Simplifying with Substitutions
Simplifying with substitutions. In this video, we discuss a sophisticated trick for simplifying some of the hardest algebra problems on the test. So this is relatively advanced material. So just be aware this is not something you're gonna need on easier problems. You're only gonna need this on some of the most advanced problems on the test.
This trick involves some right-brain pattern matching. So, the left brain, just a quick reminder, the left brain, that's the part of the brain that's really good at following rules and procedures and recipe. The right brain, that's the part that is much better at intuition and matching patterns. And so that can be harder to train.
And so sometimes it's hard to see these right brain things until they're pointed out. So here's a practice problem. Pause the video and think about this practice problem for a minute. Think about how you would go about solving this. Okay, well, obviously we have something that is quadratic-like, so we definitely wanna subtract the 24 and get everything equal to 0.
Now, we could use FOIL to multiply out 2x-1 squared, then we get a bunch of different terms. Then we distribute the 5 in the other term, we'd collect all the like terms. We'd wind up with a horribly large number of terms, it would take many steps to simplify that. And then we'd still have a quadratic we'd have to factor.
Well, think about it this way. Instead, since that same thing appears twice, what happens if we just let u equal 2x- 1. So everywhere where there's a 2x-1, we're just gonna replace that with a u. Well, then we get this remarkably simple equation, u squared + 5u- 24 = 0. Well, that's something we can solve.
We can factor that and find both values of u. All right, now we have to be very careful. Those are not the answers yet, those are the values of u. Now what do we do with this, once we have these values of u? Well, now what we're gonna do is we're gonna set everything that equals u, that's gonna equal 2x-1.
So we use these values of u to solve for x. So if u = -8, that means -8 = 2x- 1. Similarly with the other one, solve each one of those for x, we get x = negative seven halves, or x = 2. And those are the solutions. Here's another practice problem.
Pause the video and think about how you would solve this. Okay, much in the same way, let u = x squared + 1, because that's the thing that appears twice. Then that entire equation becomes u squared- 15u + 50 = 0. Very easy to factor, u- 5, u- 10. And this means that u = 5 or u = 10.
Well, now that we have these values of u, we'll use these to solve for x. So it means x squared + 1 = 5, or x squared + 1 = 10. Subtract 1 in both equations, take the square root, keeping in mind that we need both the positive and the negative. So x is positive or negative 2, or x is positive or negative 3. So it turns out that original equation is actually an equation that has four different solutions.
Finally, a very different kind of problem. Pause the video, and then we'll talk about this. All right, one way to go about this, break this into stages. Let A equal that entire denominator. So we're just gonna simplify this. So all we're dealing with is 3 divided by A = 15.
Well, we can multiply by A, divide by 15, and that simplifies to one fifth. So A equals one fifth, that whole denominator equals one fifth. All right, well, now we have that piece. Now, let B equal that ugly fraction thing. Then, of course, the denominator is just 1 minus that fraction thing, A = 1- B. Plug in the value we have for A, we can solve for B, so B equals four fifths.
So this means that 8 over 7 + k = four fifths. Well, now what I'm gonna do with this, first of all, remember, with proportions we can cancel a common factor in the two numerators. So I'm gonna cancel a factor of 4 in the two numerators, divide both sides by 4. Now I can just cross multiply, I get 7k = 10, subtract 7, I get k = 3, and that is the solution.
So that's a relatively efficient way to go through that problem. In summary, if one expression is repeated in an equation, we can choose a single variable, often we use u, for that expression, solve for its value, the value of u, and then use that to solve for the original variable. If an algebraic expression has multiple parts, such as a compound fraction, we can solve for the numerical values part by part.
Read full transcriptThis trick involves some right-brain pattern matching. So, the left brain, just a quick reminder, the left brain, that's the part of the brain that's really good at following rules and procedures and recipe. The right brain, that's the part that is much better at intuition and matching patterns. And so that can be harder to train.
And so sometimes it's hard to see these right brain things until they're pointed out. So here's a practice problem. Pause the video and think about this practice problem for a minute. Think about how you would go about solving this. Okay, well, obviously we have something that is quadratic-like, so we definitely wanna subtract the 24 and get everything equal to 0.
Now, we could use FOIL to multiply out 2x-1 squared, then we get a bunch of different terms. Then we distribute the 5 in the other term, we'd collect all the like terms. We'd wind up with a horribly large number of terms, it would take many steps to simplify that. And then we'd still have a quadratic we'd have to factor.
Well, think about it this way. Instead, since that same thing appears twice, what happens if we just let u equal 2x- 1. So everywhere where there's a 2x-1, we're just gonna replace that with a u. Well, then we get this remarkably simple equation, u squared + 5u- 24 = 0. Well, that's something we can solve.
We can factor that and find both values of u. All right, now we have to be very careful. Those are not the answers yet, those are the values of u. Now what do we do with this, once we have these values of u? Well, now what we're gonna do is we're gonna set everything that equals u, that's gonna equal 2x-1.
So we use these values of u to solve for x. So if u = -8, that means -8 = 2x- 1. Similarly with the other one, solve each one of those for x, we get x = negative seven halves, or x = 2. And those are the solutions. Here's another practice problem.
Pause the video and think about how you would solve this. Okay, much in the same way, let u = x squared + 1, because that's the thing that appears twice. Then that entire equation becomes u squared- 15u + 50 = 0. Very easy to factor, u- 5, u- 10. And this means that u = 5 or u = 10.
Well, now that we have these values of u, we'll use these to solve for x. So it means x squared + 1 = 5, or x squared + 1 = 10. Subtract 1 in both equations, take the square root, keeping in mind that we need both the positive and the negative. So x is positive or negative 2, or x is positive or negative 3. So it turns out that original equation is actually an equation that has four different solutions.
Finally, a very different kind of problem. Pause the video, and then we'll talk about this. All right, one way to go about this, break this into stages. Let A equal that entire denominator. So we're just gonna simplify this. So all we're dealing with is 3 divided by A = 15.
Well, we can multiply by A, divide by 15, and that simplifies to one fifth. So A equals one fifth, that whole denominator equals one fifth. All right, well, now we have that piece. Now, let B equal that ugly fraction thing. Then, of course, the denominator is just 1 minus that fraction thing, A = 1- B. Plug in the value we have for A, we can solve for B, so B equals four fifths.
So this means that 8 over 7 + k = four fifths. Well, now what I'm gonna do with this, first of all, remember, with proportions we can cancel a common factor in the two numerators. So I'm gonna cancel a factor of 4 in the two numerators, divide both sides by 4. Now I can just cross multiply, I get 7k = 10, subtract 7, I get k = 3, and that is the solution.
So that's a relatively efficient way to go through that problem. In summary, if one expression is repeated in an equation, we can choose a single variable, often we use u, for that expression, solve for its value, the value of u, and then use that to solve for the original variable. If an algebraic expression has multiple parts, such as a compound fraction, we can solve for the numerical values part by part.
