System-of-Equations Word Problems
Purplemath
Very commonly, system-of-equations word problems involves mixtures or combinations of some sort. For instance:
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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:
number of bushes: B
number of trees: t
With these variables, I can set up a system of equations; each equation will represent one of the transactions they've given me:
1st order: 13B + 4t = 487
2nd order: 6B + 2t = 232
Multiplying the second row by −2, I get:
13B + 4t = 487
−12B − 4t = −464
Adding down the t-terms cancel out, leaving me with 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, my solution is:
bushes: $23 each
trees: $47 each
You probably remember the "distance" word problems where you had a boat going with the current and then against the current, or a plane going with the wind (that is, having a tailwind) and then against the wind (that is, having a headwind). Once you learn how to solve systems of equations, you'll see more of these sorts of exercises.
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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?
They are asking me for "the jet's speed 'in still air'". How does it apply to my set-up?
What do they mean by "in still air" or "in still water"?
When they ask me 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.
On a very windy day, you can watch birds flapping frantically in the air, trying to cross the street from, say, the tree in your front yard to the birdfeeder in a neighbor's yard. But, no matter how hard they flap, they make little or no forward progress; sometimes, a bird will even appear to fly backwards! Does this mean that the bird isn'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) is not fast enough to usefully counteract the wind that is hitting it square in the face. The bird's "speed in still air", less the wind's speed in the opposite direction, is close to zero, or even negative. So the bird gives up and lands back in your tree.
The same concept applies to engines and their speedometers. If a boat's motor is chugging away at 10 miles an hour (according to the speedometer), but the boat is facing an opposing water current of 15 miles an hour, then the boat will end up going backwards at five miles an hour. You could be driving uphill in snowy conditions, reach a patch of ice, and feel your upward progress slow to a stop. You can rev the engine and spin your tires all you like, but your car is still gonna start sliding backwards downhill.
In other words, a motor's speedometer reading is not always the actual speed of the object that the motor is powering.
Returning to the exercise:
I'll pick variables and set up a system. In this case, I'll use:
plane's speedometer: p
windspeed: w
When the plane is going "with" the wind, the plane's powered speed and the windspeed will add together; when the plane is going "against" the wind, 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 spent 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:
p + w = 600
p − w = 450
Then, by adding down, I get:
2p = 1050
p = 525
Back-solving, I see that the windspeed w must be 75 mph.
jet's speed: 525 mph
windspeed: 75 mph
Another topic you might see (if not now, then later in calculus) is decomposing rational expressions using partial fractions.
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Find the partial fraction decomposition of the following:
The denominator of the polynomial fraction they've given me factors as:
(x + 2)(x + 1)(x − 1)
These factors will be the denominators in the partial-fraction decomposition. That is, I'll be looking for the values of A, B, and C which will complete the following:
The above expression is meant to be equal to the original fraction they gave me. Setting them equal and then 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
The standard way of solving this big messy equation is the process of "comparing coefficients". Two polynomials are equal only if the coefficients of their terms are equal. This is why I grouped my terms the way I did in the last line above; I've grouped everything that was multiplied by x2, everything that was multiplied by x, and everything that was multiplied by 1 (that is, everything that was just a constant, with no variable part).
On the left-hand side, I've got 5x + 7, which has no term with an x2, so I'll need to think of "5x + 7" as "0x2 + 5x + 7". This will allow me to create new equations, based on the fact that the coefficients on either side of the "equals" sign have to be the same. This gives me:
x2: A + B + C = 0
x: B + 3C = 5
1: −A − 2B + 2C = 7
Solving this system, I get A = −1, B = −1, and C = 2. Then the partial fraction decomposition is:
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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