Chapter 9 Basic Algebra © 2010 Pearson Education, Inc. All rights reserved.

Copyright © 2010 Pearson Education, Inc. All rights reserved.
9.1 Signed Numbers Objectives 1. Write negative numbers. 2. Graph signed numbers on a number line. 3. Use the < and > symbols. 4. Find absolute value. 5. Find the opposite of a number. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

All the numbers you have studied so far in this book have been either 0 or greater than 0. Numbers greater than 0 are called positive numbers. For example: Salary of $55,000 Temperature of 98.6 Length of 5 feet Copyright © 2010 Pearson Education, Inc. All rights reserved.

Parallel Example 1 Graphing Signed Numbers Graph these numbers on the number line. 5 b. 3 c. –2 d e. Place a dot at the correct location for each number. c d e b a Copyright © 2010 Pearson Education, Inc. All rights reserved.

When you look at the number line below, 3 is to the left of 5. Recall the following symbols for comparing two numbers. < means “is less than.” > means “is greater than.” Use these symbols to write “3 is less than 5” 3 < 5 3 is less than 5 Copyright © 2010 Pearson Education, Inc. All rights reserved.

Parallel Example 2 Using the Symbols < and > Use the number line to compare each pair of numbers. Then write > or < to make each statement true. a. Compare 3 and 3 is to the left of 5 on the number line Compare –4 and –2 –4 –2 –4 is to the left of –2 < < Copyright © 2010 Pearson Education, Inc. All rights reserved.

In order to graph a number on the number line, you need to answer two questions: Which direction is it from 0? It can be a positive direction or a negative direction. You can tell the direction by looking for a positive or negative sign (or no sign, which is positive). How far is it from 0? The distance from 0 is the absolute value of a number. Absolute value is indicated by two vertical bars. |7| is read “the absolute value of 7.” Copyright © 2010 Pearson Education, Inc. All rights reserved.

Parallel Example 3 Finding Absolute Values Simplify each absolute value expression. |5| |5| = 5 b. |–3| |–3| = 3 Distance is 5. Direction is positive. Distance is 3. Direction is negative. Copyright © 2010 Pearson Education, Inc. All rights reserved.

Parallel Example 3 Finding Absolute Values Simplify each absolute value expression. c. –|–4| First |–4| = 4 But there is also a negative sign outside the absolute value bars. So, –4 is the simplified expression. Copyright © 2010 Pearson Education, Inc. All rights reserved.

Two numbers that are the same distance from 0 on a number line but on opposite sides of 0 are called opposites of each other. To indicate the opposite of a number, write a negative sign in front of the number. Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.2 Adding and Subtracting Signed Numbers
Objectives 1. Add signed numbers by using a number line. 2. Add signed numbers without using a number line. 3. Find the additive inverse of a number. 4. Subtract signed numbers. 5. Add or subtract a series of signed numbers. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 1 Adding Signed Numbers Using a Number Line a (−2) − 2 5 4 5 3 1 -1 -5 -6 -4 -3 -2 2 6 Start at 0 and draw an arrow 5 units to the right. From the end of this arrow, draw an arrow 2 units to the left. 5 + (−2) = 3 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 1 continued Adding Signed Numbers Using a Number Line b. (−4) + (−2) − 2 − 4 4 5 3 1 -1 -5 -6 -4 -3 -2 2 6 As the arrows along the number line show: (−4) + (−2) = −6 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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Parallel Example 2 Adding Two Negative Numbers a. −8 + (−9) The absolute value of −8 is 8. The absolute value of −9 is 9. Add the absolute values. 8 + 9 = 17 Write a negative sign in front of the sum. −8 + (−9) = −17 b. −6 + (−13) = −19 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 2 continued Adding Two Negative Numbers c. The absolute value of −2/3 is 2/3. The absolute value of −1/6 is 1/6. Add the absolute values. Write a negative sign in front of the sum. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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Parallel Example 3 Adding Two Numbers with Different Signs a (−4) First find the sum with a number line. −4 6 4 5 3 1 -1 -5 -6 -4 -3 -2 2 6 Now find the sum by using the absolute value rule. Subtract the lesser absolute value from the greater absolute value. 6 − 4 = 2 The positive number 6 has the greater absolute value, so the answer is positive. 6 + (−4) = 2 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 3 continued Adding Two Numbers with Different Signs b. − = −7 Write a negative sign in front of the answer because −24 has the greater absolute value. c (−6) = 5 The answer is positive because the positive number 11 has the greater absolute value. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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Parallel Example 5 Subtracting Signed Numbers a. 15 − 19 The first number 15 stays the same. Change the subtraction sign to addition. Change the sign of the second number to its opposite. 15 − 19 15 + (−19) Now add. 15 + (−19) = −4 So 15 − 19 = −4 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 7 Combining Addition and Subtraction of Signed Numbers According to the last step in the order of operations, perform addition and subtraction from left to right. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

9.3 Multiplying and Dividing Signed Numbers
Objectives 1. Multiply or divide two numbers with opposite signs. 2. Multiply or divide two numbers with the same sign. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

In mathematics, the rules or patterns must be consistent. We can use this idea to see how to multiply two numbers with different signs. Look for a pattern in this list of problems. As the numbers in blue decrease by 1, the numbers in red decrease by 2. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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Parallel Example 1 Multiplying Numbers with Different Signs Multiply. a. b. Factors have different signs so the product is negative. Negative Positive Factors have different signs so the product is negative. Positive Negative Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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Parallel Example 3 Dividing Signed Numbers Divide. Numbers have different signs, so the quotient is negative. Numbers have the same sign (both negative), so the quotient is positive. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

9.4 Order of Operations Objectives 1. Use the order of operations. 2. Use the order of operations with exponents. 3. Use the order of operations with fractions bars. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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Parallel Example 2 Parentheses and the Order of Operations a. −5(8 – 4) – 3 b. 3 + 4(4 – 9)(20 ÷ 4) −5(4) – 3 (–5)(20 ÷ 4) −20 – 3 (–5) (5) 3 + (–20) (5) −20 + (– 3) 3 + (–100) −23 –97 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 3 Exponents and the Order of Operations a. 52 − (−2)2 b. (−7)2 − (5 − 8)2 (−4) (−7)2 − (− 3)2 (−4) 25 − 4 49 − 9(−4) 21 49 − (−36) (+36) 85 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 4 Fraction Bars and the Order of Operations Simplify. First do the work in the numerator. Then do the work in the denominator. −14 + 3(5 – 7) 6 – 42 ÷ 8 −14 + 3(– 2) 6 – 16 ÷ 8 −14 + (– 6) 6 – 2 4 −20 Denominator Numerator Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

9.5 Evaluating Expressions and Formulas
Objectives 1. Define variable and expression. 2. Find the value of an expression when values of the variables are given. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

In algebra, we often write multiplication without the multiplication dots. If there is no operation sign written between two letters, or between a letter and a number, you assume it is multiplication. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Letters (such as the l, p, r and t used on the previous slide) that represent numbers are called variables. A combination of operations on letters and numbers is an expression. Examples: 9 + p 9r 2w – 3a The value of an expression changes depending on the value of each variable. To find the value of an expression, replace the variables with their values. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 1 Finding the Value of an Expression Find the value of 6x – 2y, if x = 2 and y = 5. Replace x with 2. Replace y with 5. Using the order of operations, do all multiplications first. Then subtract. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 2 Finding the Value of an Expression What is the value of if w = –2, a = 3 and t = –1? Replace w with 2, a with 3 and t with 1. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 3 Evaluating an Expression with Negative Signs and Negative Values Find the value of x – 2y, if x = 2 and y = 5. Replace x with 2. Replace y with 5. (2) is the opposite of 2 which is (+2). Multiply 2  5. Change subtraction to addition; change (15) to its opposite, (+15). Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 4 Evaluating a Formula The formula used in a previous chapter for the area of a triangle can now be written without the multiplication dots. In this formula, b is the length of the base of the triangle and h is the height of the triangle. What is the area if b = 10 cm and h = 32 cm? Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

9.6 Solving Equations Objectives 1. Determine whether a number is a solution of an equation. 2. Solve equations using the addition property of equations. 3. Solve equations using the multiplication property of equations. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 1 Determining Whether a Number is a Solution of an Equation Is 9 a solution of either one of these equations? a. 16 = x + 7 b. 3y + 2 = 30 Replace x with 9. Replace y with 9. 3y + 2 = 30 16 = x + 7 3(9) + 2 = 30 16 = 9 + 7 = 30 16 = 16 True 29 = 30 False 9 is a solution of the equation. 9 is not a solution of the equation. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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Parallel Example 2 Solving Equations Using the Addition Property Solve each equation. a. m – 13 = 28 m – = m + 0 = 41 Check: m = 41 m – 13 = 28 The solution is 41. To check, replace m with 41 in the original equation. 41 – 13 = 28 28 = 28 The result is true, so 41 is the solution. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 2 continued Solving Equations Using the Addition Property Solve each equation. b. 5 = n + 7 5 + (−7) = n (–7) –2 = n Check: –2 = n 5 = n + 7 The solution is −2. To check, replace n with −2 in the original equation. 5 = −2 + 7 5 = 5 The result is true, so −2 is the solution. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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Parallel Example 3 Solving Equations Using the Multiplication Property Solve each equation. a. 6k = 54 Divide both sides by 6, to get k by itself. 1 1 Check: The solution is 9. To check, replace k with 9 in the original equation. 6k = 54 6 ∙ 9 = 54 54 = 54 The result is true, so 9 is the solution. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 3 continued Solving Equations Using the Multiplication Property Solve each equation. b. −8y = 32 Divide both sides by −8, to get y by itself. 1 1 Check: −8y = 32 −8(−4) = 32 The solution is −4. To check, replace y with −4 in the original equation. 32 = 32 The result is true, so −4 is the solution. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 4 Solving Equations Using the Multiplication Property Solve each equation. a. Multiply both sides by 5, to get x by itself. 1 1 Check: The solution is 35. To check, replace x with 35 in the original equation. The result is true, so 35 is the solution. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 4 continued Solving Equations Using the Multiplication Property b. Multiply both sides by −9/2, to get m by itself. 1 1 2 Check: 1 1 1 −2 The solution is −18. To check, replace m with −18 in the original equation. 1 The result is true, so −18 is the solution. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Here is a summary of the rules for using the multiplication property. In these rules, x, is the variable and a, b, and c represent numbers. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

9.7 Solving Equations with Several Steps
Objectives 1. Solve equations with several steps. 2. Use the distributive property. 3. Combine like terms. 4. Solve more difficult equations. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 1 Solving an Equation with Several Steps Solve 4w + 2 = 18. Step 1 Subtract 2 from both sides. Step 2 Divide both sides by 4. Step 3 Check the solution. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

We can use the order of operations to simplify these two expressions. Because both answers are the same, the two expressions are equivalent. This is an example of the distributive property. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

A single letter or number, or the product of a variable and a number, makes up a term. Examples of terms: 4y 7 –2 9w 5y2 Terms with exactly the same variable and the same exponent are called like terms. 5x and 3x like terms 5x and 3y not like terms; variables are different 5x2 and 3x3 not like terms; exponents are different 5x4 and 3x4 like terms Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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Parallel Example 5 Solving Equations Using the Distributive Property Solve 10 = 2(y – 5) Step 1 Use the distributive property on the right side of the equation. Step 2 Combine like terms. Check the left side of the equation. There are no like terms. Check the right side. No like terms there either, so go to Step 3. Step 3 Add 10 to both sides. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 5 Solving Equations Using the Distributive Property Solve 10 = 2(y – 5) Step 4 Divide both sides by 2. Step 5 Check. Go back to the original equation. 0 is the correct solution (not 10). Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

9.8 Using Equations to Solve Application Problems
Objectives 1. Translate word phrases into expressions with variables. 2. Translate sentences into equations. 3. Solve application problems. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 1 Translating Word Phrases into Expressions with Variables Write each word phrase in symbols, using x as the variable. Words Algebraic Expression A number plus nine 7 more than a number −12 added to a number 3 less than a number A number decreased by 1 14 minus a number x + 9 or 9 + x x + 7 or 7 + x −12 + x or x + (−12) x – 3 x – 1 14 – x Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 2 Translating Word Phrases into Expressions with Variables Write each word phrase in symbols, using x as the variable. Words Algebraic Expression 3 times a number Twice a number The quotient of 8 and a number A number divided by 15 The result is 3x 2x = Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 2 Translating a Sentence into an Equation If 8 times a number is added to 13, the result is 45. Find the number. Let x represent the unknown number. 8 times a number added to 13 is 45 8x + 13 = 45 Next, solve the equation. Check: 8x + 13 − 13 = 45 − 13 8x + 13 = 45 8x = 32 8(4) + 13 = 45 45 = 45 The solution is 4. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

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Parallel Example 5 Solving an Application Problem with One Unknown Frankie has washed 6 less than twice as many windows washed as Rita. If Frankie has washed 14 windows, how many windows has Rita washed? Step 1 Read. The problem asks for the number of windows that Rita has washed. Step 2 Assign a variable. There is only one unknown, Rita’s number of windows washed. Step 3 Write an equation. 6 less than twice Rita’s number. The number Frankie washed. 14 = 2x – 6 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 5 continued Solving an Application Problem with One Unknown Frankie has washed 6 less than twice as many windows washed as Rita. If Frankie has washed 14 windows, how many windows has Rita washed? Step 4 Solve. 14 = 2x – 6 = 2x – 6 + 6 20 = 2x Step 5 State the answer. Rita washed 20 windows. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 5 continued Solving an Application Problem with One Unknown Frankie has washed 6 less than twice as many windows washed as Rita. If Frankie has washed 14 windows, how many windows has Rita washed? Step 6 Check. 14 = 2x – 6 14 = 2(10) – 6 14 = 14 So 10 is the correct solution because it “works” in the original problem. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 6 Solving an Application Problem with Two Unknowns On a shopping spree, Yoshi spent $54 more than Lowell. The total spent by them both was $276. Find the amount spent by each person. Step 1 Read. The problem asks for the amount spent by each person. Step 2 Assign a variable. There are two unknowns. Let x represent the amounts spent by Lowell and x + 54 be the amount spent by Yoshi. Step 3 Write an equation. Amount spent by Yoshi. Amount spent by Lowell x + x + 54 = 276 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 6 continued Solving an Application Problem with Two Unknowns On a shopping spree, Yoshi spent $54 more than Lowell. The total spent by them both was $276. Find the amount spent by each person. Step 4 Solve. x + x + 54 = 276 2x + 54 = 276 2x + 54 − 54 = 276 − 54 2x = 222 1 1 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 6 continued Solving an Application Problem with One Unknown Step 5 State the answer. The amount Lowell spent is x, so Lowell spent $111. The amount Yoshi spent is x + 54, so Yoshi spent $165. Step 6 Check. Use the words in the original problem. Yoshi’s $165 is $54 more dollars than Lowell’s $111, so that checks. The total spent is $111 + $165 = $276 which also checks. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 7 Solving a Geometry Application Problem The length of a rectangle is 3 inches more than the width. The perimeter is 78 inches. Find the length and width. Step 1 Read. The problem asks for the length and width of the rectangle. Step 2 Assign a variable. There are two unknowns, length and width. Let x represent the width and x + 3 represent the length. Step 3 Write an equation. Use the formula for perimeter of a rectangle. P = 2l + 2w Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 7 continued Solving a Geometry Application Problem The length of a rectangle is 3 inches more than the width. The perimeter is 78 inches. Find the length and width. Step 4 Solve. P = 2l + 2w 78 = 2(x + 3) + 2 ∙ x 78 = 2x x 78 = 4x + 6 78 – 6 = 4x + 6 – 6 72 = 4x 1 18 = x 1 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Parallel Example 7 continued Solving a Geometry Application Problem Step 5 State the answer. The width is x, so the width is 18 inches. The length is x + 3, so the length is 21 inches. Step 6 Check. Use the words in the original problem. The original problem says that the perimeter is 78 inches. P = 2 ∙ 18 in. + 2 ∙21in. P = 36 in in. P = 78 in. checks Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide

Chapter 9 Basic Algebra © 2010 Pearson Education, Inc. All rights reserved.

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Chapter 9 Basic Algebra © 2010 Pearson Education, Inc. All rights reserved.

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