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Math Strategy: Picking Numbers - Part I
Transcript
Now, we can talk about the important strategy of picking numbers. So when you have an algebra problem, of course, there's the straight forward way to solve it. The way that your algebra teacher would want you to solve it, but there's also other possibilities and one is alternative would be to use picking numbers. When the same variable is given in both and the prompt and the answer choices, picking a number for this variable can be very helpful.
Picking numbers can also be particularly effective when variables appear in inequalities. As with mental math techniques and number sense, this can be a time saver on the ACT math. But as with those, you would have to practice it to get quick at it. When you get quick at picking numbers, it becomes an extraordinarily efficient strategy.
You may wonder why you need to solve a problem in this new way, if you already know how to solve it directly. In fact, if you can practice a problem in two different ways and get the answer in two different ways, you really understand it vert deeply. And so it's always useful to find more than one way to solve a problem. One aspect of picking numbers is to go after any low-hanging fruit and what we mean by that is sometimes a convenient choice of the variable will make all or part of the expression equal to 0, or we can just pick the 0 for one of the variables.
And just with this very easy choice, it will allow us to eliminate a few answers right away. That's very powerful and especially if you don't know the legitimate way to solve the problem, eliminating answers increases your odds of guessing correctly. So this is a very efficient way to eliminate maybe two or three answers at once with very little effort.
Here is a practice problem. Pause the video and then we'll talk about this. So here's a hard algebra problem and there would be a straightforward algebraic way to solve it. Let's pretend that we don't know that. Let's pretend that maybe you do know how to do that, but let's just pretend that we don't know the regular algebra way to solve this.
How could we use picking numbers to get an answer to this question? So one interesting choice is simply y=0. If we pick y equals 0, then that expression in the prompt just becomes, we get 0's in the numerators. So those fractions equal 0, the whole thing equals 0. And so what we need is something where if y=0, the answer equals 0.
So any answer choice that doesn't equal 0 when y=0 can't be the answer. So when we plug in, we see those 4 equals 0. So one of the four of them could be the answers. But when we plug in to E, what we get there, when y=0, we get negative 2x over 1-x squared. And so unless x also equals 0 and we're not gardened that x equals 0 at the moment, then that would not equal 0.
And so E, we can defiantly eliminate. That is definitely one that does not always equal 0, if y=0. Well, that wasn't the most spectacular choice. That one just allowed us to eliminate one answer. Another interesting choice we can try, what if we make x=0? If we make x= 0, well, then what we get is essentially y over 1 plus y over 1 and that just equals 2y.
And so we plug in x=0, the correct answer should be the one that equals 2y. So we go back to the answer choices, what happens when I plug in x=0? Well, A, that's not going to equal 2y. That's just y. So that's not gonna work. B, if I have x=0, that's gonna be y over 0.
That's gonna be undefined. So that's certainly not gonna equal 2y. C, very interesting. C, if I plug in x=0, then I get simply 2y. So that does come out to the correct answer. Put that on hold a second.
And then D, if I plug x=0 into that, I get 0. So that doesn't equal 2y. So at this point, we can eliminate A, B, D and E. And by process of elimination, we have figured out the answer is C. In that problem with a couple easy choices, we're able to eliminate four choices and isolate a single answer by the process of elimination.
Sometimes it's easy to eliminate two or three choices, but it's harder to pick numbers deciding between the remaining choices. Remember if you could allude two or three choices, your odds of guessing the right answer increases substantially and that is one of the many advantages of the picking numbers approach. In summary, picking numbers is an alternative strategy for dealing with algebra, particularly useful when variables appear in complicated equations or inequalities.
And one aspect of picking numbers is to go after low hanging fruit, that is to say easy to calculate choices that eliminate a few answers at once.
Read full transcriptPicking numbers can also be particularly effective when variables appear in inequalities. As with mental math techniques and number sense, this can be a time saver on the ACT math. But as with those, you would have to practice it to get quick at it. When you get quick at picking numbers, it becomes an extraordinarily efficient strategy.
You may wonder why you need to solve a problem in this new way, if you already know how to solve it directly. In fact, if you can practice a problem in two different ways and get the answer in two different ways, you really understand it vert deeply. And so it's always useful to find more than one way to solve a problem. One aspect of picking numbers is to go after any low-hanging fruit and what we mean by that is sometimes a convenient choice of the variable will make all or part of the expression equal to 0, or we can just pick the 0 for one of the variables.
And just with this very easy choice, it will allow us to eliminate a few answers right away. That's very powerful and especially if you don't know the legitimate way to solve the problem, eliminating answers increases your odds of guessing correctly. So this is a very efficient way to eliminate maybe two or three answers at once with very little effort.
Here is a practice problem. Pause the video and then we'll talk about this. So here's a hard algebra problem and there would be a straightforward algebraic way to solve it. Let's pretend that we don't know that. Let's pretend that maybe you do know how to do that, but let's just pretend that we don't know the regular algebra way to solve this.
How could we use picking numbers to get an answer to this question? So one interesting choice is simply y=0. If we pick y equals 0, then that expression in the prompt just becomes, we get 0's in the numerators. So those fractions equal 0, the whole thing equals 0. And so what we need is something where if y=0, the answer equals 0.
So any answer choice that doesn't equal 0 when y=0 can't be the answer. So when we plug in, we see those 4 equals 0. So one of the four of them could be the answers. But when we plug in to E, what we get there, when y=0, we get negative 2x over 1-x squared. And so unless x also equals 0 and we're not gardened that x equals 0 at the moment, then that would not equal 0.
And so E, we can defiantly eliminate. That is definitely one that does not always equal 0, if y=0. Well, that wasn't the most spectacular choice. That one just allowed us to eliminate one answer. Another interesting choice we can try, what if we make x=0? If we make x= 0, well, then what we get is essentially y over 1 plus y over 1 and that just equals 2y.
And so we plug in x=0, the correct answer should be the one that equals 2y. So we go back to the answer choices, what happens when I plug in x=0? Well, A, that's not going to equal 2y. That's just y. So that's not gonna work. B, if I have x=0, that's gonna be y over 0.
That's gonna be undefined. So that's certainly not gonna equal 2y. C, very interesting. C, if I plug in x=0, then I get simply 2y. So that does come out to the correct answer. Put that on hold a second.
And then D, if I plug x=0 into that, I get 0. So that doesn't equal 2y. So at this point, we can eliminate A, B, D and E. And by process of elimination, we have figured out the answer is C. In that problem with a couple easy choices, we're able to eliminate four choices and isolate a single answer by the process of elimination.
Sometimes it's easy to eliminate two or three choices, but it's harder to pick numbers deciding between the remaining choices. Remember if you could allude two or three choices, your odds of guessing the right answer increases substantially and that is one of the many advantages of the picking numbers approach. In summary, picking numbers is an alternative strategy for dealing with algebra, particularly useful when variables appear in complicated equations or inequalities.
And one aspect of picking numbers is to go after low hanging fruit, that is to say easy to calculate choices that eliminate a few answers at once.
