Section 2.4: Applications of Linear Equations
When we say applications, we really mean word problems. So this section introduces word problems and the
steps necessary to set up and solve. We will be going over more difficult ones later on in the course. In this
section, we only use one variable. However, there may be one or more unknown values. So we need to express
each unknown value in terms of the variable.
§1 Learn the Steps!
Step 1 – Read the problem carefully. What information is given? What are you asked to find?
Step 2 – Assign a variable to represent the unknown value (you can always use x). If necessary, express any other
unknown values in terms of the variable.
Step 3 – Write an equation using the variable expression that relates to the problem.
Step 4 – Solve the equation
Step 5 – State the answer.
Step 6 – Check your answer in the words of the original problem.
These steps can be found in your textbook as well. In this section, there are 4 types of word problems you will
see. They are 1) solving problems involving unknown numbers, 2) solving problems involving sums of quantities,
3) solve problems involving consecutive integers, and 4) solve problems involving angles. We will look at some
examples of each. Remember, the most important part of a word problems is to first determine how many
quantities you need to find and how to express these quantities in terms of the variable.
§2 Solve Problems Involving Unknown Numbers
For these problems, its very important to know the key words. We want to translate from words to algebra.
Words such as ‘sum of’, ‘added to’ and ‘less than’ all are very specific in what type of operation needs to occur.
Try the following: If 5 is added to the product of 9 and a number, the result is 19 less than the number. Find the
number. It looks like a lot of information here, but if you are familiar with the key words hopefully the problem
gets simplified. First of all, we want to find a number. That’s what the question is asking. There is only one
number to find, so there is only one quantity hence we can call the number x. Next, we use the key words to set
up the equation. The key word ‘result is’ means equal to. You should get the equation 9 x  5  x  19 . Also
remember, ’19 less than the number’ means x  19, not 19  x. Once you set up the equation, solving it should
be the easy part. The final answer is x = -3. Hence the number is equal to -3. Check this answer to make sure it is
correct!
PRACTICE
1) The product of 5 and 3 more than twice a number is 85. What is the number?
2) If 3 is added to a number and this sum is doubled, the result is 2 more than the number. Find the number.
§3 Solve Problems Involving Sums of Quantities
In these types of problems, there are usually two quantities to find. Hence we need to express both quantities in
terms of the same variable. Please make sure you understand what this means. It DOES NOT mean we call both
quantities x. One of the quantities is x, but the other quantity has some relation to x. Some examples may have
three or more quantities. The main point to remember is to use one variable.
Try the following: In a certain math class there were 7 more males than females. If there were 35 students in the
class, how many males and how many females are there? Note that there are two quantities to find – the
number of males and the number of females. As mentioned before, we DON’T let x = the number of males and x
= the number of females. This just means that the number of males and females is the same. We need to let x =
one of the quantities – it doesn’t matter which one. So let’s say that x = the number of females. Then how can
we express the number of males in terms of x? Well, the problem says that there were 7 MORE males than
females. Hence if x = the number of females, then x + 7 = the number of males. The total number of students in
the class is 35. Hence the equation would become x   x  7   35. Solve this to get x = 14. Hence there are 14
females and 21 males in the class.
PRACTICE
3) In one day, a certain coffeehouse found that the number of orders of croissants was 1/6 the number of orders
for muffins. If the total number of orders for croissants and muffins was 56, how many orders were placed for
croissants?
4) A pipe that is 50 inches long is cut into three pieces. The longest piece is 10 inches longer that the middlesized piece, and the shortest piece is 5 inches less than the middle-sized piece. How long is each piece?
§4 Solve Problems Involving Consecutive Integers
First, it would be wise to go over what consecutive integers means. Consecutive means one after the other. So if
we say two consecutive integers, then they must differ by one. Hence if we let x = the first integer, then the next
integer must be x + 1. What about when we say consecutive even and/or consecutive odd integers? How can we
express these using x?
Remember, whenever we deal with consecutive integers, one of them is always the smallest one!
Try the following: The sum of the page numbers of an open book is 95. What are the page numbers? There are
two quantities. Let x = the page number of the first page. Then x+1 = the page number of the second page. Since
the sum is 95, our equation becomes x  ( x  1)  95 . Solve this for x to get that x = 47. Hence the two page
numbers are 47 and 48. Do they add up to 95?
Another similar problem is as follows: 2 times the larger of two consecutive odd numbers is 3 less than three
times the smaller. Find the two integers. The first integer we can just say is x. Then how would we denote the
next consecutive odd integer? It should be x + 2. Now we can set up our equation. It becomes 2( x  2)  3x  3.
Solve this for x to get that x = 7. Hence our two integers are 7 and 9.
PRACTICE
5) Find two consecutive even integers such that six times the lesser added to the greater gives a sum of 86.
6) IF 5 times the lesser of two consecutive integers is added to three times the greater, the result is 59. Find the
integers.
§5 Solve Problems Involving Angles
First some terminology. Two angles whose sum is 90 degrees are said to be complementary. Two angles who
sum is 180 degrees is said to be supplementary.
Note that in both cases there are two angles in question. Hence if we know that two angles are complementary,
we can call one angle x and the other angle 90 – x. Similarly, if we know that two angles are complementary, we
can call one angle x and the other angle would be 180 – x.
Try this: An angle is 30 degrees less than twice its complement. Find the measure of each angle. If we call one
angle x, then its complement must be 90 – x. Now we can set up our equation. We get that x  2  90  x   30 .
Solve this for x to get that x = 50. Hence the angle is 50 degrees, and its complement must then be 40 degrees.
PRACTICE
7) Find the measure of an angle whose supplement is three times its measure.
8) Find the measure of an angle whose supplement measures 38 degrees less than three times its complement.