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SAT (Fall 2023)

Course: SAT (Fall 2023) > Unit 10

Lesson 2: Passport to advanced mathematics

Linear and quadratic systems — Harder example

Watch Sal work through a harder Linear and quadratic systems problem.

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Video transcript

- [Instructor] We're told if x comma y is a solution of the system of equations above and x is greater than zero, what is the value of xy? Pause this video and see if you can figure this out. All right now, this is interesting. We have a system of equations and this one looks linear. The top one looks quadratic. So what I'm going to do is on this linear one, it looks like it's solve for y and then I could, whatever Y is equal to, I could substitute back here for y and then let's let me see if I can solve for x. So just focusing on this bottom equation, we can subtract three from both sides and then that would lead us to y is equal to two times x plus three minus three and then I could distribute out this two and so this would be equal to 2x plus 6 then minus 3, which is, of course, if we just combine these two, we get it's equal to 2x + 3 and so now I can take this. This is equal to y, so we could substitute this back in for y and we would get 2x + 3 is equal to 2x squared + 13x - 3. Now this looks really complex, but we're just dealing with a quadratic if we can just combine terms onto one side, so let's get rid of the 2x and the 3 on the left hand side, so I'm gonna subtract that from both sides, so I'm gonna subtract the 2x from both sides and I'm going to subtract a three from both sides and then what that leaves me with on the left hand side, by design, it leaves me with zero and on the right hand side, I have 2x squared. Now 13x - 2x is going to be 11x, so plus 11x and then we have 2 -3, so that is -6. Now to solve for x here, we might wanna factor and there's techniques for doing that or we could just go straight to the quadratic formula. If either option does not seem familiar to you, I encourage you to review that in the algebra sections of Khan Academy, but my brain goes to quadratic formula, at least for something where the leading coefficient is not a one over here, so the quadratic formula of course tells us if this is of the form, ax squared, actually let me write it this way, if we're of the form ax squared + bx + c, we could see that two is equal to a, 11 is equal to b, and c is equal to -6 and so the quadratic formula, of course, is negative b plus or minus the square root of b squared minus four ac all of that over 2a. This will give us the xes that can satisfy this equation. Now, this would mean that this is going to be equal to negative b, b is 11, so negative 11 plus or minus the square root, 11 squared is 121, minus four times a, which is two times c, which is negative six, so I could put the six here and then make that a positive, since I'm multiplying by negative six, and then all of that over two times a, so a is two, so two times that is going to be equal to four and now what do we have in here? Four times two times six is equal to 48, so this is equal to 48 and so 121 + 48 is 169, so let me write it this way. Everything under the radicle is 169, which is convenient because that is 13 squared and so this is going to simplify. It's equal to negative 11 plus or minus the square root of 169, which is 13 over 4. Now we're going to have two roots here, but they gave us a little bit of an extra constraint. X needs to be greater than zero. If we did a -11 - 13 over 4, that's going to give us an x value less than zero, so the root that we're going to focus on is when we add the 13, so negative 11 + 13 is, let me actually do the two options, - 11 + 13 is 2 over 4, so 2/4 is equal to 1/2. This is good. That is greater than zero. The other option would have been -11 - 13, which is -24, negative 24 over 4, which is equal to -6. Either of these x values would satisfy, would be a, if we think about it, a point of intersection or the x coordinate of the point of intersection, but it tells us x is greater than zero, so we rule this one out. So we figured out the x value they care about, now to calculate xy, we have to figure out the y value that they care about and the simplest place, we just have to substitute this x value back into one of these equations. I find it a lot easier to substitute back into this one that y is equal to 2x + 3. Let me rewrite it over here maybe. Y is equal to 2x + 3, so that if x is equal to 1/2, so this is equal to two times 1/2 plus three. Well, that's just equal to one plus three, which is equal to four. So when x is 1/2, y is equal to four, we multiply the two, so if we just take 1/2, so I'll let me write this way, xy is going to be equal to 1/2 times four, which is going to be equal to two and we are done.