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1 Topic Applications of Inequalities Applications of Inequalities

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2 Topic Applications of Inequalities California Standards: 4.0 Students simplify expressions before solving linear equations and inequalities in one variable, such as 3(2 x – 5) + 4( x – 2) = Students solve multistep problems, including word problems, involving linear equations and linear inequalities in one variable and provide justification for each step. What it means for you: Youll solve real-life problems involving probabilities. Key words: inequality isolate

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3 Topic Some real-life problems include phrases like at least or at most, or deal with maximums or minimums. Applications of Inequalities If you come across these phrases, chances are youll need to model the situation as an inequality.

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4 Topic Applications of Inequalities are Real-Life Problems In the same way that applications of equations are real-life problems, applications of inequalities are real-life inequalities problems. Real-life problems involving inequalities could be about pretty much anything from finding the area of a field to figuring out how many CDs you can buy with a certain amount of money. Applications of Inequalities What they all have in common is that theyll all be word problems and youll always have to set up and solve an inequality.

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5 Topic Applications of Inequalities Solving Real-Life Inequality Problems 1. First decide how you will label the variables then write the task out as an inequality make sure you include all the information given then solve the inequality.

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6 Topic Example 1 Find the three smallest consecutive even integers whose sum is more than 60. Solution follows… Solution First you need to label the variables: Applications of Inequalities Let x = first (smallest) even integer x + 2 = next (second) even integer x + 4 = next (third) even integer Then you need to write it out as an inequality: In English: The sum of three consecutive even integers is more than 60. In math: x + ( x + 2) + ( x + 4) > 60 Solution continues…

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7 Topic Example 1 Find the three smallest consecutive even integers whose sum is more than 60. Solution (continued) Applications of Inequalities The inequality you need to solve is: x + ( x + 2) + ( x + 4) > 60 Then simplify: x + x x + 4 > 60 3 x + 6 > 60 3 x + 6 – 6 > 60 – 6 3 x > 54 x > > 3x3x 3

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8 Topic Example 1 Find the three smallest consecutive even integers whose sum is more than 60. Solution (continued) Applications of Inequalities Answer in English: The smallest even integer is more than 18. Answer in math: x > 18 So the three smallest consecutive even integers whose sum is greater than 60 are 20, 22, and 24.

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9 Topic Guided Practice 1. Find the largest three consecutive odd integers whose sum is at most 147. x + ( x + 2) + ( x + 4) x x 47 47, 49, and 51 Applications of Inequalities 2. Find the three smallest consecutive odd integers whose sum is more than 45. Solution follows… x + ( x + 2) + ( x + 4) > 45 3 x , 17, and 19

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10 Topic Guided Practice x + ( x + 2) > ( x + 4) x + 2 > x + 15 x > 13 15, 17, and 19 Applications of Inequalities 3. Find the smallest three consecutive odd integers such that the sum of the first two integers is greater than the sum of the third integer and 11. Solution follows… 4. Find the smallest three consecutive even integers whose sum is greater than 198. x + ( x + 2) + ( x + 4) > x + 6 > 198 x > 64 66, 68, and 70

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11 Topic Guided Practice 5. The difference between a number and twice that number is at least 7. Find the smallest possible integer that satisfies this criterion. Applications of Inequalities 6. Three times a number is added to 11, and the result is less than 7 plus twice the number. Find the highest possible integer the number could be. Solution follows… 2 x – x 7 x 7 The smallest possible integer is 7. 3 x + 11 < x x < –4 The highest possible integer is –5.

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12 Topic Guided Practice Applications of Inequalities 7. José scored 75 and 85 on his first and second algebra quizzes. If he wants an average of at least 83 after his third quiz, what is the least score that José must get on the third quiz? Solution follows… 8. Lorraines test scores for the semester so far are 60%, 70%, 75%, 80%, and 85%. If the cutoff score for a letter grade of B is 78%, what is the least score Lorraine must get on the final test to earn a B? ( x ) ÷ x 249 x 89 Jose must score at least 89. ( x ) ÷ x 468 x 98 Lorraine must score at least 98%.

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13 Topic Example 2 The perimeter of a rectangular field, F, is given by the formula p = 2 l + 2 w. As shown in the diagram, the length l = (10 x – 6) m and the width w = (5 x – 3) m. Find the possible values of x for which the given rectangular field would have a perimeter of at least 1182 metres. Solution follows… Solution Applications of Inequalities Solution continues… The variables are already labeled, so you just need to write it out as an inequality: Substituting the variables into the expression for the perimeter gives: p = 2(10 x – 6) + 2(5 x – 3) Youre also told that the perimeter is at least 1182 meters in math, you write that as p F (10 x – 6) m (5 x – 3) m

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14 Topic Example 2 The perimeter of a rectangular field, F, is given by the formula p = 2 l + 2 w. As shown in the diagram, the length l = (10 x – 6) m and the width w = (5 x – 3) m. Find the possible values of x for which the given rectangular field would have a perimeter of at least 1182 metres. Solution (continued) Applications of Inequalities So the inequality is: 2(10 x – 6) + 2(5 x – 3) x – x – x – x 1200 x 40 Answer in math: x 40 Answer in English: x must be greater than or equal to 40. F (10 x – 6) m (5 x – 3) m

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15 Topic Guided Practice 9. Lily wants to build a fence along the perimeter of her rectangular garden. She cannot afford to buy more than 14 meters of fencing. The length of the garden, l, is (2 x – 3) m and the width, w, is (3 x – 10) m. Write an inequality to represent the situation and solve it. What would the dimensions of the garden fence have to be to keep the fencing no longer than 14 meters? P = 2 l + 2 w and P 14 So, 2(2 x – 3) + 2(3 x – 10) 14 4 x – x – x 40, so x 4. This means l 2 × 4 – 3, l 5 m and w 3 × 4 – 10, w 2 m. The dimensions can be no more than 5 m × 2 m Applications of Inequalities Solution follows…

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16 Topic Guided Practice 10. A car travels (17 x – 5) miles in ( x – 7) hours. The car travels at a constant speed not exceeding the speed limit of 55 miles per hour. If speed = distance ÷ time, write an inequality to represent this situation and find the minimum possible number of miles traveled. Speed = (17 x – 5) ÷ ( x – 7) and speed 55 mph. So, (17 x – 5) ÷ ( x – 7) x – 5 55 x – x, so 10 x. So, the minimum distance traveled is 17 × 10 – 5 = 165 miles. Applications of Inequalities Solution follows…

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17 Topic Guided Practice Applications of Inequalities Solution follows… 11. The formula for calculating the speed of an accelerating car is v = u + at, where v is the final speed, u is the original speed, a is the acceleration, and t is the time taken. Car A starts at 5 m/s and accelerates at 4 m/s 2. At the same time, Car B starts at 10 m/s and accelerates at 2 m/s 2. Write and solve an inequality to find out how long it will be before Car A is traveling faster than Car B. Speed of Car A is t and the speed of Car B is t. You need to find t when speed of Car A > speed of Car B. So solve t > t. 4 t > t 2 t > 5 t > 2.5 s. So it will be 2.5 seconds before Car A is traveling faster than Car B.

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18 Topic Example 3 A long-distance telephone call from Los Angeles, California, to Harare, Zimbabwe, costs $9.50 for the first three minutes, plus $0.80 for each additional minute (or fraction of a minute). Colleen has $18.30 to spend on a call. What is the maximum number of additional minutes she can spend on the phone? Solution follows… Solution Applications of Inequalities Solution continues… First you need to label the variables: Let x = the additional number of minutes after the first 3 minutes. Then you need to write it out as an inequality…

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19 Solution continues… Topic Example 3 A long-distance telephone call from Los Angeles, California, to Harare, Zimbabwe, costs $9.50 for the first three minutes, plus $0.80 for each additional minute (or fraction of a minute). Colleen has $18.30 to spend on a call. What is the maximum number of additional minutes she can spend on the phone? Solution (continued) Applications of Inequalities In English: The total cost is $9.50 plus $0.80 per additional minute. The total cost must be less than or equal to $ In math: x = Total cost for the call Total cost for the call x 18.30

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20 Solution continues… Topic Example 3 A long-distance telephone call from Los Angeles, California, to Harare, Zimbabwe, costs $9.50 for the first three minutes, plus $0.80 for each additional minute (or fraction of a minute). Colleen has $18.30 to spend on a call. What is the maximum number of additional minutes she can spend on the phone? Solution (continued) Applications of Inequalities Then simplify: x x x Total cost for the call x 18.30

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21 Topic Example 3 A long-distance telephone call from Los Angeles, California, to Harare, Zimbabwe, costs $9.50 for the first three minutes, plus $0.80 for each additional minute (or fraction of a minute). Colleen has $18.30 to spend on a call. What is the maximum number of additional minutes she can spend on the phone? Solution (continued) Applications of Inequalities Answer in math: x 11 Answer in English: The number of additional minutes must be no more than 11. Colleen can spend up to 11 additional minutes on the phone.

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22 Topic Guided Practice 12. Marisa is buying a new car. Car A costs $20,000, and has an average annual fuel cost of $1000. Car B costs $22,500, and has an average annual fuel cost of $500. After how many years will Car A have cost more than Car B? Assume that all other maintenance costs are equal for both. Car A costs 20, y and Car B costs 22, y. For Car A to cost more, 20, y > 22, y. Solve to find y : 1000 y > y 500 y > 2500 y > 5 Car A will have cost more after 5 years. Applications of Inequalities Solution follows…

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23 Topic Guided Practice Applications of Inequalities Solution follows… 13. A cell phone company offers its customers either Plan A or Plan B. Plan A costs $90 per month with unlimited air time. Plan B costs $60 per month, plus 50¢ for each minute of cell phone time. How many minutes can a customer who chooses Plan B use the cell phone before the cost of the calls exceeds the amount it would have cost under Plan A? Plan A costs 90 and Plan B costs t. For Plan B to cost no more than Plan A, t 90. Solve to find t : 0.50 t 30 t 60 minutes. So, Plan B will cost more than Plan A when more than 60 minutes are used.

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24 Topic Independent Practice Solution follows… 1. Bank Ts checking account has monthly charges of an $8 service fee plus 6¢ per check written. Bank Ss checking account has monthly charges of a $10 service fee plus 4¢ per check written. A company has 150 employees, and pays them monthly by check. The companys financial adviser suggests that Bank S would be cheaper to use. Set up and solve an inequality that supports this recommendation. Applications of Inequalities Bank S is cheaper for the company if: c > c This simplifies to: c > 100, so Bank S is cheaper if the company writes more than 100 checks per month.

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25 Topic Independent Practice Solution follows… Applications of Inequalities 2. Jack is doing a sponsored swim to raise money for charity. His mom sponsors him $10, plus $1 for every length of the pool he completes. His uncle sponsors him just $1.50 for every length he completes. How many lengths will Jack have to complete for his uncle to pay more than his mom? 21 lengths

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26 Topic Independent Practice Solution follows… 3. On average Sendi uses 350 minutes of air time per month. Company M offers a cell phone plan of $70 per month plus 85¢ for each minute of air time. Company V offers a cell phone plan of $130 per month plus 65¢ for each minute of air time. Sendi chooses Company V. Use an inequality to show that this plan is cheaper for her. Applications of Inequalities Company V is cheaper for Sendi if: t > t This simplifies to: t > 300, so Company V is cheaper for anyone who uses more than 300 minutes.

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27 Topic Independent Practice Solution follows… Applications of Inequalities 4. A group of friends want to drive to a beach resort and spend 5 days there. A car rental firm offers them two rental plans; $15 a day plus 30¢ per mile traveled, or $20 a day plus 10¢ per mile. Which rental plan would be better if the beach resort is 150 miles from home, and why? The first plan is cheaper than the second for 5 days use when (15 × 5) m < (20 × 5) m. This simplifies to m < 125. The friends will drive more than 125 miles, so the second rental plan ($20 a day) will be cheaper for them.

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28 Topic Independent Practice Solution follows… 5. A bank charges a $10 monthly service fee plus 5¢ handling fee per check processed through its Gold checking account. The bank also offers a Platinum checking account and charges a $15 monthly service fee plus 3¢ handling fee per check drawn from this account. What is the highest number of checks per month for which the Gold account is cheaper than the Platinum account? Applications of Inequalities 249 checks

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29 Topic Independent Practice Solution follows… Applications of Inequalities 6. A group of executives is traveling to a meeting, so they decide to hire a car and travel together. The car rental agency rents luxury cars at $65 per day plus 65¢ per mile traveled, or $55 per day plus 85¢ per mile traveled. What is the maximum number of miles that they can drive before the $55 per day plan becomes more expensive than the $65 per day plan? 50 miles

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30 Topic Round Up This Topic is very similar to real life applications of equations, which is covered in Sections 2.4–2.7. Applications of Inequalities Always remember to give your solution as a sentence that answers the original problem.

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