Solving an Algebra EquationDate: 3/20/96 at 3:1:35 From: Anonymous Subject: linear equations Dear Dr Math: I hope you will be able to help me with this problem, I'm really stuck. 10-(k+5) = 3(k+2) I would also appreciate any explanation. Date: 3/20/96 at 15:1:58 From: Doctor Elise Subject: Re: linear equations Hi! It's a little confusing when you have the same variable on both sides of the equals sign, isn't it? We know that when the problem is solved, it will look like "k = something". So we have to try to get all the "k's" on one side of the equation, and all the "somethings" on the other, and then we'll be done! The first step is to do all the multiplication and addition and subtraction on each side of the equals sign by itself. That way we get rid of the parentheses. 10 - (k+5) = 3(k+2) On the left side, we have 10 minus (k plus 5). The parentheses mean we have to subtract BOTH k and 5 from 10. Okay. So we can rewrite it without the parentheses and the left side becomes: 10 - k - 5 And because addition and subtraction without parentheses can be done in any order (which is called the commutative property of addition - the one that says a + b = b + a) we can rewrite it as 10 - 5 - k and get: 5 - k So that's the left side. Let's look at the right side: 3(k+2) Okay, we could say this as 3 times the sum of k and 2. The distributive property of multiplication means that we can "distribute" the 3 over both the k and the 2, meaning that 3 times (k+2) is the same as 3 times k plus 3 times 2. (I'll use '*' for "times".) 3*k + 3*2 We usually write 3*k as just 3k, so this becomes: 3k + 6 Now we've simplified the left and the right sides of the original equation, so by now we have: 5 - k = 3k + 6 Doesn't that look better? It still doesn't look like "k = something" though. Okay. The key to moving terms around in a polynomial is that you can do anything to one side of the equation as long as you do the same thing to the other side. I want to add k to each side so that I don't have a 'k' on the left, I can do it this way: 5 - k + k = 3k + 6 + k And now I can simplify both sides by adding the terms with 'k's in them: 5 = 4k + 6 Well, this is looking better! Let's subtract 6 from each side so that we have only 'k's on the right: 5 - 6 = 4k + 6 - 6 -1 = 4k Great! This is the same as: 4k = -1 Now we need to get rid of that pesky 4. Let's divide each side by 4: 4k/4 = -1/4 k = -1/4 Yay! An answer! There were an awful lot of numbers flying around to get here, so I'm going to check my work by "plugging in" this value of 'k' into the original equation 10 - (k+5) = 3(k+2): 10 - (-1/4 + 5) = 3(-1/4 + 2) 10 - (4 3/4) = 3( 1 3/4) 5 1/4 = 3 * 1 + 3 * 3/4 5 1/4 = 3 + 9/4 5 1/4 = 3 + 2 1/4 5 1/4 = 5 1/4 Okay, it checks out! Good luck with your next polynomial! - Dr. Elise - The Math Forum |
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