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This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. In this chapter, the emphasis is on the mechanics of equation solving, which clearly explains how to isolate a variable. The goal is to help the student feel more comfortable with solving applied problems. Ample opportunity is provided for the student to practice translating words to symbols, which is an important part of the "Five-Step Method" of solving applied problems (discussed in modules (<link document="m21980"/>) and (<link document="m21979"/>)). Objectives of this module: be able to solve various applied problems.


  • Solving Applied Problems

Solving applied problems

Let’s study some interesting problems that involve linear equations in one variable. In order to solve such problems, we apply the following five-step method:

Five-step method for solving word problems

  1. Let x (or some other letter) represent the unknown quantity.
  2. Translate the words to mathematical symbols and form an equation.
  3. Solve this equation.
  4. Ask yourself "Does this result seem reasonable?" Check the solution by substituting the result into the original statement of the problem.

    If the answer doesn’t check, you have either solved the equation incorrectly, or you have developed the wrong equation. Check your method of solution first. If the result does not check, reconsider your equation.

  5. Write the conclusion.

If it has been your experience that word problems are difficult, then follow the five-step method carefully. Most people have difficulty because they neglect step 1.


Keep in mind what the variable is representing throughout the problem.

Sample set a

This year an item costs $ 44 , an increase of $ 3 over last year’s price. What was last year’s price?

Step 1 : Let x = last year's price . Step 2 : x + 3 = 44. x + 3 represents the $3 increase in price . Step 3 : x + 3 = 44 x + 3 3 = 44 3 x = 41 Step 4 : 41 + 3 = 44 Yes, this is correct . Step 5 : Last year's price was $ 41.

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Practice set a

This year an item costs $ 23 , an increase of $ 4 over last year’s price. What was last year’s price?

  1. Let x =
  2. Last year's price was .

Last year's price was $ 19

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Sample set b

The perimeter (length around) of a square is 60 cm (centimeters). Find the length of a side.

Step 1: Let x = length of a side . Step 2: We can draw a picture .

A square with side of length x and an equation x plus x plus x plus x equals sixty next to the square.

Step 3 : x + x + x + x = 60 4 x = 60 Divide both sides by 4. x = 15. Step 4 : 4 ( 15 ) = 60. Yes, this is correct . Step 5 : The length of a side is 15 cm .

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Practice set b

The perimeter of a triangle is 54 inches. If each side has the same length, find the length of a side.

  1. Let x =
  2. The length of a side is inches.

The length of a side is 18 inches.

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Sample set c

Six percent of a number is 54. What is the number?

Step 1 : Let x = the number Step 2 : We must convert 6 % to a decimal.
6 % = .06 .06 x = 54 .06 x occurs because we want 6 % of x . Step 3 : .06 x = 54. Divide both sides by .06. x = 54 .06 x = 900 Step 4 : .06 ( 900 ) = 54. Yes, this is correct . Step 5 : The number is 900.

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Practice set c

Eight percent of a number is 36. What is the number?

  1. Let x =
  2. The number is .

The number is 450.

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Sample set d

An astronomer notices that one star gives off about 3.6 times as much energy as another star. Together the stars give off 55.844 units of energy. How many units of energy does each star emit?

  1. In this problem we have two unknowns and, therefore, we might think, two variables. However, notice that the energy given off by one star is given in terms of the other star. So, rather than introducing two variables, we introduce only one. The other unknown(s) is expressed in terms of this one. (We might call this quantity the base quantity.)

    Let x = number of units of energy given off by the less energetic star. Then, 3.6 x = number of units of energy given off by the more energetic star.

    Step 2: x + 3.6 x = 55.844. Step 3: x + 3.6 x = 55.844 4.6 x = 55.844 Divide both sides by 4 .6 . A calculator would be useful at this point . x = 55.844 4.6 x = 12.14 The wording of the problem implies t w o numbers are needed = for a complete solution . We need the number of units of energy for the other star. 3.6 x = 3.6 ( 12.14 ) = 43.704 Step 4: 12.14 + 43.704 = 55.844. Yes, this is correct . Step 5 : One star gives off 12.14 units of energy and the other star gives off 43.704 units of energy .

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Questions & Answers

what is Nano technology ?
Bob Reply
write examples of Nano molecule?
The nanotechnology is as new science, to scale nanometric
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
what king of growth are you checking .?
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
why we need to study biomolecules, molecular biology in nanotechnology?
Adin Reply
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
what school?
biomolecules are e building blocks of every organics and inorganic materials.
anyone know any internet site where one can find nanotechnology papers?
Damian Reply
sciencedirect big data base
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
how to know photocatalytic properties of tio2 nanoparticles...what to do now
Akash Reply
it is a goid question and i want to know the answer as well
characteristics of micro business
for teaching engĺish at school how nano technology help us
Do somebody tell me a best nano engineering book for beginners?
s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
what is the actual application of fullerenes nowadays?
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
is Bucky paper clear?
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Do you know which machine is used to that process?
how to fabricate graphene ink ?
for screen printed electrodes ?
What is lattice structure?
s. Reply
of graphene you mean?
or in general
in general
Graphene has a hexagonal structure
On having this app for quite a bit time, Haven't realised there's a chat room in it.
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Berger describes sociologists as concerned with
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Source:  OpenStax, Elementary algebra. OpenStax CNX. May 08, 2009 Download for free at http://cnx.org/content/col10614/1.3
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