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Algebraic properties | Worked example

Sal Khan works through a question on algebraic properties from the Praxis Core Math test.

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  • blobby green style avatar for user ronciehaley
    In a question like this, how do you know what numbers to select so as to test your hypothesis?
    (1 vote)
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  • leaf green style avatar for user Melissa Panisse
    In the option A: If A decreases, why the value is 2 and not 1/2?, and if B remains constant, why the value is 2 not 1? Is it because the values are selected based on the assigned values at the beginning of the explanation (All values increased by a factor of 2)?
    (1 vote)
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Video transcript

- [Instructor] We're told in the equation above, A, B, and C are all positive. So, this is the equation. It says M is equal to two times A over B plus C. Which of the following will always cause M to decrease? Choose all answers that apply. So, pause the video and see if you can work through it before we work through it together. Okay, now let's work through it together. So, there's a couple of ways to approach it. We could just go into the choices and see if they make sense. So, choice A says A decreases, B remains constant, and C increases. So, we could try out some values for A, B, and C, and change them the way this is talking about it and at least see what does that do to M? So, let's pick actually, for all of these cases, let's start with A is equal to, and let's see. I want all of these things to be able to decrease or increase. Let's start 'em at two. So, A is equal to two, and to be simple, let's just make B is equal to two, and C is equal to two. And so, if this is the case, and this doesn't have to be the case, but this could be the case, M would be equal to two times two, two times two, over two plus two, over two plus two. So, this would be equal to four over four, which is equal to one. So, that's where we're starting from. Now, let's do what they're saying. If A decreases, let's say that A is equal to one now, so it's decreased. B remains constant, so B is still equal to two, and C increases, so now let's say C is equal to three. Once again, we're just trying out values that are consistent with what they're telling us. So, now what is M? M is going to be equal to two times one, two times one, over B plus C, over two plus three, which is equal to two in the numerator, over five. Well, 2/5 for sure is less than one. Now, so that makes us feel pretty good about this, so I'll put a little check here. Now, you need to keep in mind that just because we found a case where when this happens, it cause M to decrease, it's not a mathematical proof that it will always cause M to decrease, but at least it makes you feel good about this choice. Now, we could reason through it a little bit more. We could say, hey, if A is decreasing, well, that's going to decrease our numerator, which is 2A. If A decreases and A is positive, then two times A will decrease. And so, if you decrease a numerator, that's one way to decrease a fraction. Now, the other thing they tell us, B remains constant, so that by itself won't change the denominator, but they say C increases. So, if C increases, and B remains constant, then the sum, and they're both positive, then the sum will increase. And if you increase the denominator of a fraction, well, that will decrease the value of a fraction. Once again, just to review, the ways to decrease the value of a fraction are either to decrease the numerator and/or increase the denominator. That's especially the case if you're dealing with numerators and denominators that are positive, and so that's what they're describing here, so we feel pretty good about this choice. Now, let's look at choice B. A remains constant, B increases, C remains constant. So, we could try those choices out again. If A remains constant, A will still be equal to two. B increases, so maybe B goes from two to three. B is equal to three. C remains constant, so it would still be equal to two. And so, this is going to be equal to, M is going to be equal to two times A, two times now it's two, over B plus C, over three plus two. This is equal to four over five, which is still less than one. So, this is looking pretty good, at least just on this case we try. And we could see that this is consistent with what we just described. A remaining constant doesn't change the numerator. B increasing and C remaining constant does increase the denominator. So, this case, case B leaves the numerator unchanged, but it increases the denominator, which will lower the value of our fraction, so I like this choice, as well. So, choice C. A increases, B increase, and C increases. So, once again, we could try that out. Let say they all go from two to three. So, A is equal to three, B is equal to three, and C is equal to three. So, in that situation, two times A, you get two times three over three plus three, B plus C here. This is going to six over six. Notice in this particular situation, M is unchanged. So, if you happen to pick these choices where M is unchanged, well then you know that this situation will not always cause M to decrease, so we could rule that out. Now, if you're just picking values, there are some set of values, some original values, or depending on how much you want to increase A, B, and C, where you could make M decrease. For example, if you increase B by a lot, if you made B 30 and C 30, this will cause M to decrease, 'cause in that situation, you have two times three over 30 plus 30, which is a lot less than one. This is equal to six over 60. And you can also do things in choice C that would make M increase. For example, if you said A is equal to 300, while B and C are only increased to 30, well then, this would become two times 300, which would be 600 here. The general principle here is if you increase everything, you don't know whether it's actually going to increase, decrease, or make M stay the same.