If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Linear and quadratic systems — Basic example

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

Want to join the conversation?

Video transcript

- [Teacher] Which of the following represents all solutions X comma Y to the system of equations shown below? This is an interesting system of equations because this is a linear equation, this first one, but the second one is nonlinear. You have a Y squared right over here. Well this one actually can be solved with substitution because 2y plus six needs to be equal to X but then we also that X is equal to Y squared minus nine. So we can take Y squared minus nine and substitute it for X. So if you do that, you then can solve for Y, the Y of a solution to this system. So if 2y plus six is equal to Y squared minus nine. And so let's see, let's just get zeroes on the left hand side. So we could subtract 2y from both sides. I'll write that subtract 2y there. And we could subtract six from both sides. And then on the left hand side we're gonna be left with zero. And on the right hand side you're gonna have Y squared minus 15 minus, actually let me write it the other way, Y squared, let me write the minus 2y minus 15. And now to solve for the Ys that would satisfy this quadratic right over here, we could factor this quadratic, we could factor this quadratic expression. And let's see, the way I would do that is first of all kind of think of two numbers whose product is negative 15. And immediately the numbers three and five, either negative three, positive five or positive three, negative five jump out; and whose sum is negative two. Well if the sum is negative that means that the negative number has to have a larger magnitude. So it's gonna be negative five and three. Negative five and three's product is negative 15; their sum is negative two. And if what I'm doing right now looks a little bit like voodoo to you, I encourage you to review factoring quadratic expressions on Kahn Academy. Do a search on Kahn Academy for factoring quadratic expressions. But we now know how to factor, this is gonna be zero is equal to, we can write this as Y minus five times Y plus three. Well, if you have the product of two things that are equal to zero that means that you could get this to equal zero by making one or both of these equal to zero. So the solutions are gonna be how do you make this one equal zero? Well Y would be equal to five. And how do you make this one equal to zero? Well Y would be equal to negative three. Either one of these would make one of these, Y equals five would make this zero, which would make the entire expression zero, so this is a solution. And Y equals negative three would make this zero, which would make this entire expression zero, so this is also a solution. Now we could go back and try to solve for the Xs or we can immediately look at our choices and say well are either of these choices have Y equals five and Y equals negative three as at least the Y-coordinates of solutions. And these first two don't even have two choices and but this one right over here, this one is actually trying to trick us a little bit because you know we have this number five and negative three, it's like hey, maybe this one but this isn't Y equals five, this is X equals five. And when X equals five, Y is equal to negative three. That's X and that's Y. The first coordinate is X, the second coordinate is Y. So Y equals five and Y equals negative three. This one is Y is equal to five and Y is equal to negative three. So you should feel immediately pretty good about this choice right over here. This one puts the five and the negative three in for the X-coordinate. That says X is equal to five and X is equal to negative three. So if I was under a time pressure, I would just stop there and I would move on to the next question 'cause I feel good about this choice. But if you want to go back and feel good that when Y equals five, X equals 16 and Y equals negative three, X is equal to zero, you can go back and substitute into this first equation. Because we have two times five plus six is equal to X. So when Y is equal to five, you have 10 plus six. You have X is equal to 16, when Y is equal to five. And that's what we saw right over here with when Y is five, X is 16. And when Y is negative three, two times negative three plus six is equal to X. This on the left hand side is gonna be negative six plus six. X is equal to zero. So when Y is negative three, X is equal to zero. So we took a little extra time to make sure that we feel good about this. But once again, under time pressure I wouldn't have necessarily gone through to this extra length.