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Chapter 1 Section 1.2 & 1.5

Prepared by Doron Shahar. Chapter 1 Section 1.2 & 1.5. Word problems with Linear and Quadratic Equations. Prepared by Doron Shahar. Warm-up. Write each description as a mathematical statement. Page 9 #1, 3 more than twice a number

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Chapter 1 Section 1.2 & 1.5

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  1. Prepared by DoronShahar Chapter 1 Section 1.2 & 1.5 Word problems with Linear and Quadratic Equations

  2. Prepared by DoronShahar Warm-up Write each description as a mathematical statement. • Page 9 #1, 3 more than twice a number • Page 9 #2, The sum of a number and 16 is three times the number. • 1.2.1 The sum of three consecutive integers is 78. What is the smallest of the three integers? • 1.5.1 The sum of the square of a number and the square of 7 more than the number is 169. What is the number?

  3. Prepared by DoronShahar Page 9 #1 Write “3 more than twice a number” as a mathematical statement. Variable: Let x be the number. Expression:

  4. Prepared by DoronShahar Page 9 #2 Write “the sum of a number and 16 is three times the number” as a mathematical statement. Variable: Let x be the number. Equation:

  5. Prepared by DoronShahar 1.2.1 The sum of three consecutive integers is 78. What is the smallest of the three integers? Want: The smallest of the three integers. Variable: Let x be the smallest of the three integers. Examples of consecutive integers: 1, 2, 3 17, 18 ,19 25, 26, 27 x, x+1, x+2 Equation:

  6. Prepared by DoronShahar 1.5.1 The sum of the square of a number and the square of 7 more than the number is 169. What is the number? Want: The number. Variable: Let x be the number. Equation:

  7. Prepared by DoronShahar Approach to word problems Solving word problems involves two key steps: • Constructing the equations you are to solve. • Solving the equations you found. The word problems in sections 1.2 and 1.5 will lead to linear and quadratic equations, which we have just learned how to solve. Therefore, we shall not solve any of the word problems in class. Rather we will focus on constructing the equations. That is, we will translate the word problems into equations.

  8. Prepared by DoronShahar 1.5.2 Projectile Problem A stone is thrown downward from a height of 274.4 meters . The stone will travel a distance of s meters in t seconds, where s=4.9t2 +49t. How long will it take the stone to hit the ground? Want: The time it will take the stone to hit the ground. Variable: Let T seconds be the time it will take the stone to hit the ground. Equation: To find the equation, let’s try using a picture.

  9. Prepared by DoronShahar 1.5.2 Equation T is the time it will take until the stone hits the ground. Initial Height Distance traveled in T seconds 4.9T2+49T meters 274.4 meters Equation:

  10. Prepared by DoronShahar 1.5.2 Summary Want: The time it will take the stone to hit the ground. Variables: Let T seconds be the time until the stone hits the ground. Equation: Solutions to equation: Note that T=−14 cannot be the answer to the word problem, because T denotes the time it will take the stone to hit the ground. So T>0. Solution to word problem: It will take 4 seconds for the stone to hit the ground.

  11. Prepared by DoronShahar Extra Projectile Problem A cannonball is launched from a cannon. The height, h, in feet of the cannonball t seconds after it leaves the cannon is given by the equation h=−16t2+63t+4. When will the cannon ball hit the ground? Want: The time when the cannonball hits the ground. Variable: Let T seconds be the time from when the cannonball is launched until it hits the ground. Equation: To find the equation, let’s try using a picture.

  12. Prepared by DoronShahar Extra problem: Equation The height, h, in feet of the cannonball t seconds after it leaves the cannon is given by the equation h=−16t2+63t+4. T is the time when the cannonball hits the ground. After T seconds, the cannonball is on the ground, and has a height of 0 feet. Equation:

  13. Prepared by DoronShahar Extra problem: Summary Want: The time when the cannonball hits the ground. Variables: Let T seconds be the time from the cannonball is launched until it hits the ground. Equation: Solutions to equation: Note that T=−0.0625 cannot be the answer to the word problem, because T denotes the time until the cannonball hits the ground. So T>0. Solution to word problem: The cannonball will hit the ground about 4 seconds after it was launched.

  14. Prepared by DoronShahar 1.5.4 Geometry Problem The area of a square is numerically 60 more than the perimeter. Determine the length of the side of the square. Want: Length of the side of the square Variables: Let S be the length of the side of the square. Let A be the area of the square. Let P be the perimeter of the square. Equations: Information given in the problem. Facts from geometry

  15. Prepared by DoronShahar 1.5.4 Equations Want: Length of the side of the square What method can we use to solve for S given these equations? Key Variable: Let S be the length of the side of the square. Equations: Substitution Now solve for S.

  16. Prepared by DoronShahar 1.5.4 Summary Want: Length of the side of the square Key Variable: Let S be the length of the side of the square. Key Equation: Solutions to equation: Note that S=−6 cannot be the answer to the word problem, because S denotes the length of the side of the square. So S>0. Solution to word problem: The side of the square is 10 units long.

  17. Prepared by DoronShahar 1.5.5 Geometry Problem The length of a rectangle is 7 centimeters longer than the width. If the diagonal of the rectangle is 17 centimeters, determine the length and width. Want: Length and Width of the rectangle Variables: Let L cm be the length of the rectangle. Let W cm be the width of the rectangle. Let D cm be the perimeter of the square. Equations: Information given in the problem. Fact from geometry

  18. Prepared by DoronShahar 1.5.5 Equations Want: Length and width of rectangle What method can we use to solve for L and W given these equations? Let L cm be the length of the rectangle. Key Variables: Let W cm be the width of the rectangle. Equations: Substitution Now solve for W. Then solve for L.

  19. Prepared by DoronShahar 1.5.5 Summary Want: Length and width of rectangle Let L cm be the length of the rectangle. Key Variable: Let W cm be the width of the rectangle. Key Equations: Solutions to first equation: Note that W=−15 cannot be the answer to the word problem, because W denotes the widths of the rectangle. So W>0. Solution to word problem: The width of the rectangle is 8 cm. The length of the rectangle is 15 cm.

  20. Prepared by DoronShahar Key Idea for many word problems In the previous problems, the equations we are trying to find are almost given, or else come from basic facts about geometry. Next we will have to find the equations.

  21. Prepared by DoronShahar Key Idea for many word problems Once we have an equation of the form, We may need to use formulas to replace the amount when dealing with values and costs.

  22. Prepared by DoronShahar 1.2.2 General problem Tina has $6.30 in nickels and quarters in her coin purse. She has a total of 54 coins. How many of each coin does she have? Well, it depends on how many coins she has that are not in her coin purse. Want: The number of nickels and the number of quarters. Variables: Let N be the number of nickels. Let Q be the number of quarters.

  23. Prepared by DoronShahar 1.2.2 Equations What is the amount being added? Number of coins Number of nickels Number of quarters Total Number of coins Dollar Value Dollar Value of nickels Dollar Value of quarters Dollar Value of all coins

  24. Prepared by DoronShahar 1.2.2 Summary Tina has $6.30 in nickels and quarters in her coin purse. She has a total of 54 coins. How many of each coin does she have? Want: The number of nickels and the number of quarters. Variables: Let N be the number of nickels. Let Q be the number of quarters. Equations: What method might we use to solve this system of equations? Elimination

  25. Prepared by DoronShahar 1.2.3 General problem A company produces a pair of skates for $43.53 and sells a pair for $89.95. If the fixed costs are $742.72, how may pairs must the company produce and sell in order to break even? Want: The number of pairs of skates that must be produced to break even. Variables: Let S be the number of pairs of skates that must be produced to break even.

  26. Prepared by DoronShahar 1.2.3 Equation Fixed costs Break even What is the amount being added? Rent Money Earned = Total Cost Cost Production Cost Fixed Cost Total Cost

  27. Prepared by DoronShahar 1.2.3 Summary A company produces a pair of skates for $43.53 and sells a pair for $89.95. If the fixed costs are $742.72, how may pairs must the company produce and sell in order to break even? Want: The number of pairs of skates that must be produced to break even. Variables: Let S be the number of pairs of skates that must be produced to break even. Equation:

  28. Prepared by DoronShahar Warm-up • Page 9, How much pure salt is in 5 gallons of a 20% salt solution? • Page 20 #3, An alloy contains 40% gold. Represent the number of grams of gold present in G grams of the alloy. • Page 9, What is the equation involving distance, rate, and time? • Page 20 #1, A car is travelling M mph for H hours. Represent the number of miles traveled.

  29. Prepared by DoronShahar Key Idea for many word problems In the previous problems, the equations we are trying to find are almost given, or else come from basic facts about geometry. Next we will have to find the equations.

  30. Prepared by DoronShahar Key Idea for many word problems Once we have an equation of the form, We may need to use formulas to replace the amount when dealing with percents and rates.

  31. Prepared by DoronShahar Key Idea for mixture problems Below is the formula for percents. Page 9, How much pure salt is in 5 gallons of a 20% salt solution?

  32. Prepared by DoronShahar Key Idea for mixture problems Below is the formula for percents. • Page 20 #3, An alloy contains 40% gold. Represent the number of • grams of gold present in G grams of the alloy.

  33. Prepared by DoronShahar 1.2.4 Mixture problem A premium mix of nuts costs $12.99 per pound, while almonds cost $6.99 per pound. A shop owner adds almonds into the premium mix to get 90 pounds of nuts that cost $10.99 per pound. How many pounds of almonds did she add? Want: The number of pounds of almonds she added. Variables: Let A be the number of pounds of almonds she added. Let P be the number of pounds of the premium mix of nuts.

  34. Prepared by DoronShahar 1.2.4 Equations What is the amount being added? Mass (Pounds) Pounds of almonds Pounds of premium mix Total Pounds in mixture Cost Cost of almonds Cost of premium mix Total cost of mixture

  35. Prepared by DoronShahar 1.2.4 Summary A premium mix of nuts costs $12.99 per pound, while almonds cost $6.99 per pound. A shop owner adds almonds into the premium mix to get 90 pounds of nuts that cost $10.99 per pound. How many pounds of almonds did she add? Want: The number of pounds of almonds she added. Key Variable: Let A be the number of pounds of almonds. Equations: What method might we use to solve this system of equations? Elimination

  36. Prepared by DoronShahar 1.2.5 Mixture problem A chemist wants to strengthen her 40L stock of 10% acid solution to 20%. How much 24% solution does she have to add to the 40L of 10% solution in order to obtain a mixture that is 20% acid? Want: The volume of the 24% solution that the chemist needs to add to obtain a mixture that is 20% acid. Variable: Let V liters be the volume of the 24% solution that needs to be added.

  37. Prepared by DoronShahar 1.2.5 Equation (40+V) liters of solution 40 liters of solution V liters of solution What is the amount being added? 20% acid 24% acid 10% acid Beaker 3 Beaker 1 Beaker 2 Volume of acid Volume of acid in beaker 1 Total volume of acid in beaker 3 Volume of acid in beaker 2

  38. Prepared by DoronShahar 1.2.5 Summary A chemist wants to strengthen her 40L stock of 10% acid solution to 20%. How much 24% solution does she have to add to the 40L of 10% solution in order to obtain a mixture that is 20% acid? Want: The volume of the 24% solution that the chemist needs to add to obtain a mixture that is 20% acid. Variable: Let V liters be the volume of the 24% solution that needs to be added. Equation:

  39. Prepared by DoronShahar Key Idea for rate problems Below is the formula for rates. Page 20 #1, A car is travelling M mph for H hours. Represent the number of miles traveled.

  40. Prepared by DoronShahar 1.2.7 Distance-Rate-Time Two motorcycles travel towards each other from Chicago and Indianapolis (350km apart). One is travelling 110 km/hr, the other 90 km/hr. If they started at the same time, when will they meet? Want: The time when they will meet. Variable: Let T hours be the time from when they started until they meet.

  41. Prepared by DoronShahar 1.2.7 Equation 90 km/hr 110 km/hr Distance traveled by Motorcycle 1 Distance traveled by Motorcycle 2 What is the amount being added? Chicago Indianapolis Motorcycle 1 Motorcycle 2 Where they meet 350 km Distance Distance traveled by motorcycle 1 Distance traveled by motorcycle 2 Total distance traveled by both

  42. Prepared by DoronShahar 1.2.7 Summary Two motorcycles travel towards each other from Chicago and Indianapolis (about 350km apart). One is travelling 110 km/hr, the other 90 km/hr. If they started at the same time, when will they meet? Want: The time when they will meet. Variable: Let T hours be the time from when they started until they meet. Equation:

  43. Prepared by DoronShahar 1.5.3 Distance-Rate-Time Amy travels 450 miles in her car at a certain speed. If the car had gone 15 mph faster, the trip would have taken 1 hour less. Determine the speed of Amy’s car. Want: The speed of Amy’s car. Variables: Let A mph be the speed of Amy’s car. Let T hours by the time it takes Amy to drive 450 miles.

  44. Prepared by DoronShahar 1.5.3 Equations A mph for T hours Amy’s car 450 miles (A+15)mph for(T-1)hours Amy’s car 450 miles

  45. Prepared by DoronShahar 1.5.3 Summary Amy travels 450 miles in her car at a certain speed. If the car had gone 15 mph faster, the trip would have taken 1 hour less. Determine the speed of Amy’s car. Want: The speed of Amy’s car. Key Variable: Let A mph be the speed of Amy’s car. Equations: What method might we use to solve this system of equations? Substitution

  46. Prepared by DoronShahar 1.2.9 Shared Work Problem Suppose a journeyman and apprentice are working on making cabinets. The journeyman is twice as fast as his apprentice. If they complete one cabinet in 14 hours, how many hours does it take for the journeyman working alone to make one cabinet? Want: The time it takes the journeyman working alone to make one cabinet. Variables: Let J hours be the time it takes the journeyman to make one cabinet working alone. Let A hours be the time it takes the apprentice to make one cabinet working alone.

  47. Prepared by DoronShahar 1.2.9 Equations The journeyman is twice as fast as his apprentice. 2 × Apprentice’s rate Journeyman’s rate Multiply both sides of the equation by AJ

  48. Prepared by DoronShahar 1.2.9 Equations What is the amount being added? Amount of the cabinet built The answer is NOT time!! Amount of the cabinet built by both in 14 hours Amount of the cabinet built by the apprentice in 14 hours Amount of the cabinet built by the journeyman in 14 hours

  49. Prepared by DoronShahar 1.2.9 Summary Want: The time it takes the journeyman working alone to make one cabinet. Key Variable: Let J be the time it takes the journeyman to make one cabinet working along. Equations: What method might we use to solve this system of equations? Substitution

  50. Prepared by DoronShahar 1.5.6 Shared Work Problem It take Julia 16 minutes longer to chop vegetables than it takes Bob. Working together, they are able to chop the vegetables in 15 minutes. How long will it take each of them if they work by themselves? Want: The time it takes each of them working alone to chop the vegetables. Variables: Let J minutes be the time it takes Julia to chop the vegetables working alone. Let B minutes be the time it takes Bob to chop the vegetables working alone.

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