Clearing Fractions in an EquationDate: 01/13/2005 at 01:24:10 From: Karen Subject: it's about algebra; clearing an equation of fractions, decim I have this math problem that I understand how to do but it's just that there's this one that I'm really stuck on. I get confused because it has parentheses and the way my teacher explains it we need to use more parentheses. Here is the problem: 1/4(8y + 4) - 17 = -1/2(4y - 8) Here's what I tried: 1/4(8y + 4) - 17 = -1/2(4y - 8) 4[1/4(8y + 4) - 17] = 4[-1/2(4y - 8)] 4(1/4)4[(8y) + 4(4) - 4(17) = 4[(-1/2)4(4y) - 4(8) This was the most I could do. I hope I'm doing it right because I was really struggling. On other problems I don't get as confused as I did on this one. Date: 01/13/2005 at 02:09:03 From: Doctor Ian Subject: Re: it's about algebra; clearing an equation of fractions, decim Hi Karen, To be honest, in this case I'd go ahead and distribute the fractions before clearing them, because that actually _will_ clear them: 1/4(8y + 4) - 17 = -1/2(4y - 8) (8y/4 + 4/4) - 17 = (-4y/2 + 8/2) (2y + 1) - 17 = -2y + 4 Look, no fractions. :^D It's easy to fall into the trap of trying to classify situations in terms of rules, e.g., "There are fractions, so I need to multiply both sides of the equation by the common denominator to clear them". It's true that often, that's the best way to go. But there are times-- like this one--when it's not. The key thing to notice here is that it's easy to take 1/4 of each of the numbers in the left hand binomial and also easy to take 1/2 of each of the numbers in the right hand binomial. Your job is to stay flexible, consider all your options, and choose the one that is going to make your life easiest in whatever situation you happen to be in. If you decide that you really want to multiply both sides by 4, there is one key idea that you seem to be missing in your work, which is making it much harder than it needs to be. Suppose you have two numbers that are multiplied together, like 2(3). We know that's 6, right? So now if you want to multiply that by 4, the answer should be 24 since 4*6 is 24. Let's show the multiplication by 4 this way: 4[2(3)] which is the same as 4[6] or 24 The mistake you are making in your work is that you are multiplying the 4 times the 2 and times the 3, so you are getting: 4(2)4(3) But that makes 8*12, which is 96, not the 24 it should be. You've multiplied by an extra 4! The important rule is that when you multiply through an equation to clear a fraction or decimal, you need to be sure to multiply each _term_ of the equation by that number. The terms are separated by addition and subtraction. Any group of items that are being multiplied together are all part of one term. So, in your equation, there are really only two terms on the left and one on the right side: 1/4(8y + 4) - 17 = -1/2(4y - 8) ----------- -- ------------ 1 term 1 term 1 term You might have been told that in a binomial, like (8y + 4), there are two terms, and that's correct. If we distribute the 1/4 like we did above, that 1/4 multiplies against each term in the binomial. But if we step back another level and look at the equation, that whole first piece is just one term because the 1/4 and the binomial are being multiplied together, jus as with 2(3) in the example above. So when we multiply by 4, we really only need to multiply each of those three terms by 4 once: (4)(1/4)(8y + 4) - (4)(17) = (4)(-1/2)(4y - 8) ------------- ---- -------------- Now, just as with 4(2)(3) above, we only use the 4 once. So on the left the (4)(1/4) cancels and just leaves (1)(8y + 4). On the right, the (4)(-1/2) makes -2 left to multiply against the (4y - 8): (1)(8y + 4) - (4)(17) = (-2)(4y - 8) Distributing the (1) and the (-2): 8y + 4 - 68 = -8y + 16 Here's a general look at what went on, replacing the binomials with "this" and "that" to make it clearer how multiplying each side by 4 works: 1/4*this - 17 = -1/2*that Now you can go ahead and clear the fractions, 4(1/4*this - 17) = 4(-1/2*that) this - 4*17 = -2*that this - 68 = -2*that Replacing "this" and "that" with the original binomials, we have: (8y + 4) - 68 = -2(4y - 8) Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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