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Equations, Inequalities and Problem Solving

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

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Equations, Inequalities and Problem Solving

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  1. Chapter 9 Equations, Inequalities and Problem Solving

  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

  3. The Addition Property of Equality § 9.1

  4. 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

  5. 8 + z = – 8 a.) 8 + (– 8) + z = – 8 + – 8 (Add –8 to each side) Addition Property of Equality Addition Property of Equality a = b and a + c = b + c are equivalent equations Example z = – 16 (Simplify both sides)

  6. 3p + (– 2p) – 11 = 2p + (– 2p) – 18 (Add –2p to both sides) p – 11 + 11 = – 18 + 11 (Add 11 to both sides) Solving Equations Example 4p – 11 – p = 2 + 2p – 20 3p – 11 = 2p – 18 (Simplify both sides) p – 11 = – 18 (Simplify both sides) p = – 7 (Simplify both sides)

  7. 6 – 3z + 4z = – 4z + 4z(Add 4z to both sides) 6 + (– 6) + z = 0 +( – 6) (Add –6 to both sides) Solving Equations Example 5(3 + z) – (8z + 9) = – 4z 15 + 5z – 8z – 9 = – 4z(Use distributive property) 6 – 3z = – 4z(Simplify left side) 6 + z = 0 (Simplify both sides) z = – 6 (Simplify both sides)

  8. The Multiplication Property of Equality § 9.2

  9. Multiplication Property of Equality (–1)(–y) = 8(–1) (Multiply both sides by –1) Multiplication property of equality a = b and ac = bc are equivalent equations Example –y = 8 y = –8 (Simplify both sides)

  10. (Multiply both sides by 7) (Simplify both sides) Solving Equations Example

  11. (Multiply both sides by fraction) (Simplify both sides) Solving Equations Example

  12. 3z – 1 + 1 = 26 + 1 (Add 1 to both sides) (Divide both sides by 3) Solving Equations Recall that multiplying by a number is equivalent to dividing by its reciprocal Example 3z – 1 = 26 3z = 27 (Simplify both sides) z = 9 (Simplify both sides)

  13. 20x + 24 + (–24) = 10 + (–24) (Add –24 to both sides) (Divide both sides by 20) (Simplify both sides) Solving Equations Example 12x + 30 + 8x – 6 = 10 20x + 24 = 10 (Simplify left side) 20x = –14 (Simplify both sides)

  14. Further Solving Linear Equations § 9.3

  15. Solving Linear Equations Solving linear equations in one variable • 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

  16. (Multiply both sides by 5) (Simplify) (Add –3y to both sides) (Simplify; add –30 to both sides) (Simplify; divide both sides by 7) (Simplify both sides) Solving Linear Equations Example

  17. 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.”

  18. 3x + (– 3x) – 7 = 3x + (– 3x) + 3 (Add –3x to both sides) Solving Linear Equations Example 3x – 7 = 3(x + 1) 3x – 7 = 3x + 3 (Use distributive property) – 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.”

  19. An Introduction to Problem Solving § 9.4

  20. Strategy for Problem Solving General Strategy for Problem Solving • 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

  21. 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

  22. The product of the difference of is the same as twice a number 5 times the number and ¾ and 3 2x · 3 = 5x – ¾ Finding an Unknown Number Example continued 2.) Translate Continued

  23. 6x + (– 5x) = 5x + (– 5x) – ¾ (Add –5x to both sides) Finding an Unknown Number Example continued 3.) Solve 2x· 3 = 5x – ¾ 6x = 5x – ¾ (Simplify left side) x = – ¾ (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.

  24. 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

  25. Daily costs mileage costs maximum budget plus is equal to 2(24.95) + 0.29x = 100 Solving a Problem Example continued 2.) Translate Continued

  26. 49.90 – 49.90 + 0.29x = 100 – 49.90 (Subtract 49.90 from both sides) (Divide both sides by 0.29) Solving a Problem Example continued 3.) Solve 2(24.95) + 0.29x= 100 49.90 + 0.29x = 100 (Simplify left side) 0.29x = 50.10 (Simplify both sides) x 172.75 (Simplify both sides) Continued

  27. 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.

  28. Formulas and Problem Solving § 9.5

  29. Formulas Aformula 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)

  30. 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

  31. (Divide both sides by 4) 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) Continued 18 = x(Simplify both sides)

  32. 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.

  33. 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 • 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 • Isolate the specified variable

  34. (Divide both sides by mr) (Simplify right side) Solving Equations for a Specific Variable Example Solve for n.

  35. (Subtract P from both sides) (Simplify right side) (Divide both sides by PR) (Simplify right side) Solving Equations for a Specific Variable Example Solve for T.

  36. (Factor out P from both terms on the right side) (Divide both sides by 1 + RT) (Simplify the right side) Solving Equations for a Specific Variable Example Solve for P.

  37. Percent and Mixture Problem Solving § 9.6

  38. 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

  39. Solving a Percent Problem: Amount Unknown n = 9% 65 · n = (0.09) (65) n = 5.85 amount = percent · base What is 9% of 65? 5.85 is 9% of 65

  40. Solving a Percent Problem: Base Unknown 36 = 6% n · amount = percent ·base 36 is 6% of what? 36 = 0.06n 36 is 6% of 600

  41. Solving a Percent Problem: Percent Unknown 24 = n  144 amount = percent· base 24 is what percent of 144?

  42. 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 + tip of 20% of the cost = $66 100% of n + 20% of n = $66 120% of n = $66 Mark and Peggy can spend up to $55 on the meal itself.

  43. 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 = 420 The discount was $420. Amount paid = list price – discount = 1200 – 420 = 780 Julie paid $780 for the sofa.

  44. 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%.

  45. 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%.

  46. 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

  47. # of pounds of $6 candy # of pounds of $8 candy # of pounds of $7.50 candy Solving Mixture Problems Example continued 2.) Translate Use a table to summarize the information. 6n + 8(144  n) = 144(7.5) Continued

  48. (144  n) = 144  36 = 108 Solving Mixture Problems Example continued 3.) Solve 6n + 8(144  n) = 144(7.5) (Eliminate the parentheses) 6n + 1152  8n = 1080 (Combine like terms) 1152  2n = 1080 2n = 72 (Subtract 1152 from both sides) n = 36 (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. Continued

  49. ? 6(36) + 8(108) = 144(7.5) ? 216 + 864 = 1080 ? 1080 = 1080 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? State: She should use 36 pounds of the $6 per pound candy and 108 pounds of the $8 per pound candy. 

  50. Solving Linear Inequalities § 9.7

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