid in the water. However, this is exactly what we want. The basic equation tells us to look at how
much of the secondary liquid is in the water.
So, this is the correct wording.
When we plug in the percentages and volumes we will think of the water
as a 0% percent solution since that is in fact what it is. So, the new word equation is,
Do not get excited about the zero in the first term. This is okay and will not be a
problem. Let’s now plug in the volumes
and solve for x.
So, we need to add in 46.875 liters of water to 28.125
liters of a 40% solution to get 75 liters of a 15% solution.


