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9.6 Rational Equation word problems

Work problems. Suppose Peter can paint a house in 8 hours and Danny can paint the same house in 11 hours. Assuming they work with maximum efficiency, how long would it take them to paint the house if they work together?The reasoning for this problem involves looking at the fraction of the job each

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9.6 Rational Equation word problems

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    1. 9.6 Rational Equation word problems Work problems Motion problems

    2. Work problems Suppose Peter can paint a house in 8 hours and Danny can paint the same house in 11 hours. Assuming they work with maximum efficiency, how long would it take them to paint the house if they work together? The reasoning for this problem involves looking at the fraction of the job each person completes (hence rational equations), and we generally want these fractions to add up to “1” whole job. For example, in 1 hr Pete paints 1/8 of the house In 2 hrs he paints 2/8 of the house In 3 hrs he paints 3/8 of the house, and so on Similarly, in 1 hr Danny paints 1/11 of the house In 2 hrs he paints 2/11 of the house, and so on

    3. Painting example, continued Let X = the # of hours it will take Danny and Peter to complete the job together In x hrs, Danny completes x/11 of the job In x hrs, Peter completes x/8 of the job And together these fractions should add up to “1” whole job That is, (x/8) + (x/11) = 1 Now multiply by the LCM (88) 11x + 8x = 88 19x = 88 X = 88/19 = 4.63 hours or 4 hours 37 min 48 sec :P

    4. Example 6-3a

    5. Example 6-3b

    6. Example 6-3c

    7. Example 6-3d

    8. Example 6-3e

    9. Motion problems We represent motion at a constant speed using the formula d = r * t Where d = distance r = rate (speed: mph, ft per sec..) t = time Generally given two of these quantities you can determine the 3rd

    10. Usually 9.6 motion problems are more complicated You may have the motion of 2 (or sometimes more objects) Sometimes you will need to use formulas like Time 1 + time 2 = total time Or sometimes… Distance 1 + Distance 2 = total distance Usually you will then replace the terms in the sum on the left side with an expression involving the other variables For example, since t = d / r Time 1 + time 2 = total time Can become (distance 1 / rate 1) + (distance 2 / rate 2) = total time

    11. Example 6-4a

    12. Example 6-4b

    13. Example 6-4c

    14. Example 6-4d

    15. Example 6-4e

    16. Example 6-4f

    17. Other motion problems Sometimes you will have a motion problem with 2 objects where one of the quantities for the 2 objects is the same For example, distance 1 = distance 2 Then you can again substitute the other variables by using d = r * t So that distance 1 = distance 2 becomes… (rate 1) (time 1) = (rate 2) (time 2)

    18. Train example The JT Express leaves the station 3.125 hours after a freight train leaves the same depot. The freight is travelling 25 mph slower than the JT Express. Find the rate of each train if the passsenger train overtakes the freight train in 5 hours.

    19. Making a table Sometimes it helps where we write what we know about each object in a table: distance rate time JT same x 5 Freight same x – 25 8.125 Note that since the distance is the same for each object, distancejt = distancefr

    20. Finishing… Distancejt = distancefr Ratejt * timejt = ratefr * timefr X * 5 = (x – 25) 8.125 5x = 8.125x – 203.125 -3.125x = -203.125 X = 65 So JT EXPRESS travels at 65 mph, and the freight train travels at 40 mph Check: 65 (5) = 40 (8.125) both equal 325 miles

    21. HOMEWORK! In the text try pg. 509 # 10 and 36 Also try the ODD problems on the worksheet.. Some are challenging You will have the even problems as part of your homework next time!!

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