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This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses how to solve algebraic problems. By the end of the module students should be more familiar with the five-step method for solving applied problems and be able to use the five-step method to solve number problems and geometry problems.

Section overview

  • The Five-Step Method
  • Number Problems
  • Geometry Problems

The five step method

We are now in a position to solve some applied problems using algebraic methods. The problems we shall solve are intended as logic developers. Although they may not seem to reflect real situations, they do serve as a basis for solving more complex, real situation, applied problems. To solve problems algebraically, we will use the five-step method.

Strategy for reading word problems

When solving mathematical word problems, you may wish to apply the following " reading strategy ." Read the problem quickly to get a feel for the situation. Do not pay close attention to details. At the first reading, too much attention to details may be overwhelming and lead to confusion and discouragement. After the first, brief reading, read the problem carefully in phrases . Reading phrases introduces information more slowly and allows us to absorb and put together important information. We can look for the unknown quantity by reading one phrase at a time.

    Five-step method for solving word problems

  1. Let x size 12{x} {} (or some other letter) represent the unknown quantity.
  2. Translate the words to mathematical symbols and form an equation. Draw a picture if possible.
  3. Solve the equation.
  4. Check the solution by substituting the result into the original statement, not equation, of the problem.
  5. Write a conclusion.

If it has been your experience that word problems are difficult, then follow the five-step method carefully. Most people have trouble with word problems for two reasons:

  1. They are not able to translate the words to mathematical symbols. (See [link] .)
  2. They neglect step 1. After working through the problem phrase by phrase, to become familiar with the situation,

INTRODUCE A VARIABLE

Number problems

Sample set a

What number decreased by six is five?

  1. Let n size 12{n} {} represent the unknown number.
  2. Translate the words to mathematical symbols and construct an equation. Read phrases.

    What number: n decreased by: six: 6 is: = five: 5 } n 6 = 5 size 12{ left none matrix { "What number:" {} # n {} ##"decreased by:" {} # - {} {} ## "six:" {} # 6 {} ##"is:" {} # ={} {} ## "five:" {} # 5{}} right rbrace n - 6=5} {}

  3. Solve this equation.

    n 6 = 5 size 12{n - 6=5} {} Add 6 to both sides.
    n 6 + 6 = 5 + 6 size 12{n - 6+6=5+6} {}
    n = 11 size 12{n="11"} {}

  4. Check the result.

    When 11 is decreased by 6, the result is 11 6 size 12{"11" - 6} {} , which is equal to 5. The solution checks.

  5. The number is 11.

When three times a number is increased by four, the result is eight more than five times the number.

  1. Let x = size 12{x={}} {} the unknown number.
  2. Translate the phrases to mathematical symbols and construct an equation.

    When three times a number: 3 x is increased by: + four: 4 the result is: = eight: 8 more than: + five times the number: 5 x } 3 x + 4 = 5 x + 8 size 12{ left none matrix { "When three times a number:" {} # 3x {} ##"is increased by:" {} # +{} {} ## "four:" {} # 4 {} ##"the result is:" {} # ={} {} ## "eight:" {} # 8 {} ##"more than:" {} # +{} {} ## "five times the number:" {} # 5x{}} right rbrace 3x+4=5x+8} {}


  3. 3 x + 4 = 5 x + 8 size 12{3x+4=5x+8} {} . Subtract 3 x from  both  sides. 3 x + 4 3 x = 5 x + 8 3 x size 12{3x+4 - 3x=5x+8 - 3x} {} 4 = 2x + 8 size 12{4=2x+8} {} Subtract 8 from  both  sides. 4 8 = 2x + 8 8 size 12{4 - 8=2x+8 - 8} {} 4 = 2x size 12{ - 4=2x} {} Divide  both  sides by 2. 2 = x size 12{ - 2=x} {}
  4. Check this result.
    Three times - 2 is - 6 . Increasing - 6 by 4 results in 6 + 4 = 2 size 12{ - 6+4= - 2} {} . Now, five times - 2 is - 10 .
    Increasing - 10 by 8 size 12{g} {} results in 10 + 8 = 2 size 12{ - "10"+8= - 2} {} . The results agree, and the solution checks.
  5. The number is - 2

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Source:  OpenStax, Algebra i for the community college. OpenStax CNX. Dec 19, 2014 Download for free at http://legacy.cnx.org/content/col11598/1.3
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