University:
California State University, Northridge
Course:
MATH 320 | Foundations of Higher Mathematics
Academic year:

2024
Views:

172

Pages:

24
Author:

Megan Lawrence

Chapter 9 Equations, Inequalities and Problem Solving Part 3 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 2 § 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) 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 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 + 30 102 – 30 = 4x + 30 – 30 72 = 4x 72 4 x = 4 4 18 = x (Simplify right side) (Subtract 30 from both sides) (Simplify both sides) (Divide both sides by 4) (Simplify both sides) 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. 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 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) 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) 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) 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 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 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 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 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. 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. 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 = 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%. 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 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 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 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. 49

Equations, Inequalities and Problem Solving, Part 3

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