Copyright © 2010 Pearson Education, Inc Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 1- 1

Equations, Inequalities, and Problem Solving 2 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 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2.1 Solving Equations Equations and Solutions The Addition Principle The Multiplication Principle Selecting the Correct Approach Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Determine whether 8 is a solution of x + 12 = 21. Solution x + 12 = 21 Writing the equation 8 + 12 | 21 Substituting 8 for x 20  21 False Since the left-hand and right-hand sides differ, 8 is not a solution. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Using the Principles Together 2.2 Using the Principles Together Applying Both Principles Combining Like Terms Clearing Fractions and Decimals Contradictions and Identities Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve: 9 + 8x = 33 Solution 9 + 8x = 33 9 + 8x  9 = 33  9 Check: 9 + 8x = 33 9 + 8(3) | 33 9 + 24 | 33 33 = 33 The solution is 3. Subtracting 9 from both sides Dividing both sides by 8 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve: Solution Adding 5 to both sides Multiplying both sides by 3/2. Simplifying Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve: 36  t = 17 Solution 36  t = 17 36  t  36 = 17  36 Check: 36  t = 17 36  19 | 17 17 = 17 The solution is 19. Subtracting 36 from both sides Multiplying both sides by  1 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve. 5x + 4x = 36 Solution 5x + 4x = 36 9x = 36 x = 4 Check: 5x + 4x = 36 5(4) + 4(4) | 36 20 + 16 | 36 36 = 36 The solution is 4. Combining like terms Dividing both sides by 9 Simplifying Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve. 4x + 7  6x = 10 + 3x + 12 Solution: 4x + 7  6x = 10 + 3x + 12 2x + 7 = 22 + 3x 2x + 7  7 = 22 + 3x  7 2x = 15 + 3x 2x  3x = 15 + 3x  3x 5x = 15 x = 3 Combining like terms Subtracting 7 from both sides Simplifying Subtracting 3x from both sides Dividing both sides by 5 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Check: 4x + 7  6x = 10 + 3x + 12 4(3) + 7 6(3) | 10 + 3(3) + 12 12 + 7 + 18 | 10  9 + 12 13 = 13 The solution is 3. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve. Solution: Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution: The equation is true regardless of the choice for x, so all real numbers are solutions. The equation is an identity. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution: The equation is false for any choice of x, so there is no solution for this equation. The equation is a contradiction. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution: There is one solution, 7. For all other choices of x, the equation is false. The equation is a conditional equation. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Evaluating Formulas Solving for a Variable 2.3 Formulas Evaluating Formulas Solving for a Variable Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Applications with Percent 2.4 Applications with Percent Converting Between Percent Notation and Decimal Notation Solving Percent Problems Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Percent Notation n% means Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Convert to decimal notation: a) 34% b) 7.6% Solution = 0.34 b) 7.6% = 7.6  0.01 = 0.076 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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: 3.24. and then write a % symbol: 324% Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 = 0.375. We move the decimal point two places to the right: 0.37.5 and then write a % symbol: 37.5% Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Key Word in Percent Translations “Of” translates to “•” or “”. “What” translates to a variable. “Is” or “Was” translates to “=”. % translates to Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example What is 13% of 72? Solution Translate: Thus, 9.36 is 13% of 72. The answer is 9.36. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example 9 is 12 percent of what? Solution Translate: Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example What percent of 60 is 27? Solution Translate: Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 = 0.75  45 a = 33.75 Catherine has completed 33.75 hours of instruction. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Five Steps for Problem Solving Applying the Five Steps 2.5 Problem Solving Five Steps for Problem Solving Applying the Five Steps Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. Translate. Rewording: Translating: 3. Carry out. x + (x + 1) = 723 If x is 361, then x + 1 is 362. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

5. State. The apartment numbers are 361 and 362. 4. Check. Our possible answers are 361 and 362. These are consecutive integers and the sum is 723. 5. State. The apartment numbers are 361 and 362. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Digicon prints digital photos for $0.12 each plus $3.29 shipping and handling. Your weekly budget for the school yearbook is $22.00. How many prints can you have made if you have $22.00? Solution 1. Familiarize. Suppose the yearbook staff takes 220 digital photos. Then the cost to print them would be the shipping charge plus $0.12 times 220. $3.29 + $0.12(220) which is $29.69. Our guess of 220 is too large, but we have familiarized ourselves with the way in which the calculation is made. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. $3.29 + 155(0.12) = $21.89, which is less than $22.00. 5. State. The yearbook staff can have 155 photos printed per week. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 180 degrees. 3x x x + 10 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 = 34 + 10 = 44 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 102 degrees. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 ) ( Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Interval Notation The closed interval [a, b] is defined as the set of all numbers x for which Thus, a b [a, b] [ ] Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Interval Notation There are two types of half-open intervals, defined as follows: (a, b] a b ( ] a b [a, b) [ ) Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Solve and graph x – 2 > 7. x – 2 > 7 The solution set is ( 5 6 7 8 9 10 11 12 13 14 15 16 17 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 . Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 ] Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 ( Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 ( Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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! Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 ] Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solving Applications with Inequalities 2.7 Solving Applications with Inequalities Translating to Inequalities Solving Problems Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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.) Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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 4 hours. The cost would be $65 + 4($45), or $245. This shows that the copier was worked on for less than 4 hours. Let h = the number of hours. 2. Translate. Rewording: Translating: Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

5. State. Jonas’ copier was worked on for less than two hours. 3. Carry out. 65 + 45h  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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. Translate. Rewording: Translating: 3. Carry out. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley