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Objective 2 Days The learner will solve real-life problems using equations (d=r*t)
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Equations and Problem Solving Pages 103 - 110
Lesson 2-5 Equations and Problem Solving Pages
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Some problems contain two or more unknown quantities.
To solve such problems, first decide which unknown quantity the variable will represent. Then express the other unknown quantity or quantities in terms of that variable.
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The length of a rectangle is 6in. more than its width
The length of a rectangle is 6in. more than its width. The perimeter of the rectangle is 24in. What is the length of the rectangle? Define: Let w be the width. Then w + 6 = length. Write: P =2l + 2w 24 = 2(w+6) + 2w (substitute) Distribute & Solve
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Now you try: The width of a rectangle is 2cm less than its length. The perimeter of the rectangle is 16cm. What is the length of the rectangle?
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Consecutive Integers differ by 1
Consecutive Integers differ by 1. The integers 50 and 51 are consecutive. So are -10, -9, and -8. For consecutive integer problems, it may help to define a variable before describing the problem in words. Let a variable represent one of the unknown integers. Then define the other unknown integers in terms of the first one.
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The sum of three consecutive integers is 147. Find the integers.
Restate: 1st integer plus 2nd integer plus 3rd integer equals 147. Illustrate: Let n be the first integer. Let n + 1 be the second integer. Let n + 2 be the third integer. n + n n + 2 = 147 Calculate & Examine
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Now you try: The sum of three consecutive integers is 48.
a. define a variable for one of the integers b. write expressions for the other two integers c. write and solve an equation to find the three integers. d. check your answer
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An object that moves at a constant rate is said to be in uniform motion.
The formula d = rt gives the relationship between distance, rate, and time. Uniform motion problems may involve objects going the same direction, opposite directions, or round trip.
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Video Speed & Distance
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For the vehicles shown, 40 * 5 = 50 * 4.
In the diagram, the two vehicles are traveling the same direction at different rates. The distances the vehicles travel are the same. Since the distances are equal, the products of rate and time for the two cars are equal. For the vehicles shown, 40 * 5 = 50 * 4.
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A train leaves a train station at 1pm
A train leaves a train station at 1pm. It travels at an average rate of 60mph. A high-speed train leaves the same station an hour later. It travels at an average rate of 96mph. The second train follows the same route as the first train on a track parallel to the first. In how many hours will the second train catch up with the first?
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A group of campers and one group leader left a campsite in a canoe
A group of campers and one group leader left a campsite in a canoe. They traveled at an average rate of 10km/h. Two hours later, the other group leader left the campsite in a motorboat. He traveled at an average rate of 22km/h. a. how long after the canoe left the campsite did the motorboat catch up with it? b. How long did the motorboat travel?
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For uniform motion problems that involve a round trip, it is important to remember that the distance going is equal to the distance returning. In the diagram, the distances are equal. So, the products of rate and time for traveling in both directions are equal. That is, 20 * 3 = 30 * 2
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Noya drives into the city to buy a software program at a computer store. Because of traffic conditions, she averages only 15mph. On her drive home she averages 35mph. If the total travel time is 2 hours, how long does it take her to drive to the computer store?
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Now you try: On his way to work from home, your uncle averaged only 20mph. On his way home, he averaged 40mph. If the total travel time was 1 and ½ hours, how long did it take him to drive to work?
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For uniform motion problems involving two objects moving in opposite directions, you can write equations using the fact that the sum of their distances is the total distance.
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Jane and Peter leave their home traveling in opposite directions on a straight road. Peter drives 15mph faster than Jane. After 3 hours, they are 225 miles apart. Find Peter’s rate and Jane’s rate.
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Solve: 3r + 3(r + 15) = 225 3r + 3r + 45 = 225 (distribute) 6r + 45 = 225 (combine like terms) 6r + 45 – 45 = 225 – 45 (subtract – both sides) 6r = 180 (simplify) 6r/6 = 180/6 (divide – both sides) r = 30 (simplify) Jane’s rate is 30mph and Peter’s rate is 15mph faster, which is 45mph.
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Now you try: Sarah and John leave Perryville traveling in opposite directions on a straight road. Sarah drive 12mph faster than John. After 2 hours, they are 176 miles apart. Find Sarah’s speed and John’s speed.
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