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3-6 Equations & Problem Solving
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  1. 3-6 Equations & Problem Solving

  2. The first step is to define the variables… There is no magic formula for doing this because each problem is a little different. You need to identify the variable based on the problem. Usually you will need to identify one variable in terms of another. Ex. The perimeter of a rectangle is 24 in. One side is 6 in. longer than the other side. The variables are side #1 and side #2. Identify one of them as s, If Side #1 = s, then Side #2 = s + 6

  3. There are 5 typical types of problems: Perimeter problems Consecutive integer problems Combined money problem Age/Comparison problems Distance problems Let’s do an example of each!

  4. Perimeter problems: These are usually pretty easy! You can always draw it… The perimeter of a rectangle is 24 in. One side is 6 in. longer than the other side. What are the dimensions? s The variables are side #1 and side #2 Identify one of them as s, If Side #1 = s, then Side #2 = s + 6 Since the perimeter is 24…add up all sides And set them equal to 24. s + 6 s + s + (s + 6) + (s + 6) = 24 4s + 12 = 24 4s = 12 s = 3 This isn’t the complete answer! The answer is 3 units by 9 units

  5. Consecutive integer problems: These are usually pretty easy as well! The sum of 3 consecutive integers is 258, find them! The variables are integer #1, integer #2, and integer #3 Identify one of them as x, The next one would be 1 more, or x + 1 The next one would be 2 more or x + 2 Remember that their sum is 258. x + (x + 1) + (x + 2) = 258 3x + 3 = 258 3x = 255 x = 85 This isn’t the complete answer! The answer is 85, 86 & 87

  6. Combined money problems: Tony has twice as much money as Alicia. She has $16 less than Ralph. Together they have $200. How much money does each person have? The variables are Tony’s $, Alicia’s $, and Ralph’s $. The key to these problems is picking the right variable! Look at the problem, and identify which one is somehow related to the other two. I know the relationship b/t Tony’s & Alicia’s Alicia’s & Ralph’s Since Alicia is connected to both, make her $ the variable. So, let Alicia’s $ = a That means Tony’s $ = 2a and Ralph’s $ = a + 16 And together they have a sum of 200…so the equation would be: a + 2a + (a + 16) = 200

  7. Age or comparison problems: These are very similar to the combined money problems. The key to these problems is picking the right variable! State College has 620 students. There are 20 more women than men. How many men are there? How many women are there? If w = the number of women, then the number of men = w – 20 Alternative method If m = the number of men, then the number of women = m + 20 It’s a good idea to avoid subtraction if possible, but either method will work! The number of men plus the number of women = 620 m + (m + 20) = 620 2m + 20 = 620 2m = 600 m = 300 This isn’t the complete answer! There are 300 men and 320 women

  8. Distance problems tend to be the toughest. The problem is usually translating them from English into Algebra.

  9. Start with the formula for speed (aka rate) distance Rate = time This formula can be manipulated….. times

  10. Start with the formula for speed (aka rate) distance Rate = time This formula can be manipulated…..

  11. You end up with 3 different (similar) formulas: d = rt r = t = d t d r

  12. There are also three types of distance problems: • Motion in opposite directions • Motion in the same direction • Roundtrip

  13. Each type of problem is solved a little differently…depending on what information you have. • Motion in opposite directions problems usually use • d1 + d2 = total d • Motion in the same direction usually use • d1 = d2 • Roundtrip problems usually use • d1 = d2

  14. No matter which type of problem it is, you should set up a chart like this: time distance rate

  15. Ex. Motion in opposite directions problems usually use • d1 + d2 = total d Jane and Peter leave their home traveling in opposite directions on a straight road. Peter drives 15 mph faster than Jane. After 3 hours, they are 225 miles apart. Find Peter’s rate and Jane’s rate. Jane Peter 225 miles What should the variable be? Look at the question…find the rate…there’s the clue!

  16. Ex. Jane and Peter leave their home traveling in opposite directions on a straight road. Peter drives 15 mph faster than Jane. After 3 hours, they are 225 miles apart. Find Peter’s rate and Jane’s rate. If Jane’s rate = r, what is Peter’s rate? The time and the distance are both given! t = 3 hours and d = 225 miles How do I put this all together using a formula? Peter’s rate = r + 15 time distance rate r 3 Jane r + 15 Peter 3 Think about the distance travelled…it’s their TOTAL distance…I have to add Jane’s distance plus Peter’s distance to get 225.

  17. Jane’s distance + Peter’s distance = 225 miles • Jane’s distance = Jane’s rate * Jane’s time • Peter’s distance = Peter’s rate * Peter’s time Peter’s rate = r + 15 Peter’s time = 3 So Peter’s distance = 3(r + 15) Jane’s rate = r Jane’s time = 3 So Jane’s distance = 3r Jane’s distance + Peter’s distance = 225 miles, so: • 3r + 3(r + 15) = 225 • 3r + 3r + 45 = 225 • 6r + 45 = 225 • 6r = 180 • r = 30 Look back at how you defined the variables…. Jane’s rate = 30 mph and Peter’s rate = 45 mph

  18. Motion in the same direction problems usually use • d1 = d2 Ex. Motion in opposite directions Jane and Peter leave their home traveling in the same direction. Peter drives 55 mph,Jane drives 40mph. Jane leaves at noon, Peter leaves at 1 P.M. When will Peter catch up to Jane? time What should the variable be? distance rate Look at the question..”when”..there’s the clue…find the time! NB! time = time spent traveling, NOT time of day! If Peter’s time = t What is Jane’s time? t + 1 Jane 40 Since she left 1 hour earlier, her time is t + 1 Peter t 55

  19. Motion in the same direction problems usually use • d1 = d2 Jane and Peter leave their home traveling in the same direction. Peter drives 55 mph,Jane drives 40mph. Jane leaves at noon, Peter leaves at 1 P.M. When will Peter catch up to Jane? Jane’s d = Peter’s d Use the formula d = r * t 40(t+1) = 55t 40t + 40= 55t 40 = 15t t = 2 t = 2 time distance rate 40(t+1) t + 1 Jane 40 Peter 55t t 55 What does this mean? When Peter has been traveling 2 hours and Jane has been traveling 3 , they will have travelled the same distance…or to put it another way, Peter will have caught up to Jane. The time will be 3:40 P.M.

  20. Roundtrip problems usually use d1 = d2 Ex. Round trip problem Jane goes to the mall. She drives 35mph. On the way home, there is lots of traffic. She averages 15 mph. Her total travel time was 2 hours. How long did it take her to get to the mall? What should the variable be? time distance rate Look at the question…”how long”...there’s the clue…find the time! If Jane’s time there = t What is her time home? t 35 there Since her total time is 2 and it takes her t hours to get there, her time home is 2 – t 2 - t 15 back You can’t avoid subtraction on this one.

  21. Jane goes to the mall. She drives 35mph. On the way home, there is lots of traffic. She averages 15 mph. Her total travel time was 2 hours. How long did it take her to get to the mall? Use the formula d = r * t d there = d back 35t = 15(2-t) 35t = 30 – 15t 50t = 30 t = 35t time distance rate 15(2-t) t 35 there 2 - t 15 back What does this mean? It took Jane of an hour to get to the mall and it took her 1 hours to get home. If you wanted to convert this to minutes, just multiply by 60.