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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 beware, 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 2 x minus 1 squared, then we get a bunch of different terms. Then we’d distribute the 5 in the other wo, the other term. We’d collect all the like terms. We’d end 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 minus one? So everywhere where there’s a 2x minus one, we’re just gonna replace that with a u.
Well then we get this remarkably simple equation, u squared plus 5u minus 24 equals 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 those, everything that equals u, that’s gonna equal 2x minus 1. So we use these values of u to solve for x. So if u equals negative 8, that means negative 8 equals 2x minus 1.
And similarly with the other one, solve each one of those for x, we get x equals negative 7 halves, or x equals 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 equal x squared plus 1, because that’s the thing that appears twice.
Then that entire equation becomes u squared minus 15u plus 50 equals 0. Very easy to factor, u minus 5, u minus 10. And this means that u equals 5, or u equals 10. Well now that we have these values of u, we’ll use these to solve for x. So it means, x squared plus 1 equals 5, or x squared plus 1 equals 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 4 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 equals 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 equals 1 minus 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 plus k equals 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 and the two numerators divide both sides by 4, now I can just cross multiply. I get 7k equals 10, subtract 7 I get k equals 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 numerical values part by part.
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