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

2024
Views:

465

Pages:

16
Author:

GaivaLiarma0w6

Chapter 9 Equations, Inequalities and Problem Solving Part 2 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 § 9.3 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 15 Solving Linear Equations Example 3( y + 3) = 2y + 6 5 5 ⋅ 3( y + 3) = 5(2 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) 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.” 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.” 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 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 21 Finding an Unknown Number Example continued 2.) Translate The product of twice a number 2x and 3 · 3 is the same as = the difference of 5 times the number 5x and ¾ – ¾ Continued 22 Finding an Unknown Number Example continued 3.) Solve 2x · 3 = 5x – ¾ 6x = 5x – ¾ 6x + (– 5x) = 5x + (– 5x) – ¾ x=–¾ (Simplify left side) (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. 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 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 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) 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. 27