Equations Inequalities and Problem Solving Chapter Sections 1

Equations, Inequalities and Problem Solving

Chapter Sections 1 – The Addition Property of Equality 2 – The Multiplication Property of Equality 3 – Further Solving Linear Equations 4 – Introduction to Problem Solving 5 – Formulas and Problem Solving 6 – Percent and Mixture Problem Solving 7 – Solving Linear Inequalities Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 2

§ 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, Developmental Introductory Algebra, Mathematics 3 ed 4

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, Developmental Introductory Algebra, Mathematics 3 ed 5

Solving Equations Example 4 p – 11 – p = 2 + 2 p – 20 3 p – 11 = 2 p – 18 (Simplify both sides) 3 p + (– 2 p) – 11 = 2 p + (– 2 p) – 18 p – 11 = – 18 (Add – 2 p to both sides) (Simplify both sides) p – 11 + 11 = – 18 + 11 (Add 11 to both sides) p=– 7 (Simplify both sides) Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 6

Solving Equations Example 5(3 + z) – (8 z + 9) = – 4 z 15 + 5 z – 8 z – 9 = – 4 z 6 – 3 z + 4 z = – 4 z + 4 z 6+z=0 6 + (– 6) + z = 0 +( – 6) z=– 6 (Use distributive property) (Simplify left side) (Add 4 z to both sides) (Simplify both sides) (Add – 6 to both sides) (Simplify both sides) Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 7

§ 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, Developmental Introductory Algebra, Mathematics 3 ed 9

Solving Equations Example (Multiply both sides by 7) (Simplify both sides) Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 10

Solving Equations Example (Multiply both sides by fraction) (Simplify both sides) Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 11

Solving Equations Recall that multiplying by a number is equivalent to dividing by its reciprocal Example 3 z – 1 = 26 3 z – 1 + 1 = 26 + 1 3 z = 27 (Add 1 to both sides) (Simplify both sides) (Divide both sides by 3) z=9 (Simplify both sides) Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 12

Solving Equations Example 12 x + 30 + 8 x – 6 = 10 20 x + 24 = 10 (Simplify left side) 20 x + 24 + (– 24) = 10 + (– 24) (Add – 24 to both sides) 20 x = – 14 (Simplify both sides) (Divide both sides by 20) (Simplify both sides) Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 13

§ 3 Further Solving Linear Equations

Solving Linear Equations Solving linear equations in one variable 1) 2) 3) 4) 5) 6) 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) Get variable alone (multiplication property of equality) Check solution by substituting into original problem Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 15

Solving Linear Equations Example (Multiply both sides by 5) (Simplify) (Add – 3 y to both sides) (Simplify; add – 30 to both sides) (Simplify; divide both sides by 7) (Simplify both sides) Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 16

Solving Linear Equations Example 5 x – 5 = 2(x + 1) + 3 x – 7 5 x – 5 = 2 x + 2 + 3 x – 7 (Use distributive property) 5 x – 5 = 5 x – 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, Developmental Introductory Algebra, Mathematics 3 ed 17

Solving Linear Equations Example 3 x – 7 = 3(x + 1) 3 x – 7 = 3 x + 3 (Use distributive property) 3 x + (– 3 x) – 7 = 3 x + (– 3 x) + 3 (Add – 3 x 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, Developmental Introductory Algebra, Mathematics 3 ed 18

§ 4 An Introduction to Problem Solving

Strategy for Problem Solving General Strategy for Problem Solving 1) 2) 3) 4) 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 Translate the problem into an equation Solve the equation Interpret the result • Check proposed solution in problem • State your conclusion Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 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 2 x, “the product of twice a number and three” translates to 2 x · 3, “five times the number” translates to 5 x, and “the difference of five times the number and ¾” translates to 5 x – ¾. Continued Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 21

Finding an Unknown Number Example continued 2. ) Translate The product of is the same as twice a number 2 x the difference of and 3 · 3 = 5 times the number 5 x and ¾ – ¾ Continued Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 22

Finding an Unknown Number Example continued 3. ) Solve 2 x · 3 = 5 x – ¾ 6 x = 5 x – ¾ (Simplify left side) 6 x + (– 5 x) = 5 x + (– 5 x) – ¾ x=–¾ (Add – 5 x 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, Developmental Introductory Algebra, Mathematics 3 ed 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. 29 x = the cost for mileage driven Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed Continued 24

Solving a Problem Example continued 2. ) Translate Daily costs mileage costs plus 2(24. 95) + maximum budget is equal to 0. 29 x = 100 Continued Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 25

Solving a Problem Example continued 3. ) Solve 2(24. 95) + 0. 29 x = 100 49. 90 + 0. 29 x = 100 (Simplify left side) 49. 90 – 49. 90 + 0. 29 x = 100 – 49. 90 (Subtract 49. 90 from both sides) 0. 29 x = 50. 10 (Simplify both sides) (Divide both sides by 0. 29) x 172. 75 (Simplify both sides) Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 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, Developmental Introductory Algebra, Mathematics 3 ed 27

§ 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, Developmental Introductory Algebra, Mathematics 3 ed 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 2 x = the length of the second side, and x + 30 = the length of the third side Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed Continued 30

Using Formulas Example continued 2. ) Translate Formula: P = a + b + c Substitute: 102 = x + 2 x + 30 3. ) Solve 102 = x + 2 x + 30 102 = 4 x + 30 (Simplify right side) 102 – 30 = 4 x + 30 – 30 72 = 4 x (Subtract 30 from both sides) (Simplify both sides) (Divide both sides by 4) 18 = x (Simplify both sides) Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 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, Developmental Introductory Algebra, Mathematics 3 ed 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, Developmental Introductory Algebra, Mathematics 3 ed 33

Solving Equations for a Specific Variable Example Solve for n. (Divide both sides by mr) (Simplify right side) Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 34

Solving Equations for a Specific Variable Example Solve for T. (Subtract P from both sides) (Simplify right side) (Divide both sides by PR) (Simplify right side) Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 35

Solving Equations for a Specific Variable Example Solve for P. (Factor out P from both terms on the right side) (Divide both sides by 1 + RT) (Simplify the right side) Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 36

§ 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, Developmental Introductory Algebra, Mathematics 3 ed 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 is 9% of 65 Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 39

Solving a Percent Problem: Base Unknown amount = percent · base 36 is 6% of what? 36 = 6% · n 36 = 0. 06 n 36 is 6% of 600 Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 40

Solving a Percent Problem: Percent Unknown amount = percent · base 24 is what percent of 144? 24 = n 144 Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 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 = = = $66 $66 Mark and Peggy can spend up to $55 on the meal itself. Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 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, Developmental Introductory Algebra, Mathematics 3 ed 43

Solving Increase Problems 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 The car’s cost increased by 8%. Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 44

Solving Decrease Problems 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 Patrick’s weight decreased by 40%. Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 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, Developmental Introductory Algebra, Mathematics 3 ed 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 Price per Pound 6 8 7. 50 Value of Candy 6 n 8(144 n) 144(7. 50) 6 n + 8(144 n) = 144(7. 5) # of pounds of of $8 candy of $7. 50 candy $6 candy Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed Continued 47

Solving Mixture Problems Example continued 3. ) Solve 6 n + 8(144 n) = 144(7. 5) 6 n + 1152 8 n = 1080 1152 2 n = 1080 2 n = 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, Developmental Introductory Algebra, Mathematics 3 ed 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, Developmental Introductory Algebra, Mathematics 3 ed 49

§ 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 Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 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 {x x 7} Represents the set {x x > – 4} Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 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] Represents the set (– 4, ) Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 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, Developmental Introductory Algebra, Mathematics 3 ed 54

Solving Linear Inequalities Solving linear inequalities in one variable 1) 2) 3) 4) 5) 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) Isolate variable (multiplication property of equality) Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 55

Solving Linear Inequalities Example 3 x + 9 5(x – 1) 3 x + 9 5 x – 5 (Use distributive property on right side) 3 x – 3 x + 9 5 x – 3 x – 5 (Subtract 3 x from both sides) 9 2 x – 5 (Simplify both sides) 9 + 5 2 x – 5 + 5 (Add 5 to both sides) 14 2 x (Simplify both sides) 7 x (Divide both sides by 2) Graph of solution (– , 7] Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 56

Solving Linear Inequalities Example 7(x – 2) + x > – 4(5 – x) – 12 7 x – 14 + x > – 20 + 4 x – 12 8 x – 14 > 4 x – 32 8 x – 4 x – 14 > 4 x – 32 4 x – 14 > – 32 4 x – 14 + 14 > – 32 + 14 4 x > – 18 (Use distributive property) (Simplify both sides) (Subtract 4 x from both sides) (Simplify both sides) (Add 14 to both sides) (Simplify both sides) (Divide both sides by 4 and simplify) Graph of solution ( , ) Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 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, Developmental Introductory Algebra, Mathematics 3 ed 58

Solving Compound Inequalities Example 0 4(5 – x) < 8 0 20 – 4 x < 8 (Use the distributive property) 0 – 20 – 4 x < 8 – 20 (Subtract 20 from each part) – 20 – 4 x < – 12 (Simplify each part) 5 x>3 (Divide each part by – 4) Remember that the sign direction changes when you divide by a number < 0! Graph of solution (3, 5] Martin-Gay, Developmental Introductory Algebra, Mathematics 3 ed 59