Solving Linear Equations by Using InversesDate: 08/13/2003 at 18:51:12 From: Chrissy Subject: Solve Solve for a in b = 5/7(a-8) Date: 08/15/2003 at 14:35:35 From: Doctor Ian Subject: Re: Solve Hi Chrissy, 5 b = --------, solve for a 7(a - 8) The basic idea is that you can do anything you want (except divide by zero) to either side of the equation, so long as you do the same thing to the other side, too. That way, the truth of the equation is unchanged. Does that rule sound familiar? Now, there are lots of things we _might_ do, but not all of them get us closer to our goal. For example, we could add 50 to both sides to get 5 b + 50 = -------- + 50 7(a - 8) There's no good reason to do this, but it's important to understand that there's nothing _wrong_ about it, either. It's sort of like if you're doing a maze puzzle. Sometimes you make turns that don't take you where you want to go, but as long as you don't cross a line, it's okay. All that happens is that you take longer to find your way out of the maze. Anyway, we'd like to end up with something that looks like a = [a bunch of stuff with no a's in it] How can we do that? Well, one thing we might try is multiplying both sides of the equation by 7. That will give us 5 * 7 7b = -------- 7(a - 8) The 7 in the numerator cancels the 7 in the denominator, leaving us with 5 7b = ------- (a - 8) Now, here's a step that might seem unintuitive, but what if we multiply both sides by (a-8)? That gets rid of the fraction: 5 * (a - 8) 7b(a - 8) = ------------- (a - 8) The (a-8)'s cancel, just as the 7's did: 7b(a - 8) = 5 You know what? Maybe the fraction wasn't so bad after all! If we divide both sides by (7b), we get 7b(a - 8) 5 ----------- = -- 7b 7b which simplifies to 5 (a - 8) = -- 7b Now we can add 8 to both sides to get 5 a - 8 + 8 = -- + 8 7b 5 a = -- + 8 7b which is the kind of equation we wanted. So we're done. Of course, every problem is different, so these exact steps won't work for other problems. But what are some general techniques that you can use when you're trying to isolate one variable? At heart, all the techniques are based on the idea of inverses. If we have something like b = a + 5 we want to invert the addition. We invert addition by subtraction: b - 5 = a + 5 - 5 b - 5 = a + 0 b - 5 = a Similarly, if we have something like b = 3a we want to invert the multiplication. We invert multiplication with division: b 3a - = -- 3 3 b 3 - = - * a 3 3 b - = 1 * a 3 b - = a 3 What makes things messy is that you often get additions and multiplications and subtractions and divisions all mixed up together in the same equation. When this happens, doing the steps is pretty simple, but finding the right _order_ for the steps can be difficult. One trick is to let the parentheses show you the way. Note that the horizontal fraction bar acts like two pairs of parentheses, e.g., a + b ------- = (a + b) / (c + d) c + d So looking back at our example, we could have written it this way: b = (5) / (7(a - 8)) Now, this is a little uglier, but it has one very nice feature: it suggests what the next thing to do should be. In this case, we have a division. To invert a division, we multiply: b(7(a - 8)) = 5 Now we have a multiplication. To invert a multiplication, we divide: 7(a - 8) = 5 / b Again, we have a multiplication. To invert a multiplication, we divide: a - 8 = (5 / b) / 7 Now we have a subtraction. To invert a subtraction, we add: a = ((5 / b) / 7) + 8 Again, this isn't the prettiest expression in the world! But it's correct, and we can either leave it like this, or tidy it up: ((5 / b) / 7) + 8 5 / b = -------- + 8 7 5 = ---- + 8 7b There's a subtle but important point to be learned from this, which is that the reason we sometimes have several different ways of writing the same thing is that some ways are good for some tasks, while others are good for other tasks. I hope this helps. Write back if you'd like to talk more about this, or anything else. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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