Very commonly, system-of-equations word problems involves mixtures or combinations of some sort. For instance:
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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.
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 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 meant 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.
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Returning to the exercise:
I'll pick variables and set up a system. In this case, I'll use:
plane's speedometer: p
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.
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.