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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.

Overview

  • 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.

Always start by INTRODUCING A VARIABLE!

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

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
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John Reply
what is physics
Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
what is the dimension formula of energy?
David Reply
what is viscosity?
David
what is inorganic
emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
what's motion
Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
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Muhammad Reply
fine, how about you?
Mohammed
hi
Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
Who can show me the full solution in this problem?
Reofrir Reply
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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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