Solving Linear Equations and the Scale MetaphorDate: 08/26/2004 at 21:53:01 From: Lisa Subject: The logic? behind the procedure used to isolate variables As a 30-something adult I am trying to conquer a life-long math phobia and I am hoping that by understanding exactly what is happening when we go through certain steps (isolating x for example) in algebra to solve a problem it will help. There is nothing intuitive about the process to me and I've heard math-types say that it is. For example, when isolating X in linear equations, I know that you have to perform the same operation on both sides of the equal sign, but I don't understand why. Hope this isn't a stupid question and thanks. Date: 08/26/2004 at 23:15:58 From: Doctor Peterson Subject: Re: The logic? behind the procedure used to isolate variables Hi, Lisa. It's not a stupid question, but it can be hard to answer! The trouble is that there are many ways to explain it, and without knowing what explanations you have seen and been unimpressed by, I can't be sure what explanation will be most likely to help you. But trying to find it can be a good exercise for both of us--for you to find a way to understand, and for me to find a way to explain something that to me is indeed "natural". (By the way, nothing is "intuitive" until your intuition is trained to see it that way! To an adult, it is intuitive that someone who leaves the room still exists, but that has to be learned by babies. So we think of it as intuitive only because we have gotten used to it, and connected it to our experience. I want to learn how to do that for people who are resistant to having their intuitions trained; I suspect that it may be because you are more skeptical and less gullible, rather than more "stupid".) Let's start with one of the more basic explanations, using a balance scale as a model. We'll use the simple equation x + 7 = 12 Imagine that the two sides represent the total weight on each side of a scale: x+7 12 --- --- | | +---+---+ | -+- The 7 and the 12 can stand for the weights of two known objects, which weigh 7 and 12 grams respectively. The x is some object whose weight we don't know; I like to think of it as a black box we can't look inside of, but whose contents we are trying to figure out. What can we do to find the weight? Well, we wish we just had the x balancing something known, because then we'd know its weight! And we can accomplish that. Think of the 7 and the 12 now as 7 1-gram weights and 12 1-gram weights respectively. Then if we take off 7 of them from each side, the scale should still balance. Do you see why that is true? Once we do that, we have only the x left on the left, and 12-7, or 5, 1-gram weights on the right: x 5 --- --- | | +---+---+ | -+- So the weight of object x is 5 grams. What we do with the equation is identical. We subtract 7 from both sides, and the equation will remain true. Consider what is really happening now that we know that x is 5. On the left we have 5+7 to start with, which is 12; on the right we have 12. Subtracting 7 from 12, no matter how it is represented, always gives the same answer, 5. So subtracting 7 from x+7 (by just dropping the 7) on the left, and subtracting 7 from 12 (by actually getting 5) on the right, will both give the same answer. Does that help? If not, let me know where you have trouble--which sentence doesn't convince you, and why (if you can tell)--and I'll see what I can do to make the explanation better, or try something altogether different. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 08/27/2004 at 07:50:43 From: Lisa Subject: Thank you (The logic? behind the procedure used to isolate variables) Excellent! The scale metaphor was perfect--if you subtract 7 grams from both sides, the scale's reading will remain exactly the same so you haven't altered the outcome of the problem. Thanks again! |
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