Solving Problems With and Without AlgebraDate: 05/14/2002 at 23:26:42 From: M Hanlon Subject: Math equation In a basketball game, Harold and Isaac scored a total of 19 points. Isaac and Jacob scored a total of 14 points. Isaac scored as many ponts as Harold and Jacob together. How many points did each player score? I have not taken algebra yet and do not know how to do this equation. I used various methods but could not come up with the answer. Thanks for the help. Date: 05/17/2002 at 12:43:17 From: Doctor Ian Subject: Re: Math equation Hi, Just for fun, let me show you how you would solve the problem 'by algebra'. Then I'll show you how to solve it without algebra. If we let H stand for Harold's points, and so on, we get H + I = 19 I + J = 14 I = H + J which is a set of three equations with three unknowns. Note that since I is the sum of H and J, we can just get rid of it by substituting (H+J) wherever we see I: H + (H+J) = 19 (H+J) + J = 14 (H+J) = H + J The third equation no longer tells us anything useful, so we can get rid of it: 2H + J = 19 H + 2J = 14 Now we can use the same trick again: If H + 2J = 14, then H = 14 - 2J, so 2(14 - 2J) + J = 19 (14 - 2J) + 2J = 14 Again, the final equation tells us nothing useful, so we can drop it. Now we have one equation, with one variable, which is good, because that's the kind of thing we can solve: 2(14 - 2J) + J = 19 28 - 4J + J = 19 28 - 3J = 19 28 - 19 = 3J 9 = 3J 3 = J So Jacob scored 3 points, and you can use that information to figure out what the other players scored. But what if you _haven't_ had algebra yet? Can you still solve a problem like this? Let's start with the fact that Harold and Isaac scored 19 points together. Here are the possibilities: Harold Isaac ----- ----- 19 0 18 1 17 2 16 3 15 4 14 5 13 6 12 7 11 8 10 9 9 10 8 11 7 12 6 13 5 14 4 15 3 16 2 17 1 18 0 19 Now, Isaac and Jacob scored a total of 14 points. So we can add another column to our table: Harold Isaac Jacob ----- ----- ----- 19 0 14 18 1 13 17 2 12 16 3 11 15 4 10 14 5 9 13 6 8 12 7 7 11 8 6 10 9 5 9 10 4 8 11 3 7 12 2 6 13 1 5 14 0 4 15 -1 3 16 -2 2 17 -3 1 18 -4 0 19 -5 Now assuming that Jacob can't score negative points, we can eliminate the final 6 rows of the table! Now for our final piece of information: Isaac scored as many points as Harold and Jacob put together. Let's add one more column to the table: Harold Isaac Jacob Harold+Jacob ----- ----- ----- ------------ 19 0 14 33 18 1 13 31 17 2 12 29 16 3 11 ? 15 4 10 ? 14 5 9 ? 13 6 8 ? 12 7 7 ? 11 8 6 ? 10 9 5 ? 9 10 4 ? 8 11 3 ? 7 12 2 9 6 13 1 7 I'll leave the other values for you to fill in. But notice that in only _one_ of these rows will the values in the 2nd and 4th columns be the same. That row satisfies all the conditions in the problem. The thing about a problem like this is that it doesn't really illustrate the power of algebra, because the table we had to build to work the problem without algebra wasn't all that big. But suppose that we had been working with numbers like 196 and 147 instead of numbers like 19 and 14? The great thing about a solution using algebra is that it's about the same size regardless of how large or small the numbers in the problem are. What that means in practice is that for small problems, it often makes more sense to give trial-and-error a chance to work before going to the trouble of setting up equations. A big part of becoming a good problem solver is learning to recognize when you need steam roller to crush something, and when you can just use a hammer... or your hand. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 05/17/2002 at 14:46:32 From: M Hanlon Subject: Math equation Thanks so much for your help and time indoing this for me. I can now understand both ways of doing it. M Hanlon |
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