In Exercises 19–28, solve each system by the addition method. y^2...

Question

In Exercises 19–28, solve each system by the addition method. y^2−x=4, x^2+y^2=4

Accepted Answer

Hey everyone welcome back in this problem. We are asked to solve the system of equations given below. Using the addition method. The system of equations we are given is x squared minus Y equals five and x squared plus y squared equals seven. Okay so we have a system of two equations and their nonlinear. Alright, the first thing we wanna do is just going to label these equations so we can refer to them. So we have equation one, X squared minus Y equals five. An equation to x squared plus y squared equals seven. Okay, so we're told to use the addition method and in the addition method, what we want to do is we want to add equation one and two. Okay we wanted to cancel out one of the variables. So either variable ax or variable. Why? In order to do that we need the coefficient of one of these terms to be equal in magnitude but opposite in sign. Okay so here we have X squared. Here we have x squared. So if we multiply equation to by negative one then the addition method is going to do exactly what we wanted to. Okay, so we're gonna take equation one, we're gonna add it to equation too But equation two is going to be multiplied by -1. Alright, so we have x squared minus y equals five and equation two times negative one is going to be minus x squared minus y squared equals minus seven. Okay, adding these together our addition method, we have x squared plus negative x squared. That's gonna give us zero. Okay negative y plus negative y squared. So we get negative y minus y squared on the right side, five plus negative seven. We're going to have negative two. And what you'll see is that we've eliminated these X terms. Okay, So we only have um variable Why now? So we can solve for y. Okay. So sometimes the addition method is called the elimination method and for that reason we eliminated one of the variables. Alright, rearranging what we have here. We get y squared plus y -2 is equal to zero. All right. I'm gonna give us some more room to write down here. Alright, we can factor this. So we get Y -1 times. Y plus two is equal to zero. Okay? And this gives us y value of one or y value of negative two. Really? Great. So now we have our y values y values here but we need to find our corresponding X values values alone are not a solution to the system. Okay, well, to find our corresponding Y values. Unlike the linear case where we just have one option, we have two options here. So let's start with if y is equal to one. Ok, so let's substitute y equals one and we can substitute it into either of our equations. Okay so we have equation one equation to we can choose either one. It's got to satisfy both In this case let's choose one. We're gonna have x squared minus Y Which is one. So minus one is equal to five. We get x squared is equal to six and X is equal to plus or minus the square of six. Okay? And don't forget the plus or minus when you take the square root. Okay. Alright so this gives us two points. Now we have the one Y value we found and it corresponds with two different X values. Okay so that gives us two points at this stage. We have minus the square root of six and one. Okay And we have plus the square root of six and one. Okay so so far we found two points. And now what we want to do is we've checked the white with one value. Let's do the same for the wife was too so if y is equal to two and again we can sub into either equation one or two. We're gonna substitute into equation one again just like we did last time. Okay so we're gonna have sub y equals and not to negative two negative two into equation one and we get x squared plus or minus Y. So that in this case it's going to be plus two Is equal to five. X squared is equal to three And X is going to be positive or negative square root of three. Alright so now we have more X values. Okay so we found X or y values. Now we have the x values corresponding with the Y value of minus two. And this gives us two more points. Okay? We have the X value minus the square root of three with our negative two. Now we have square root of three with a Y. Value of negative two. And so if we're writing our solution set, well we have four points that we found. Okay, so we have negative square root of six and one. We have squared of six and 1 negative squared of three negative two and squared of three negative two. Okay. And so this is our solution set. It contains those four points that we found. And so going back up to our options, we are going to have answers. Okay, so we have this solution set here with four points. Thanks everyone for watching See you in the next video.

In Exercises 19–28, solve each system by the addition method. y^2−x=4, x^2+y^2=4

Key Concepts

Systems of Equations

Addition Method (Elimination Method)

Quadratic Equations

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