Solve the system

Question

Solve the system

Two equations representing a system of linear equations to solve.

Accepted Answer

Hey, everyone. Welcome back in this problem. We're asked to solve the given system of equations and the equations we're given are X minus two Y equals three and Y squared is equal to two X minus nine. OK? So we have two equations, two variables and these are nonlinear equations. OK? We have a Y squared term in the second. So we have a nonlinear equation. Now we're told to solve, we're not told what method to use. So we can choose any method. OK? Let's go ahead and use substitution. OK? Because I think it just looks like it's going to be easy to isolate this X in the first equation. Um So we'll use substitution method. But if you would like to choose a different method, you can go ahead and do that as well. OK. So first thing we're gonna do is just label our equations. So first equation X minus two, Y equals three, an equation two is going to be Y squared is equal to two X minus nine. OK. This just helps us keep track of those equations. We can refer to them later and it makes our work more clear. OK. So again, we're using the substitution method in the substitution method. What we wanna do is take one of the equations, choose one of the variables and isolate it and then substitute that into the other equation. OK. Substituting it into the other equation. Lets us find solutions that solve both equations. OK. So we're solving that system rather than just solving one of the equations. All right. So again, let's start with equation one. OK? And let's do it down here. We'll start with equation one. We have X minus two Y is equal to three. We're gonna isolate for X. OK. It already has a coefficient of one. So we won't need to do any dividing. Um So that's gonna be easy. So we have X is equal to moving the two Y over three plus two Y. OK? We're gonna call this one star. OK? It's still equation one. It still has the exact same information as equation one. It's just been rearranged. So we'll call it one star. And now we want to do the actual substitution. OK. So we want to substitute equation one star. So this equation for X into two and again, this allows us to find solutions that solve both of those equations. So we're solving that system. OK? So we're gonna move down. So we have some more room. OK? So we're so substituting one star into equation two. OK? Now equation two is Y squared is equal to two X. So we're gonna have two times three plus two. Y mine is not so expanding, we get Y squared is equal to six plus four, Y minus nine. OK? And we see here we have this quadratic equation which means we're gonna get two answers for Y OK. We're gonna have two roots for Y possibly. So we have to make sure we check for both of them. OK? So let's move everything to one side. We're gonna get Y squared. OK? We have six minus nine. That's gonna be negative three. When we move it over, it's gonna become positive. OK? So we get Y squared minus four, Y plus three is equal to zero. And we can go ahead and factor this to find those values for Y. OK? We can factor this as Y minus three times Y minus one is equal to zero. And again, if you're ever in a situation where you're on a test or something and you need to factor or you need to find these values of Y and you're having a hard time factoring, you can always use a quadratic equation, OK? Use a quadratic formula to find those Y values if you're having a hard time factoring in a situation like this. OK? And so this gives us Y values one and three. OK? So we have two potential Y values one and three, but that's not a solution. OK. Remember our equations have X and Y. So in order to have a full solution. We need X values and Y values. So we found the Y values. Let's find the corresponding X values. OK? Let's start with the first Y value. OK? So we're gonna start with Y is equal to one. OK? And you can substitute this into either equation one or equation two. Let me go back up just to remind you of what those equations are because we have equation one and equation two, you can substitute into either equation one or two. At this point, we're gonna choose equation one. Just because the X is already on its own, it'll be a little bit simpler, but either one will work. No problem. So we're gonna substitute into equation one and we get X minus two Y so minus two times one is equal to three. OK? And so X is equal to five. OK? So when Y is one, X is five, so we get the 0.51 highlight that. So we don't lose it. And now we're gonna do the second Y value. OK. The second Y value we found was three. So what happens when we have Y equals three? Again, we're gonna substitute it into equation one, but you could choose equation two. If you'd want, we get X minus two times Y. So minus two times three is equal to three and this gives us X is equal to nine. Yeah. So we get the 0.93. OK? When we have the Y value of three, we have an X value of nine. And so we found two points that satisfy this system of equations. So if we wanted to write this solution, we can write it in set notation with the two points, we found 51 and 93. This is gonna be our final solution. And that is going to correspond with answer C the solution set is the set containing the 0.51 and the 0.93. That's it for this one. I hope this video helped see you in the next one.

Solve the system

Key Concepts

Systems of Linear Equations

Quadratic Equations

Substitution Method

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