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Chapter 9 Equations, Inequalities and Problem Solving Chapter Sections 9.1 – The Addition Property of Equality 9.2 – The Multiplication Property of Equality 9.3 – Further Solving Linear Equations 9.4 – Introduction to Problem Solving 9.5 – Formulas and Problem Solving 9.6 – Percent and Mixture Problem Solving 9.7 – Solving Linear Inequalities Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 2 § 9.1 The Addition Property of Equality Linear Equations Linear equation in one variable can be written in the form ax + b = c, a 0 Equivalent equations are equations with the same solutions in the form of variable = number, or number = variable Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 4 Addition Property of Equality Addition Property of Equality a = b and a + c = b + c are equivalent equations Example a.) 8 + z = – 8 8 + (– 8) + z = – 8 + – 8 z = – 16 (Add –8 to each side) (Simplify both sides) Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 5 Solving Equations Example 4p – 11 – p = 2 + 2p – 20 3p – 11 = 2p – 18 (Simplify both sides) 3p + (– 2p) – 11 = 2p + (– 2p) – 18 p – 11 = – 18 (Simplify both sides) p – 11 + 11 = – 18 + 11 p=–7 (Add –2p to both sides) (Add 11 to both sides) (Simplify both sides) Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 6 Solving Equations Example 5(3 + z) – (8z + 9) = – 4z 15 + 5z – 8z – 9 = – 4z 6 – 3z = – 4z (Use distributive property) (Simplify left side) 6 – 3z + 4z = – 4z + 4z (Add 4z to both sides) 6+z=0 6 + (– 6) + z = 0 +( – 6) z=–6 (Simplify both sides) (Add –6 to both sides) (Simplify both sides) Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 7 § 9.2 The Multiplication Property of Equality Multiplication Property of Equality Multiplication property of equality a = b and ac = bc are equivalent equations Example –y=8 (– 1)(– y) = 8(– 1) y=–8 (Multiply both sides by –1) (Simplify both sides) Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 9 Solving Equations Example 1 5 x 7 9 1 5 7 x 7 7 9 35 x 9 (Multiply both sides by 7) (Simplify both sides) Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 10 Solving Equations Example 8 x6 3 8 3 8 x 6 38 3 x 16 (Multiply both sides by fraction) (Simplify both sides) Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 11 Solving Equations Recall that multiplying by a number is equivalent to dividing by its reciprocal Example 3z – 1 = 26 3z – 1 + 1 = 26 + 1 (Add 1 to both sides) 3z = 27 3 z 27 3 3 (Simplify both sides) z=9 (Simplify both sides) (Divide both sides by 3) Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 12 Solving Equations Example 12x + 30 + 8x – 6 = 10 20x + 24 = 10 20x + 24 + (– 24) = 10 + (– 24) (Simplify left side) (Add –24 to both sides) 20x = – 14 (Simplify both sides) 20 x 14 20 20 (Divide both sides by 20) 7 x 10 (Simplify both sides) Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 13 § 9.3 Further Solving Linear Equations Solving Linear Equations Solving linear equations in one variable 1) 2) 3) 4) Multiply to clear fractions Use distributive property Simplify each side of equation Get all variable terms on one side and number terms on the other side of equation (addition property of equality) 5) Get variable alone (multiplication property of equality) 6) Check solution by substituting into original problem Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 15 Solving Linear Equations Example 3( y 3) 2y 6 5 5 3( y 3) 52 y 6 5 3 y 9 10 y 30 (Multiply both sides by 5) (Simplify) 3 y (3 y ) 9 10 y (3 y ) 30 (Add –3y to both sides) 9 (30) 7 y 30 (30) (Simplify; add –30 to both sides) 21 7 y 7 7 3 y (Simplify; divide both sides by 7) (Simplify both sides) Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 16 Solving Linear Equations Example 5x – 5 = 2(x + 1) + 3x – 7 5x – 5 = 2x + 2 + 3x – 7 (Use distributive property) 5x – 5 = 5x – 5 (Simplify the right side) 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.” Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 17 Solving Linear Equations Example 3x – 7 = 3(x + 1) 3x – 7 = 3x + 3 (Use distributive property) 3x + (– 3x) – 7 = 3x + (– 3x) + 3 (Add –3x to both sides) –7=3 (Simplify both sides) Since no value for the variable x can be substituted into this equation that will make this a true statement, there is “no solution.” Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 18 § 9.4 An Introduction to Problem Solving Strategy for Problem Solving General Strategy for Problem Solving 1) Understand the problem • Read and reread the problem • Choose a variable to represent the unknown • Construct a drawing, whenever possible • Propose a solution and check 2) Translate the problem into an equation 3) Solve the equation 4) Interpret the result • Check proposed solution in problem • State your conclusion Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 20 Finding an Unknown Number Example The product of twice a number and three is the same as the difference of five times the number and ¾. Find the number. 1.) Understand Read and reread the problem. If we let x = the unknown number, then “twice a number” translates to 2x, “the product of twice a number and three” translates to 2x · 3, “five times the number” translates to 5x, and “the difference of five times the number and ¾” translates to 5x – ¾. Continued Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 21 Finding an Unknown Number Example continued 2.) Translate The product of is the same as twice a number 2x the difference of and 3 · 3 = 5 times the number 5x and ¾ – ¾ Continued Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 22 Finding an Unknown Number Example continued 3.) Solve 2x · 3 = 5x – ¾ 6x = 5x – ¾ (Simplify left side) 6x + (– 5x) = 5x + (– 5x) – ¾ x=–¾ (Add –5x to both sides) (Simplify both sides) 4.) Interpret Check: Replace “number” in the original statement of the problem with – ¾. The product of twice – ¾ and 3 is 2(– ¾)(3) = – 4.5. The difference of five times – ¾ and ¾ is 5(– ¾) – ¾ = – 4.5. We get the same results for both portions. State: The number is – ¾. Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 23 Solving a Problem Example A car rental agency advertised renting a Buick Century for $24.95 per day and $0.29 per mile. If you rent this car for 2 days, how many whole miles can you drive on a $100 budget? 1.) Understand Read and reread the problem. Let’s propose that we drive a total of 100 miles over the 2 days. Then we need to take twice the daily rate and add the fee for mileage to get 2(24.95) + 0.29(100) = 49.90 + 29 = 78.90. This gives us an idea of how the cost is calculated, and also know that the number of miles will be greater than 100. If we let x = the number of whole miles driven, then 0.29x = the cost for mileage driven Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed Continued 24 Solving a Problem Example continued 2.) Translate Daily costs mileage costs plus 2(24.95) + maximum budget is equal to 0.29x = 100 Continued Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 25 Solving a Problem Example continued 3.) Solve 2(24.95) + 0.29x = 100 49.90 + 0.29x = 100 (Simplify left side) 49.90 – 49.90 + 0.29x = 100 – 49.90 (Subtract 49.90 from both sides) 0.29x = 50.10 0.29 x 50.10 0.29 0.29 x 172.75 (Simplify both sides) (Divide both sides by 0.29) (Simplify both sides) Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed Continued 26 Solving a Problem Example continued 4.) Interpret Check: Recall that the original statement of the problem asked for a “whole number” of miles. If we replace “number of miles” in the problem with 173, then 49.90 + 0.29(173) = 100.07, which is over our budget. However, 49.90 + 0.29(172) = 99.78, which is within the budget. State: The maximum number of whole number miles is 172. Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 27 § 9.5 Formulas and Problem Solving 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) Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 29 Using Formulas Example 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. 1.) 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 Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed Continued 30 Using Formulas Example continued 2.) Translate Formula: P = a + b + c Substitute: 102 = x + 2x + x + 30 3.) Solve 102 = x + 2x + x + 30 102 = 4x + 30 (Simplify right side) 102 – 30 = 4x + 30 – 30 (Subtract 30 from both sides) 72 = 4x (Simplify both sides) 72 4 x 4 4 (Divide both sides by 4) 18 = x (Simplify both sides) Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed Continued 31 Using Formulas Example continued 4.) 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. Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 32 Solving Formulas It is often necessary to rewrite a formula so that it is solved for one of the variables. This is accomplished by isolating the designated variable on one side of the equal sign. Solving Equations for a Specific Variable 1) 2) 3) 4) Multiply to clear fractions Use distributive to remove grouping symbols Combine like terms to simply each side Get all terms containing specified variable on the same time, other terms on opposite side 5) Isolate the specified variable Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 33 Solving Equations for a Specific Variable Example Solve for n. T mnr T mnr mr mr (Divide both sides by mr) T n mr (Simplify right side) Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 34 Solving Equations for a Specific Variable Example Solve for T. A P PRT A P P P PRT (Subtract P from both sides) A P PRT (Simplify right side) A P PRT PR PR (Divide both sides by PR) A P T PR (Simplify right side) Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 35 Solving Equations for a Specific Variable Example Solve for P. A P PRT A P(1 RT ) (Factor out P from both terms on the right side) A P(1 RT ) 1 RT 1 RT (Divide both sides by 1 + RT) A P 1 RT (Simplify the right side) Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 36 § 9.6 Percent and Mixture Problem Solving Solving a Percent Problem A percent problem has three different parts: amount = percent · base Any one of the three quantities may be unknown. 1. When we do not know the amount: n = 10% · 500 2. When we do not know the base: 50 = 10% · n 3. When we do not know the percent: 50 = n · 500 Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 38 Solving a Percent Problem: Amount Unknown amount = percent · base What is 9% of 65? n = 9% · 65 n = (0.09) (65) n = 5.85 5.85 is 9% of 65 Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 39 Solving a Percent Problem: Base Unknown amount = percent · base 36 is 6% of what? 36 = 6% · n 36 = 0.06n 36 0.06n = 0.06 0.06 600 = n 36 is 6% of 600 Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 40 Solving a Percent Problem: Percent Unknown amount = percent · base 24 is what percent of 144? 24 = n 144 24 = 144n 24 144n = 144 144 0.16 = n 2 16 % = n 3 2 24 is 16 % of 144 3 Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 41 Solving Markup Problems Example Mark is taking Peggy out to dinner. He has $66 to spend. If he wants to tip the server 20%, how much can he afford to spend on the meal? Let n = the cost of the meal. Cost of meal n 100% of n + + tip of 20% of the cost 20% of n 120% of n 1.2n 66 1.2n 66 1.2 1.2 n 55 = = = $66 $66 $66 Mark and Peggy can spend up to $55 on the meal itself. Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 42 Solving Discount Problems Example Julie bought a leather sofa that was on sale for 35% off the original price of $1200. What was the discount? How much did Julie pay for the sofa? Discount = discount rate list price = 35% 1200 The discount was $420. = 420 Amount paid = list price – discount = 1200 – 420 = 780 Julie paid $780 for the sofa. Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 43 Solving Increase Problems Percent of increase = amount of increase original amount Example The cost of a certain car increased from $16,000 last year to $17,280 this year. What was the percent of increase? Amount of increase = original amount – new amount = 17,280 – 16,000 = 1280 amount of increase original amount 1280 = = 0.08 The car’s cost increased by 8%. 16000 Percent of increase = Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 44 Solving Decrease Problems Percent of decrease = amount of decrease original amount Example Patrick weighed 285 pounds two years ago. After dieting, he reduced his weight to 171 pounds. What was the percent of decrease in his weight? Amount of decrease = original amount – new amount = 285 – 171 = 114 amount of decrease Percent of decrease = original amount 114 = = 0.4 285 Patrick’s weight decreased by 40%. Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 45 Solving Mixture Problems Example The owner of a candy store is mixing candy worth $6 per pound with candy worth $8 per pound. She wants to obtain 144 pounds of candy worth $7.50 per pound. How much of each type of candy should she use in the mixture? 1.) Understand Let n = the number of pounds of candy costing $6 per pound. Since the total needs to be 144 pounds, we can use 144 n for the candy costing $8 per pound. Continued Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 46 Solving Mixture Problems Example continued 2.) Translate Use a table to summarize the information. $6 candy $8 candy $7.50 candy Number of Pounds n 144 n 144 Price per Pound 6 8 7.50 Value of Candy 6n 8(144 n) 144(7.50) 6n + 8(144 n) = 144(7.5) # of pounds of $6 candy # of pounds of $8 candy # of pounds of $7.50 candy Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed Continued 47 Solving Mixture Problems Example continued 3.) Solve 6n + 8(144 n) = 144(7.5) 6n + 1152 8n = 1080 1152 2n = 1080 2n = 72 n = 36 (Eliminate the parentheses) (Combine like terms) (Subtract 1152 from both sides) (Divide both sides by 2) She should use 36 pounds of the $6 per pound candy. She should use 108 pounds of the $8 per pound candy. (144 n) = 144 36 = 108 Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed Continued 48 Solving Mixture Problems Example continued 4.) Interpret Check: Will using 36 pounds of the $6 per pound candy and 108 pounds of the $8 per pound candy yield 144 pounds of candy costing $7.50 per pound? ? 6(36) + 8(108) = 144(7.5) ? 216 + 864 = 1080 ? 1080 = 1080 State: She should use 36 pounds of the $6 per pound candy and 108 pounds of the $8 per pound candy. Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 49 § 9.7 Solving Linear Inequalities Linear Inequalities A linear inequality in one variable is an equation that can be written in the form ax + b < c • a, b, and c are real numbers, a 0 • < symbol could be replaced by > or or Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 51 Graphing Solutions Graphing solutions to linear inequalities in one variable (using circles) • 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 {xx 7} 7 Represents the set {xx > – 4} -4 Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 52 Linear Inequalities Graphing solutions to linear inequalities in one variable (using interval notation) • Use a number line • Use a bracket at the endpoint of a interval if you want to include the point • Use a parenthesis at the endpoint if you DO NOT want to include the point Interval Notation ] Represents the set (– , 7] 7 ( Represents the set (– 4, ) -4 Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 53 Properties of Inequality Addition Property of Inequality • a < b and a + c < b + c are equivalent inequalities Multiplication Property of Inequality • a < b and ac < bc are equivalent inequalities, if c is positive • a < b and ac > bc are equivalent inequalities, if c is negative Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 54 Solving Linear Inequalities Solving linear inequalities in one variable 1) 2) 3) 4) Multiply to clear fractions Use distributive property Simplify each side of equation Get all variable terms on one side and numbers on the other side of equation (addition property of equality) 5) Isolate variable (multiplication property of equality) Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 55 Solving Linear Inequalities Example 3x + 9 5(x – 1) 3x + 9 5x – 5 (Use distributive property on right side) 3x – 3x + 9 5x – 3x – 5 (Subtract 3x from both sides) 9 2x – 5 (Simplify both sides) 9 + 5 2x – 5 + 5 (Add 5 to both sides) 14 2x (Simplify both sides) 7x (Divide both sides by 2) Graph of solution (– ,7] ] 7 Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 56 Solving Linear Inequalities Example 7(x – 2) + x > – 4(5 – x) – 12 7x – 14 + x > – 20 + 4x – 12 8x – 14 > 4x – 32 8x – 4x – 14 > 4x – 4x – 32 4x – 14 > –32 4x – 14 + 14 > –32 + 14 4x > –18 x 9 2 (Use distributive property) (Simplify both sides) (Subtract 4x from both sides) (Simplify both sides) (Add 14 to both sides) (Simplify both sides) (Divide both sides by 4 and simplify) Graph of solution ( -9 ,) 2 ( -9 2 Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 57 Compound Inequalities A compound inequality is two inequalities joined together. 0 4(5 – x) < 8 To solve the compound inequality, perform operations simultaneously to all three parts of the inequality (left, middle and right). Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 58 Solving Compound Inequalities Example 0 4(5 – x) < 8 0 20 – 4x < 8 (Use the distributive property) 0 – 20 20 – 20 – 4x < 8 – 20 (Subtract 20 from each part) – 20 – 4x < – 12 (Simplify each part) 5x>3 (Divide each part by –4) Remember that the sign direction changes when you divide by a number < 0! Graph of solution (3,5] ( ] 3 5 Martin-Gay, Martin-Gay,Developmental Introductory Algebra, Mathematics 3ed 59