Date: 7 Feb 1995 19:54:13 -0500 From: Binford M.S. Gifted Program Subject: a question Please help me answer this question: If 10y = 1 and 10x = 1, what is the value of x+y? Thanks.
Date: 8 Feb 1995 00:18:28 -0500 From: Dr. Ken Subject: Re: a question from Jeanice Hello there! Well, I can think of a couple of different ways to do this question. I'll give them both to you. (A) We can try to find out what x is, and what y is, and then add them together. Let's do it. We know that 10y = 1 so let's ask ourselves "what number can you multiply by 10 to get 1?" Well, if you think about the number 1/10, that's exactly what it is. It's one divided into ten parts. So if you multiply 1/10 by 10, you'll get 1. So we'll say y = 1/10. Here's another way of figuring out what y is. We have the equation 10y = 1. Think about what that means. If you read it out loud, you'll say "ten times y equals one." In other words, 10y and 1 are EXACTLY THE SAME THING. If we do the same thing to each of 10y and 1, the two things we get will be equal, too. Let's divide both sides by 10. On the left side, we'll get plain old y, and on the right side we'll get 1 divided by 10, or 1/10. So y = 1/10, just as we found in the previous example. We could use the exact same methods to find x, and again we would find out that x = 1/10. So to find out what x + y is, we can add the two values for x and y. We get 1 1 1+1 2 1 x + y = --- + --- = ----- = ----- = --- 10 10 10 10 5 (B) Here's the other way to find out what x+y is. We have the two equations 10x = 1 and 10y = 1 We can add them together to get the new equation 10x + 10y = 1 + 1 = 2 Now look what happens when we divide both sides by 10, as we did in method (A). We get 10x + 10y 2 ---------- = --- 10 10 So we can simplify each side, and we get x+y = 1/5. Just as before. I hope these methods make sense to you. Write back if you have more questions! -Ken "Dr." Math
Date: 08/06/98 at 23:17:33 From: Eileen Subject: I don't understand the Algebra problem you have given In the math problem that you have given, I am stuck on the part about x + y. I do not understand how you add them together on either example (A) or (B). This is my first year for algebra, but I still don't understand how it works. So if you could explain it to me a little more, you would be my hero. Eileen
Date: 08/07/98 at 09:58:51 From: Doctor Peterson Subject: Re: I don't understand the Algebra problem you have given Hi, Eileen. I hope I can help you feel better about algebra. Sometimes reading what someone else says doesn't help as much as really interacting with someone who can answer your specific problems. If what I have to say still doesn't help, let's keep talking back and forth until we figure out what it takes to help you understand. It may take a few tries, but you'll make it! Above, where the question is: If 10y = 1 and 10x = 1, what is the value of x + y? I think the explanation is very good, so I'll try explaining things just a little differently, then suggest that you look again and think through everything very carefully. If it still doesn't make sense, maybe you can point to some particular statement that troubles you, so we can focus on that. In method (A), we're hardly doing algebra at all. You have probably seen questions like "10 times x is 50; what is x?" for years now. What's probably confusing is the use of letters, and the fact that there are two different letters, x and y, in the same problem. You have to focus on one thing at a time and not let the other stuff scare you - this is really just several simple problems stuck together, so if you take it one step at a time you should be able to handle it. The first step is just to solve "10x = 1". If the answer, x = 1/10, is not obvious, let me know and I can help you through that. The second step is practically the same: solve "10y = 1". The answer is that y = 1/10. Notice that the names of variables don't matter, so whatever you can do with x, you can do with y, and you don't even have to do the work over again to solve this. The third step is just to figure out what x + y is, now that you know that x and y are both 1/10. That's not algebra, just addition: x + y = 1/10 + 1/10 = 2/10 = 1/5 Now, method (B) is real algebra. You don't have to do it this way, but it is important to understand the idea of adding two equations (and I think that's your main question), so let's look at that. We know that: 10x = 1 and: 10y = 1 Picture a balance scale, where you can put things in two pans, one on each side, and if they weigh the same the pans will balance. The first equation tells us that if we put 10x (that is, 10 little boxes labelled "x") in the left pan, and 1 (that is, a known weight labelled "1" - maybe one pound?) on the right, the pans will balance. They have the same value. Likewise, the second equation tells us we can put 10y on the left and 1 on the right. Now suppose I put BOTH 10x and 10y on the left, and BOTH 1 and another 1 on the right. Then the pans should still balance, because I still have the same amount of stuff on both sides. I've "added" the two equations, showing that 10x + 10y = 2 The rest is a little tricky to think about, and I don't expect you to have thought of this at your level. What's being done is to think of "x + y" as if it were a single variable you want to solve for. 10x + 10y is the same as 10(x + y). Picture taking the 10 x's and 10 y's on the left side of our scale and gluing together pairs of one x and one y, so we have 10 "x + y"'s. (You've probably heard of that as the distributive rule, but never thought of it as saying that things multiply the same whether they're glued together or not.) If 10 of these things weigh 2 pounds, then one of them must weigh 2/10 of a pound, or 1/5. So that's the answer. I'm guessing that your main problem is with working with equations - adding them together, moving things from one side to the other, and so on. Thinking in terms of a scale can be very helpful when you're getting started on this. Let me know if you have some other specific problems that we can look at. - Doctor Peterson, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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