VICs - رویکرد جبری

سرفصل: بخش ریاضی / سرفصل: مسائل کلمه / درس 23

VICs - رویکرد جبری

توضیح مختصر

  • زمان مطالعه 6 دقیقه
  • سطح خیلی سخت

دانلود اپلیکیشن «زوم»

این درس را می‌توانید به بهترین شکل و با امکانات عالی در اپلیکیشن «زوم» بخوانید

دانلود اپلیکیشن «زوم»

فایل ویدیویی

متن انگلیسی درس

VICs - Algebraic Approach

In the last video, we said that we could solve problems with variables in the answer choices either using a direct algebraic approach or by picking numbers. So those are the two big strategies. In this lesson, we’re gonna focus on the algebraic approach and we will discuss number picking in the next lesson. So, just to remind you, the algebraic approach, so we’re gonna take the variables given in the problem, and just do all the work, all the algebra with those variables.

Of course, the advantage of that is that if you do all the algebra correctly, you will get to a single unambiguous answer. The disadvantage is, of course, you have to be comfortable with algebra. So, the first idea is that just because you are following the algebraic method that doesn’t mean you have to think through every last detail in letters. Many folks find it helpful to reason out some particular algebraic step by temporarily plugging in some numbers, and that’s perfectly fine.

So, for example, in a problem, suppose we’re told there are H hammers and they have a weight ratio of Q pounds per hammer. Suppose that in the course of the problem we would need an expression for the total weight of these hammers. So this would not be the total problem, this might be one aspect or one piece of a larger problem.

We need the weight of these hammers and then we’re gonna combine them with something else. Well, suppose you’re wondering well exactly what do I do with the H and the Q to get the total weight of the H hammers? Well, think about it this way. If the hammers are five pounds per hammer and I have eight of these hammers well that would be 40 pounds.

So in other words the total weight 40 equals five times eight. And so what that means if I want the total weight, of course what I’m gonna do is multiply Q times H. So even if you’re reasonably sure that that’s what you should do, sometimes it’s helpful just to plug in some sample numbers and think, okay. Just to verify yourself.

Okay. I do multiply those two. That’s how I get the total weight. That makes sense. And now we can use QH, we’ll go on to the rest of the problem using the variables, not the number. But that just helps you to verify one step.

That can be very helpful in a problem. Another issue, this is a big one, especially with more complex expressions, is that there’s more than one way, more than one correct way to write the expression. This means it’s perfectly possible to do all the algebra correctly to get to your answer, and then have something that doesn’t match any of the answer choices.

And this really freaks people out, because they do all their algebra, they get to the answer choices, it doesn’t match, and they think oh my God, I’ve done something wrong, I’ve made some big mistake. And what’s going on is maybe you haven’t made a mistake. Maybe it’s just that what you have is correct but you have to change the form of it so it looks like one of the answer choices.

So very important to think about this. Here’s a sample problem. Well, I’m gonna say pause the video, read this problem then we’ll talk about it. Okay, so this is a classic variable in the answer choice problem. And we want to know the average speed of the second leg of the trip.

Well, we’re gonna do this with algebra. So, the first leg, we know the distance is A, the speed is p, so the time of that first leg is A divided by p. That’s, that’s our good old, time equals rate divided, distance divided by rate formula. For the whole trip, the time is capital T, the speed is capital V.

So the distance of the whole trip is V times T. And so this means that the distance of the second leg is the total distance, VT, minus the distance of the first leg A, VT minus A. The time of the second leg is the time of the whole trip, T minus the time of the first leg, A over p. And so we find the velocity just by doing distance divided by time.

So the average velocity for the second leg is this complex fraction. And the numerator, VT minus A and then the denominator, T minus A over P. Okay, so we’ve done our correct algebra. We have our answer. Everything is correct and here are the answer choices. And we have a problem because what we have here does not match the answer choices.

And again when this happens don’t go into panic mode, don’t automatically think, oh my god I’ve done something wrong. Just keep in mind, is there a way that I might rearrange the answer I found to make it match one of the answer choices. The answer we found is not listed as one of the answers but it appears close to answer choice c.

And in particular, the change I’m gonna make is I’m gonna divide the numerator and the denominator by T. So it’s like I’m gonna multiply by a one over T over one over T. And so in the numerator when I divide, I’ll get the T will cancel. In the first term, I’ve get V minus A over T. In the denominator, the T divided by T will be 1, and then I’ll wind up with a T in the denominator of that A over p fraction.

So I wind up with something like this. And this matches answer choice C. This is exactly the form in which answer choice C is written. Frequently, algebraic expressions can be written in more than one form. Either by multiplying or dividing the numerator and denominator of a fraction, or expanding and simplifying an expression with an exponent.

It’s not enough to be able to find an algebraically correct answer. It’s an important skill to match what you find with the answer choice listed. And, for that reason, it’s very important to be comfortable especially with a complex expression being able to change it into equivalent forms. Move things around and make, make other forms that look different, but are mathematically equivalent because you may have to do that to find the correct answer.

In summary, an algebraic solution to a variable in answer choice problem can be an efficient means of solution. It really can narrow down a very particular solution, there’s no guess work. When figuring out the right operation to create expressions or variables it can be helpful to explore relationship with numbers that can help you think things through and verify your algebra.

And remember, even if all your algebraic work is correct, you may have to rearrange your correct answer to match what is listed among the answer choices.

مشارکت کنندگان در این صفحه

تا کنون فردی در بازسازی این صفحه مشارکت نداشته است.

🖊 شما نیز می‌توانید برای مشارکت در ترجمه‌ی این صفحه یا اصلاح متن انگلیسی، به این لینک مراجعه بفرمایید.