Download presentation

Presentation is loading. Please wait.

Published byAugustus Gilmore Modified over 9 years ago

1
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Chapter 9 Equations, Inequalities, and Problem Solving

2
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 9.1 Symbols and Sets of Numbers

3
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 33 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Set of Numbers Natural numbers – {1, 2, 3, 4, 5, 6...} Whole numbers – {0, 1, 2, 3, 4...} Integers – {... –3, -2, -1, 0, 1, 2, 3...} Rational numbers – the set of all numbers that can be expressed as a quotient of integers, with denominator ≠ 0 Irrational numbers – the set of all numbers that can NOT be expressed as a quotient of integers Real numbers – the set of all rational and irrational numbers combined

4
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 44 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Equality and Inequality Symbols SymbolMeaning a = b a ≠ b a < b a > b a ≤ b a ≥ b a is equal to b. a is not equal to b. a is less than b. a is greater than b. a is less then or equal to b. a is greater than or equal to b.

5
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 55 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. The Number Line A number line is a line on which each point is associated with a number. 2– 201345– 1– 3– 4– 5 Negative numbers Positive numbers – 4.81.5

6
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 66 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. For any two real numbers a and b, a is less than b if a is to the left of b on the number line. a < b means a is to the left of b on a number line. a > b means a is to the right of b on a number line. Order Property for Real Numbers Insert between the following pair of numbers to make a true statement. Example

7
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 77 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Absolute Value The absolute value of a real number a, denoted by |a|, is the distance between a and 0 on the number line. 2– 201345– 1– 3– 4– 5 | – 4| = 4 Distance of 4 Symbol for absolute value |5| = 5 Distance of 5

8
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 9.2 Properties of Real Numbers

9
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 99 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Commutative and Associative Property Associative property of addition: (a + b) + c = a + (b + c) of multiplication: (a · b) · c = a · (b · c) Commutative property of addition: a + b = b + a of multiplication: a · b = b · a

10
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 10 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Distributive property of multiplication over addition a(b + c) = ab + ac Identities for addition: 0 is the identity element since a + 0 = a and 0 + a = a. for multiplication: 1 is the identity element since a · 1 = a and 1 · a = a. Distributive Property

11
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 11 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Inverses For multiplication: b and (for b ≠ 0) are reciprocals or multiplicative inverses of each other because their product is 1; that is, Inverses For addition: a and –a are additive inverses or opposites of each other because their sum is 0; that is, a + (– a) = 0 = 1

12
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 9.3 Further Solving Linear Equations

13
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 13 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. To Solve Linear Equations in One Variable Solving linear equations in one variable Step 1:If an equation contains fractions, multiply both sides by the LCD to clear the equation of fractions. Step 2: Use the distributive property to remove parentheses if they are present. Step 3:Simplify each side of the equation by combining like terms. Continued

14
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 14 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. To Solve Linear Equations in One Variable Step 4:Get all variable terms on one side and all numbers on the other side by using the addition property of equality. Step 5: Get the variable alone by using the multiplication property of equality. Step 6:Check the solution by substituting it into the original equation.

15
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 15 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Simplify. Divide both sides by 7. Simplify. Solving Linear Equations Example Multiply both sides by 5. Add –3y to both sides. Add –30 to both sides. Solve :

16
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 16 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving Linear Equations Example 5x – 5 = 2(x + 1) + 3x – 7 5x – 5 = 2x + 2 + 3x – 7 Apply the distributive property. 5x – 5 = 5x – 5 Combine like terms. Both sides of the equation are identical. Since this equation will be true for every x that is substituted into the equation, the solution is “all real numbers.” Solve: 5x – 5 = 2(x + 1) + 3x – 7

17
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 17 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving Linear Equations Example Since no value for the variable x can be substituted into this equation that will make this a true statement, there is “no solution.” 3x – 7 = 3(x + 1) 3x – 7 = 3x + 3 Apply the distributive property. – 7 = 3 Simplify. 3x + ( – 3x) – 7 = 3x + ( – 3x) + 3 Add –3x to both sides. Solve : 3x – 7 = 3(x + 1)

18
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 9.4 Further Problem Solving

19
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 19 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 1. UNDERSTAND the problem. During this step, become comfortable with the problem. Some ways of doing this are: Read and reread the problem. Choose a variable to represent the unknown. Construct a drawing. Propose a solution and check it. Pay careful attention to how you check your proposed solution. This will help when writing an equation to model the problem. Problem-Solving Steps Continued

20
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 20 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 2. TRANSLATE the problem into an equation. 3. SOLVE the equation. 4. INTERPRET the results. Check the proposed solution in the stated problem and state your conclusion. Problem-Solving Steps Martin-Gay, Prealgebra, 5ed

21
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 21 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Angles in a Triangle What is the sum of the measures of the interior angles of a triangle? A B C A C + B + The sum of the measures of the angles of any triangle is 180°. 21 Martin-Gay, Prealgebra & Introductory Algebra, 2ed

22
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 22 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Translating and Solving Problems The measure of the second angle of a triangle is twice the measure of the smallest angle. The measure of the third angle of the triangle is three times the measure of the smallest angle. Find the measures of the angles. UNDERSTAND: Let x = degree measure of smallest angle 2x = degree measure of second angle 3x = degree measure of third angle Draw a diagram. Continued Example

23
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 23 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Translating and Solving Problems TRANSLATE: Continued Recall that the sum of the measures of the angles of a triangle equals 180. x2x2x+3x3x+=180 measure of first angle measure of second angle measure of third angleequals180 Example continued

24
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 24 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Translating and Solving Problems SOLVE:x + 2x + 3x = 180 6x = 180 Combine like terms. Divide both sides by 6. 66 x = 30 INTERPRET: If x = 30, then 2x = 2(30) = 60 and 3x = 3(30) = 90 The sum of the angles is 30 + 60 + 90 = 180. Check: State:The smallest angle is 30º, the second angle is 60º, and the third angle is 90º. 6x = 180 Example continued

25
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 9.5 Formulas and Problem Solving

26
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 26 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Formulas A formula is an equation that states a known relationship among multiple quantities (has more than one variable in it) A = lw(Area of a rectangle = length · width) I = PRT(Simple Interest = Principal · Rate · Time) P = a + b + c(Perimeter of a triangle = side a + side b + side c) d = rt(distance = rate · time) V = lwh (Volume of a rectangular solid = length · width · height)

27
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 27 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Using Formulas A flower bed is in the shape of a triangle with one side twice the length of the shortest side, and the third side is 30 feet more than the length of the shortest side. Find the dimensions if the perimeter is 102 feet. UNDERSTAND: Read and reread the problem. Recall that the formula for the perimeter of a triangle is P = a + b + c. If we let x = the length of the shortest side, then 2x = the length of the second side, and x + 30 = the length of the third side Example Continued

28
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 28 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Using Formulas Example continued TRANSLATE: Continued Formula: P = a + b + cSubstitute: 102 = x + 2x + x + 30 SOLVE: 102= x + 2x + x + 30 102= 4x + 30Simplify right side. 102 – 30= 4x + 30 – 30Subtract 30 from both sides. 72= 4xSimplify both sides. Divide both sides by 4. 18 = x Simplify both sides.

29
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 29 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Using Formulas Example continued INTERPRET: Check: If the shortest side of the triangle is 18 feet, then the second side is 2(18) = 36 feet, and the third side is 18 + 30 = 48 feet. This gives a perimeter of P = 18 + 36 + 48 = 102 feet, the correct perimeter. State: The three sides of the triangle have a length of 18 feet, 36 feet, and 48 feet.

30
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 30 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving a Formula for a Variable It is often necessary to rewrite a formula so that it is solved for one of the variables. To solve a formula or an equation for a specified variable, we use the same steps as for solving a linear equation except that we treat the specified variable as the only variable in the equation.

31
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 31 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Step 1:Multiply on both sides to clear the equation of fractions if they appear. Step 2:Use the distributive property to remove parentheses if they appear. Step 3:Simplify each side of the equation by combining like terms. Step 4:Get all terms containing the specified variable on one side and all other terms on the other side by using the addition property of equality. Step 5:Get the specified variable alone by using the multiplication property of equality. Solving Equations for a Specified Varia ble

32
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 32 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving Equations for a Specified Variable Example Divide both sides by mr.Simplify. Solve T = mnr for n.

33
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 33 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving Equations for a Specified Variable Example Solve for A = PRT for T. Subtract P from both sides. Simplify. Divide both sides by PR. Simplify.

34
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 34 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving Equations for a Specified Variable Example Solve Factor out P on the right side. Divide both sides by 1 + RT.Simplify. for P.

35
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 9.6 Linear Inequalities and Problem Solving

36
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 36 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Linear Inequalities A linear inequality in one variable is an equation that can be written in the form ax + b < c where a, b, and c are real numbers, a ≠ 0. (the or ≤ or ≥.)

37
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 37 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Graphing solutions to linear inequalities in one variable Use a number line Use a closed circle at the endpoint of a interval if you want to include the point Use an open circle at the endpoint if you DO NOT want to include the point Represents the set {x x 7} Represents the set {x x > – 4} Graphing Solutions

38
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 38 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Addition Property of Inequality If a, b, and c are real numbers, then a < b and a + c < b + c are equivalent inequalities.

39
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 39 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Multiplication Property of Inequality 1.If a, b, and c are real numbers, and c is positive, then a < b and ac < bc are equivalent inequalities. 2.If a, b, and c are real numbers, and c is negative, then a bc are equivalent inequalities.

40
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 40 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. To Solve Linear Inequalities in One Variable Step 1: If an inequality contains fractions, multiply both sides by the LCD to clear the inequality of fractions. Step 2: Use distributive property to remove parentheses if they appear. Step 3: Simplify each side of inequality by combining like terms. Step 4: Get all variable terms on one side and all numbers on the other side by using the addition property of inequality. Step 5: Get the variable alone by using the multiplication property of inequality. Solving Linear Inequalities

41
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 41 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solve: 3x + 9 ≥ 5(x – 1). Graph the solution set. 3x + 9 ≥ 5x – 5 Apply the distributive property. 3x – 3x + 9 ≥ 5x – 3x – 5 Subtract 3x from both sides. 9 ≥ 2x – 5 Simplify. 14 ≥ 2x Simplify. 7 ≥ x Divide both sides by 2. 9 + 5 ≥ 2x – 5 + 5 Add 5 to both sides. The graph of solution set {x|x ≤ 7} is shown. Solving Linear Inequalities Example 3x + 9 ≥ 5(x – 1)

42
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 42 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 7(x – 2) + x > – 4(5 – x) – 12 7x – 14 + x > – 20 + 4x – 12 Apply the distributive property. 8x – 14 > 4x – 32 Combine like terms. 8x – 4x – 14 > 4x – 4x – 32 Subtract 4x from both sides. 4x – 14 > –32 Simplify. 4x – 14 + 14 > –32 + 14 Add 14 to both sides. 4x > –18 Simplify. Divide both sides by 4. Solving Linear Inequalities Example Solve: 7(x – 2) + x > – 4(5 – x) – 12. Graph the solution set. The graph of solution set {x|x > –9/2}is shown.

Similar presentations

© 2024 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google