Solving Rational Equations by Using the LCDDate: 09/27/2007 at 19:32:27 From: Nikki Subject: Literal Equations/Quadratic Equations How do you solve the equation 3/(n-2) + 6/(n^2-5n+6) = 4/(n-3)? Whenever I try to solve the equation and try to put it in quadratic form, the solution I get from the quadratic formula doesn't work when I substitute it back into the equation. Please, I also need the steps to solving the equation right from the beginning. Solving these types of equations are entirely new to me and it is a school homework assignment and I have no clue where to begin. The only thing I know is that I have to put it in quadratic form. Date: 09/27/2007 at 22:49:39 From: Doctor Peterson Subject: Re: Literal Equations/Quadratic Equations Hi, Nikki. You say you got a solution, but it didn't work. What was that solution? I'd want to be sure you didn't just make a mistake in the check before saying too much else. I would approach this by first factoring the quadratic denominator, and then finding the LCD, which will be a lot simpler than it would have looked without having factored. Then I'd multiply each term by that LCD and cancel before doing any actual multiplying. That would leave me with an equation, possibly quadratic, to solve. But it is entirely possible to get an extraneous root when you do this, because you have multiplied by a variable expression, which could be worth zero. So you do have to check, and if each solution you found would result in division by zero, then they are not really solutions of the original equation. That's perfectly normal. In this case, though, you must have made a mistake, because when I solve it I don't get a quadratic equation at all. It will help if you can show me what you did, so I can identify what went wrong. My guess is that you didn't cancel correctly, or you used the wrong LCD. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 09/28/2007 at 16:44:15 From: Nikki Subject: Literal Equations/Quadratic Equations Hi, Thanks for your answer. I didn't solve it your way. I solved it by first multiplying all the denominators together. Then I multiplied them to the numerator. I got 3n^3 - 15n^2 + 18n - 9n^2 + 45n - 54 + 6n^2 - 18n - 12n + 36 = 4n^3 - 20n^2 + 24n - 8n^2 + 40n - 48, and that is all over a common denominator that I got when I multiplied them together. Then I combined the like terms: -n^3 + 10n^2 - 31n + 30 = 0 I'm not sure my solution is right because I then divided 0 by -n^3 to get a quadratic equation. Are my steps to solving correct? If not, could you please direct me in the right steps? And I'm not sure what you mean by finding the LCD of the denominators. Do I have to find the LCD of each of the denominators, or just the one big denominator? Thanks, Nikki Date: 09/28/2007 at 21:20:36 From: Doctor Peterson Subject: Re: Literal Equations/Quadratic Equations Hi, Nikki. >I didn't solve it your way. I solved it by first multiplying all the >denominators together. Then I multiplied them to the numerator. You can do that, but it makes the problem a LOT harder. You're finding a common denominator by just multiplying everything together, where I'm finding the LEAST common denominator, which means a lot less multiplication and a lot less factoring to get things to simplify again. You really need to learn my method, rather than practicing the hard way. Math is not about making life hard; it's about learning how to simplify things and save work. >I got 3n^3 - 15n^2 + 18n - 9n^2 + 45n - 54 + 6n^2 - 18n - 12n + 36 = >4n^3 - 20n^2 + 24n - 8n^2 + 40n - 48, and that is all over a common >denominator that I got when I multiplied them together. It looks like you got this from 3(n-3)(n^2-5n+6) + 6(n-2)(n-3) = 4(n-2)(n^2-5n+6) This is the correct numerator if you use your denominator. >Then I combined the like terms: > >-n^3 + 10n^2 - 31n + 30 = 0 That is what I get. >I'm not sure my solution is right because I then divided 0 by -n^3 >to get a quadratic equation. No, you can't do that. The -n^3 is ADDED to the rest; if you divide the left side by -n^3, it won't just remove that term. Try it! Make sure you always use the inverse operation to undo something, and check that it really does what you want to do. When something is added, you can only remove it by subtracting, not by dividing; when something is multiplied, you can only remove it by dividing, not by subtracting. >Are my steps to solving correct? If not, could you please direct me >in the right steps? And I'm not sure what you mean by finding the >LCD of the denominators. Do I have to find the LCD of each of the >denominators, or just the one big denominator? Now you have to solve a cubic equation. That's not easy; you pretty much have to guess at a factor, and it can take a long time. It's not incorrect, really, but it wastes a lot of time. What you've done is like adding two ordinary fractions without using the LCD: 7 5 7*84 + 5*120 1188 --- + --- = ------------ = ----- 120 84 120*84 10080 Now you've got to simplify that monstrosity! You can do it, but it's a lot of unnecessary work, considering that you just MADE these huge numbers by multiplying, and now you have to undo them by factoring. That's hard with numbers; with polynomials, it can be impossible. The smart thing to do is to use the LCD, which in this case is 840: 7 5 7*7 + 5*10 99 --- + --- = ---------- = --- 120 84 840 840 Now you have a lot less simplifying; I get 33/280. Presumably before you would be given a problem like yours, you were shown how to find the LCD, because that is THE method both for solving rational equations like this, and for adding rational expressions, just as it is for adding fractions. There is no such thing as the LCD of ONE denominator; the term LCD means Least COMMON Denominator, that is, the smallest number that can be used as a denominator by ALL of the fractions. It is the least common multiple of the denominators. The way to find it is to factor each denominator, and then choose only those factors that you need in order to get a multiple of each. In your case, the denominators are n-2 (n-2)(n-3) n-3 To make a common multiple, you need at least one (n-2), and at least one (n-3); the LCM is therefore (n-2)(n-3). Use this, and the whole thing will be incredibly easier. You can either do the addition and then multiply by the (new) denominator, or just multiply by the LCD immediately and cancel, so you never have to write the denominator again. Your text probably shows how to do this. It's probably a good experience for you to have done all that work, and then to see how much easier it can be. Hopefully you'll learn well that this is a useful method! - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 09/28/2007 at 22:55:48 From: Nikki Subject: Thank you (Literal Equations/Quadratic Equations) Thanks! The solution was 5! This was a really big help to me! Nikki Date: 09/28/2007 at 23:15:21 From: Doctor Peterson Subject: Re: Thank you (Literal Equations/Quadratic Equations) Hi, Nikki. Let's do one final step, which you always have to do: check your answer. Plugging in 5, we have 3 6 4 ----- + ------------ = ----- n - 2 n^2 - 5n + 6 n - 3 3 6 4 ----- + ------------- = ----- 5 - 2 5^2 - 5*5 + 6 5 - 3 3 6 4 --- + --- = --- 3 6 2 1 + 1 = 2 Yes, it does work. And by doing the check, we made sure that we didn't end up dividing by 0, which would have made it an extraneous solution, which means not really a solution at all. Good work. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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