1. Copyright © 2010 Pearson Education, Inc
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2. Equations, Inequalities, and Problem Solving2
Equations, Inequalities, and Problem Solving
2.1 Solving Equations
2.2 Using the Principles Together
2.3 Formulas
2.4 Applications with Percent
2.5 Problem Solving
2.6 Solving Inequalities
2.7 Solving Applications with Inequalities
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3. 2.1 Solving Equations Equations and Solutions The Addition Principle
The Multiplication Principle
Selecting the Correct Approach
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4. Solution of an Equation
Any replacement for the variable that makes an equation true is called a solution of the equation. To solve an equation means to find all of its solutions.
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5. Example Determine whether 8 is a solution of x + 12 = 21.
Solution x + 12 = Writing the equation
| Substituting 8 for x
20  False
Since the left-hand and right-hand sides differ, 8 is not a solution.
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6. Equivalent Equations The Addition Principle
Equations with the same solutions are called equivalent equations.
The Addition Principle
For any real numbers a, b, and c,
a = b is equivalent to a + c = b + c.
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7. Example Solve: 8.3 = y  17.9 Solution 8.3 = y  17.9
Check: 8.3 = y  17.9
8.3 | 9.6  17.9
8.3 = 8.3
The solution is 9.6.
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8. The Multiplication Principle
For any real numbers a, b, and c with c  0,
a = b is equivalent to a • c = b • c.
Example
Solve:
Solution
Multiplying both sides by 4/3.
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9. Example Solve: 7x = 84 Solution 7x = 84 Check: 7x = 84 7(12) | 84
Dividing both sides by 7.
Check: 7x = 84
7(12) | 84
84 = 84
The solution is –12.
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10. Using the Principles Together
2.2
Using the Principles Together
Applying Both Principles
Combining Like Terms
Clearing Fractions and Decimals
Contradictions and Identities
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11. Example Solve: 9 + 8x = 33 Solution 9 + 8x = 33 9 + 8x  9 = 33  9
Check: x = 33
9 + 8(3) | 33
| 33
33 = The solution is 3.
Subtracting 9 from both sides
Dividing both sides by 8
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12. Example Solve: Solution Adding 5 to both sides
Multiplying both sides by 3/2.
Simplifying
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13. Example Solve: 36  t = 17 Solution 36  t = 17 36  t  36 = 17  36
Check:  t = 17
36  19 | 17
17 = 17
The solution is 19.
Subtracting 36 from both sides
Multiplying both sides by  1
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14. Combining Like Terms
If like terms appear on the same side of an equation, we combine them and then solve.
Should like terms appear on both sides of an equation, we can use the addition principle to rewrite all like terms on one side.
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15. Example Solve. 5x + 4x = 36 Solution 5x + 4x = 36 9x = 36 x = 4
Check: x + 4x = (4) + 4(4) | 36
| 36
36 = 36
The solution is 4.
Combining like terms
Dividing both sides by 9
Simplifying
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16. Example Solve. 4x + 7  6x = 10 + 3x + 12
Solution: 4x + 7  6x = x + 12
2x + 7 = x
2x + 7  7 = x  7
2x = x
2x  3x = x  3x
5x = 15
x = 3
Combining like terms
Subtracting 7 from both sides
Simplifying
Subtracting 3x from both sides
Dividing both sides by 5
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17. Check: 4x + 7  6x = 10 + 3x + 12 4(3) + 7 6(3) | 10 + 3(3) + 12
 | 10 
13 = 13
The solution is 3.
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18. Example Solve. Solution:
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19. An Equation-Solving Procedure
1. Use the multiplication principle to clear any fractions or decimals. (This is optional, but can ease computations.
2. If necessary, use the distributive law to remove parentheses. Then combine like terms on each side.
3. Use the addition principle, as needed, to isolate all variable terms on one side. Then combine like terms.
4. Multiply or divide to solve for the variable, using the multiplication principle.
5. Check all possible solutions in the original equation.
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20. Contradictions and Identities
An identity is an equation that is true for all replacements that can be used on both sides of the equation.
A contradiction is an equation that is never true.
A conditional equation is sometimes true and sometimes false, depending on what the replacement is.
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21. Example
Solution:
The equation is true regardless of the choice for x, so all real numbers are solutions. The equation is an identity.
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22. Example
Solution:
The equation is false for any choice of x, so there is no solution for this equation. The equation is a contradiction.
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23. Example
Solution:
There is one solution, 7. For all other choices of x, the equation is false. The equation is a conditional equation.
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24. Evaluating Formulas Solving for a Variable
2.3
Formulas
Evaluating Formulas
Solving for a Variable
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25. Many applications of mathematics involve relationships among two or more quantities. An equation that represents such a relationship will use two or more letters and is known as a formula.
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26. Example The formula can be used to determine how far
M, in miles, you are from lightening when its thunder takes t seconds to reach your ears. If it takes 5 seconds for the sound of thunder to reach you after you have seen the lightening, how far away is the storm?
Solution We substitute 5 for t and calculate M.
The storm is 1 mile away.
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27. Circumference of a circle.
Example
The formula C = d gives the circumference C of a circle with diameter d. Solve for d.
Solution C = d d
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28. To Solve a Formula for a Given Variable
1. If the variable for which you are solving appears in a fraction, use the multiplication principle to clear fractions.
2. Isolate the term(s), with the variable for which you are solving on one side of the equation.
3. If two or more terms contain the variable for which you are solving, factor the variable out.
4. Multiply or divide to solve for the variable in question.
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29. Example Solve for y: x = wy + zy  6 Solution x = wy + zy  6
x + 6 = y(w + z)
We want this letter alone.
Adding 6 to both sides
Factoring
Dividing both sides by w + z
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30. Applications with Percent
2.4
Applications with Percent
Converting Between Percent Notation and Decimal Notation
Solving Percent Problems
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31. Percent Notation n% means
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32. Example Convert to decimal notation: a) 34% b) 7.6% Solution
= 0.34
b) 7.6% = 7.6  0.01
= 0.076
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33. To convert from percent notation to decimal notation, move the decimal point two places to the left and drop the percent symbol.
Example
Convert the percent notation in the following sentence to decimal notation: Marta received a 75% on her first algebra test.
Solution
Move the decimal point two places to the left.
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34. To convert from decimal notation to percent notation, move the decimal point two places to the right and write a percent symbol.
Example
Convert to percent notation:
a) 3.24 b) 0.2 c)
Solution
a) We first move the decimal point two places to the right: and then write a % symbol: 324%
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35. Solution continued
b) We first move the decimal point two places to the right (recall that 0.2 = 0.20): 0.20.
and then write a % symbol: 20%
c) Note that 3/8 = We move the decimal point two places to the right:
and then write a % symbol: 37.5%
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36. Key Word in Percent Translations “Of” translates to “•” or “”.
“What” translates to a variable.
“Is” or “Was” translates to “=”.
% translates to
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37. Example What is 13% of 72? Solution Translate:
Thus, 9.36 is 13% of 72. The answer is 9.36.
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38. Example 9 is 12 percent of what? Solution Translate:
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39. Example What percent of 60 is 27? Solution Translate:
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40. Example
To complete her water safety course instruction, Catherine must complete 45 hours of instruction. If she has completed 75% of her requirement, how many hours has Catherine completed?
Solution
Rewording: What is 75% of 45?
Translating: a =  45
a =
Catherine has completed hours of instruction.
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41. Five Steps for Problem Solving Applying the Five Steps
2.5
Problem Solving
Five Steps for Problem Solving
Applying the Five Steps
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42. Five Steps for Problem Solving in Algebra
1. Familiarize yourself with the problem.
2. Translate to mathematical language. (This often means writing an equation.)
3. Carry out some mathematical manipulation (This often means solving an equation.)
4. Check your possible answer in the original problem.
5. State the answer clearly, using a complete English sentence.
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43. To Become Familiar with a Problem
1. Read the problem carefully. Try to visualize the problem.
2. Reread the problem, perhaps aloud. Make sure you understand all important words.
3. List the information given and the question(s) to be answered. Choose a variable (or variables) to represent the unknown and specify what the variable represents. For example, let L = length in centimeters, d = distance in miles, and so on.
4. Look for similarities between the problem and other problems you have already solved.
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44. To Become Familiar with a Problem (continued)
5. Find more information. Look up a formula in a book, at the library, or online. Consult a reference librarian or an expert in the field.
6. Make a table that uses all the information you have available. Look for patterns that may help in the translation.
7. Make a drawing and label it with known and unknown information, using specific units if given.
8. Think of a possible answer and check the guess. Note the manner in which the guess is checked.
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45. Example
The apartments in Wanda’s apartment house are consecutively numbered on each floor. The sum of her number and her next door neighbor’s number is 723. What are the two numbers?
Solution
1. Familiarize. The apartment numbers are consecutive integers.
Let x = Wanda’s apartment
Let x + 1 = neighbor’s apartment
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46. 2. Translate. Rewording: Translating: 3. Carry out. x + (x + 1) = 723
If x is 361, then x + 1 is 362.
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47. 5. State. The apartment numbers are 361 and 362.
4. Check. Our possible answers are 361 and These are consecutive integers and the sum is
5. State. The apartment numbers are 361 and 362.
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48. Example
Digicon prints digital photos for $0.12 each plus $3.29 shipping and handling. Your weekly budget for the school yearbook is $ How many prints can you have made if you have $22.00?
Solution
1. Familiarize. Suppose the yearbook staff takes digital photos. Then the cost to print them would be the shipping charge plus $0.12 times 220.
$ $0.12(220) which is $ Our guess of 220 is too large, but we have familiarized ourselves with the way in which the calculation is made.
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49. 5. State. The yearbook staff can have 155 photos printed per week.
2. Translate.
Rewording:
Translating:
3. Carry out.
4. Check. Check in the original problem. $ (0.12) = $21.89, which is less than $22.00.
5. State. The yearbook staff can have 155 photos printed per week.
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50. Example
You are constructing a triangular kite. The second angle of the kite is three times as large as the first. The third angle is 10 degrees more than the first. Find the measure of each angle.
Solution
1. Familiarize. Make a drawing and write in the given information.
2. Translate. To translate, we need to recall that the sum of the measures of the angles in a triangle is degrees. 3x x
x + 10
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51. 2. Translate (continued).
3. Carry out.
The measures for the angles appear to be:
first angle: x = 34
second angle: 3x = 3(34) = 102;
third angle: x + 10 = = 44
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52. 5. State. The measures of the angles are 34, 44 and 102 degrees.
4. Check. Consider 34, 102 and 44 degrees. The sum of these numbers is 180 degrees and the second angle is three times the first angle. The third angle is 10 degrees more than the first. These numbers check.
5. State. The measures of the angles are 34, 44 and degrees.
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53. 2.6 Solving Inequalities Solutions of Inequalities
Graphs of Inequalities
Set Builder and Interval Notation
Solving Inequalities Using the Addition Principle
Solving Inequalities Using the Multiplication Principle
Using the Principles Together
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54. Solutions of Inequalities
An inequality is a number sentence containing > (is greater than), < (is less than),  (is greater than or equal to), or  (is less than or equal to).
Determine whether the given number is a solution of x < 5: a) 4 b) 6
Solution
a) Since 4 < 5 is true, 4 is a solution.
b) Since 6 < 5 is false, 6 is not a solution.
Example 6 -3 -1 1 3 5 -4 4 -2 2
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55. Solutions of Inequalities
An inequality is any sentence containing
Examples:
Any value for a variable that makes an inequality true is called a solution. The set of all solutions is called the solution set. When all solutions of an inequality are found, we say that we have solved the inequality.
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56. Example Solution Determine whether 5 is a solution to
We substitute to get 3(5) + 2 > 7, or 17 >7, a true statement. Thus, 5 is a solution.
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57. The graph of an inequality is a visual representation of the inequality’s solution set. An inequality in one variable can be graphed on a number line.
Example
Graph x < 2 on a number line.
Solution ) 4 5 3 1 -1 -5 -6 -4 -3 -2 2 6
Note that in set-builder notation the solution is
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58. Interval Notation
We can write solutions of an inequality in one variable using interval notation. Interval notation uses parentheses, ( ), and brackets, [ ].
If a and b are real numbers such that a < b, we define the open interval (a, b) as the set of all numbers x for which a < x < b. Thus,
(a, b)
a b ) (
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59. Interval Notation
The closed interval [a, b] is defined as the set of all numbers x for which
Thus,
a b
[a, b] [ ]
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60. Interval Notation
There are two types of half-open intervals, defined as follows:
(a, b]
a b ( ]
a b
[a, b) [ )
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61. Interval Notation
We use the symbols to represent positive and negative infinity, respectively. Thus the notation (a, ) represents the set of all real numbers greater than a, and ( , a) represents the set of all numbers less than a. a ( )
The notations (– , a] and [a, ) are used when we want to include the endpoint a.
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62. The Addition Principle for Inequalities
For any real numbers a, b, and c:
a < b is equivalent to a + c < b + c;
a  b is equivalent to a + c  b + c;
a > b is equivalent to a + c > b + c;
a  b is equivalent to a + c  b + c.
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63. Example Solution Solve and graph x – 2 > 7. x – 2 > 7
The solution set is (
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64. Example Solve 4x  1  x  10 and then graph the solution. Solution
4x  x  x  x  9
3x  9
x  3
The solution set is {x|x  3}.
Adding 1 to both sides
Simplifying
Subtracting x from both sides
Simplifying
Dividing both sides by 3
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65. The Multiplication Principle for Inequalities
For any real numbers a and b, and for any positive number c:
a < b is equivalent to ac < bc, and
a > b is equivalent to ac > bc.
For any real numbers a and b, and for any negative number c:
a < b is equivalent to ac > bc, and
a > b is equivalent to ac < bc.
Similar statements hold for  and .
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66. Example Solve and graph each inequality: a) b) 4y < 20 Solution a)
The solution set is {x|x  28}. The graph is shown below.
Multiplying both sides by 4
Simplifying 30 15 25 5 20 10 ]
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67. The solution set is {y|y > 5}. The graph is shown below.
b) 4y < 20
The solution set is {y|y > 5}. The graph is shown below.
Dividing both sides by 4
At this step, we reverse the inequality, because 4 is negative. 4 -5 -3 -1 1 3 -6 -2 2 -4 (
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68. Example Solve. 3x  3 > x + 7 Solution 3x  3 > x + 7
3x  x > x  x + 10
2x > 10
x > 5
The solution set is {x|x > 5}.
Adding 3 to both sides
Simplifying
Subtracting x from both sides
Simplifying
Dividing both sides by 2
Simplifying 8 -1 1 3 5 7 -2 4 2 6 (
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69. Example Solve. 15.4  3.2x < 6.76 Solution 15.4  3.2x < 6.76
The solution set is {x|x > 6.925}.
Remember to reverse the symbol!
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70. Example Solve: 5(x  3)  7x  4(x  3) + 9 Solution
2x + 2x  12  4x + 2x
12  6x
2  x
The solution set is {x|x  2}.
Using the distributive law to remove parentheses
Simplifying
Adding 3 to both sides
Adding 2x to both sides
Dividing both sides by 6 2 -7 -5 -3 -1 1 -8 -6 -2 -4 ]
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71. Solving Applications with Inequalities
2.7
Solving Applications with Inequalities
Translating to Inequalities
Solving Problems
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72. Important Words Sample Sentence Translation
is at least
Brian is at least 16 years old
b  16
is at most
At most 3 students failed the course
s  3
cannot exceed
To qualify, earnings cannot exceed $5000
e  $5000
must exceed
The speed must exceed 20 mph
s > 20
is less than
Nicholas is less than 60 lb.
n < 60
is more than
Chicago is more than 300 miles away.
c > 300
is between
The movie is between 70 and 120 minutes.
70 < m < 120
no more than
The calf weighs no more than 560 lb.
w  560
no less than
Carmon scored no less than 9.4.
c  9.4
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73. Translating “at least” and “at most”
The quantity x is at least some amount q: x  q.
(If x is at least q, it cannot be less than q.)
The quantity x is at most some amount q: x  q.
(If x is at most q, it cannot be more than q.)
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74. Example
Lazer Line charges $65 plus $45 per hour for copier repair. Jonas remembers being billed less than $150. How many hours was Jonas’ copier worked on?
Solution
1. Familiarize. Suppose the copier was worked on for hours. The cost would be $65 + 4($45), or $ This shows that the copier was worked on for less than 4 hours. Let h = the number of hours.
2. Translate.
Rewording:
Translating:
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75. 5. State. Jonas’ copier was worked on for less than two hours.
3. Carry out.
h  150
45h  85
4. Check. Since the time represents hours, we round down to one hour. If the copier was worked on for one hour, the cost would be $110, and if worked on for two hours the cost would exceed $150.
5. State. Jonas’ copier was worked on for less than two hours.
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76. Average, or Mean
To find the average or mean of a set of numbers, add the numbers and then divide by the number of addends.
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77. Example
Samantha has test grades of 86, 88, and 78 on her first three math tests. If she wants an average of at least 80 after the fourth test, what possible scores can she earn on the fourth test?
Solution
1. Familiarize. Suppose she earned an 85 on her fourth test Her test average would be
This shows she could score an 85. Let’s let x represent the fourth test score.
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78. 2. Translate. Rewording: Translating: 3. Carry out.
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79. 4. Check.
As a partial check, we show that Samantha can earn a 68 on | the fourth test and average 80 for the four tests.
5. State.
Samantha’s test average will not drop below 80 if she earns at least a 68 on the fourth test.
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