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System-of-Equations Word Problems (page 2 of 2)


  • A landscaping company placed two orders with a nursery. The first order was for 13 bushes and 4 trees, and totalled $487. The second order was for 6 bushes and 2 trees, and totalled $232. The bills do not list the per-item price. What were the costs of one bush and of one tree?

    I could try to add the bushes and trees, to get 19 bushes and 6 trees, but this wouldn't get me anywhere, because I don't have subtotals for the bushes and trees. So I'll pick variables ("b" for the number of bushes and "t" for the number of trees) and set up a system of equations:

      first order: 13b + 4t = 487
      second order: 6b + 2t = 232

    Multiplying the second row by 2, I get:

      13b + 4t = 487
      12b 4t = 464

    This says that b = 23. Back-solving, I get that t = 47. Of course, the exercise didn't ask for the values of the two variables. Translating back into English, the solution is:

      Bushes cost $23 each; trees cost $47 each.


You probably remember the "distance" problems where you had a boat going with and against the current or a plane going with and against the wind. Once you learn how to solve systems of equations, you'll see more of these sorts of exercises.

  • A passenger jet took three hours to fly 1800 miles in the direction of the jetstream. The return trip against the jetstream took four hours. What was the jet's speed in still air and the jetstream's speed?
  • When they ask you about the speed "in still air" (for planes) or "in still water" (for boats), they are referring to the speedometer reading; they are referring only to the powered input, irrespective of outside influences.

 

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    On some very windy day, you can watch birds flapping frantically in the air, trying to cross the street, say, from the tree in your front yard. No matter how hard they flapped, they made little or no forward progress; sometimes, a bird will even appear to "fly" backwards! Did this mean that the bird wasn't actually flapping? No; it means that the bird's attempted speed (how fast the flapping would have moved the bird on a windless day) was not fast enough to usefully counteract the wind hitting it in the face. The bird's "speed in still air", less the wind's speed in the opposite direction, was close to zero, or even negative.

    The same concept applies to machinery. If a boat's motor is chugging away at 10 miles an hour (according to the speedometer), but the boat is facing a water current of 15 miles an hour in the opposite direction, then the boat will end up going backwards at five miles an hour. In other words, the speedometer reading is not always the actual speed.

    Returning to the exercise, I'll pick variables and set up a system. In this case, I'll use "p" for "the plane's speedometer reading (apparent speed)" and "w" for "the windspeed". When the plane is going "with" the wind (when it has a "tailwind"), the two speeds will add together; when the plane is going "against" the wind (when it has a "headwind"), the windspeed will be subtracted from the plane's speedometer reading (that is, from the engines' actual output).

    In each case, the "distance" equation will be "(the combined speed) times (the time at that speed) equals (the total distance travelled)":

      with the jetstream:  (p + w)(3) = 1800

      against the jetstream:  (p w)(4) = 1800

    Rather than multiply through, I notice that, if I divide off the 3 and the 4, I'll have a system that's already set for solving by addition:Copyright Elizabeth Stapel 2003-2011 All Rights Reserved

      p + w = 600
      p w = 450

    Then, by adding down, 2p = 1050 so p = 525, and w must then be 75.

      The jet's speed was 525 mph and the jetstream windspeed was 75 mph.


Another topic you might see (if not now, then later in calculus) is decomposing rational expressions using partial fractions.

  • Find the partial fraction decomposition of the following:
    • (5x + 7) / (x^3 + 2x^2 – x – 2)

    The denominator factors as (x + 2)(x + 1)(x 1), so these will be the denominators in the decomposition. That is, I'm looking for A, B, and C in the following:

      A/(x + 2) + B/(x + 1) + C/(x – 1)

    Multiplying both sides by the common denominator, I get:

      5x + 7 = A(x + 1)(x 1) + B(x + 2)(x 1) + C(x + 2)(x + 1)

                 = A(x2 1) + B(x2 + x 2) + C(x2 + 3x + 2)

                 = (A + B + C)x2 + (B + 3C)x + (A 2B + 2C)1

    Comparing coeffients (by thinking of "5x + 7" as "0x2 + 5x + 7"), I get:

      A + B + C = 0
      B + 3C = 5

      A 2B + 2C = 7

    Solving this system, I get A = 1, B = 1, and C = 2.  Then the partial fraction decomposition is:

      –1/(x + 2) – 1/(x + 1) + 2/(x – 1)


Whatever you do, don't panic when you face a systems-of-equations word problem. If you take them step-by-step, they're usually pretty do-able. That said, it would probably be to your benefit if you did extra practice problems, just to help you get in the swing of things. With any luck, your tests will then go a little faster.

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Cite this article as:

Stapel, Elizabeth. "System-of-Equations Word Problems." Purplemath. Available from
    http://www.purplemath.com/modules/systprob2.htm. Accessed
 

 



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