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

Question

In Exercises 19–28, solve each system by the addition method. x^2+y^2=13,x^2−y^2=5

Accepted Answer

Hey everyone welcome back in this problem. We are asked to solve the system of equations using the addition method. Okay. And the system of equations were given here is x squared plus, sorry, X squared minus Y squared is equal to 16 and x squared plus y squared is equal to 34. Okay, now first thing we're gonna do is just rewrite these equations and label them so that we can refer to them later. So we have equation one, X squared minus y squared equals 16 and equation two X squared plus Y squared equals 34. Okay, Alright, so we're told to use the addition method and with the addition method, what we want to do is we want to add these two equations in a way that allows us to eliminate one of the variables. In this case we have a system of two nonlinear equations. Okay, we have squared terms. We have nonlinear equations. We have to be a little bit careful um that we don't have x squared and x is and that kind of thing. But in this problem we have here we see that if we add these as they are, we are able to eliminate the y squared term. Okay, so in this case we can just go ahead and say add equation one plus the equation too. Okay, we're going to have x squared minus y squared equals 16 and we're gonna add it to x squared plus Y squared equals 34. Okay, X squared plus x squared is gonna give us two X squared minus y squared plus y squared. Well that's gonna give us zero, which is exactly what we were hoping for. Okay and 16 plus 34 is equal to 15. And so by adding these terms together we've eliminated the y squared term. And sometimes this edition method is called the elimination method for that reason. Alright, dividing by two X squared is equal to 50 and taking the square or sorry, divided by two X squared is equal to 25 And taking the square root X is equal to plus or -5. Okay? And don't forget when you take the square root that you need to consider both the positive and negative route. Okay? Alright. So we found our X value. Now we need to find our corresponding Y value. And what we wanna do is we want to take each X value and substitute back into our equation. Okay, We can choose either equation one or equation too because the solution is going to solve both of them. Okay, it has to satisfy both. So let's substitute X equals will choose the positive route first. Okay? So x equals five into let's choose equation to so if we substitute X equals five we get five squared in the five squared outside plus y squared equals 34. Okay And you might notice here when we are taking this X value we're squaring it. So whether we use the positive or the negative X value, we're gonna get the same Y value. So we can save ourselves a step and evaluate both the positive and negative X values in one step. Okay? Now if you didn't notice that right off the bat that's okay. You substitute the X. Value the positive X value first. You get y values, you substitute the negative X. Value and you get you'll get the same Y values. Okay? But you can do it in two steps. Um This just saves us a little bit of work. All right. So we're going to get 25 again whether we take the positive or negative five we get five squared is 25. Ok. Plus y squared equals 34. Subtracting 25 we get y squared is equal to nine and taking the square root Y. Is equal to plus or minus three. Okay so we have two X values and we have two corresponding y values. Okay? So whether we use X equals five or X equals negative five. We get two Y values. So our solution set looks like this, it looks like the point. The set containing the points. -5 -3. 8 -5 positive three Positive 5 -3 and positive five positive three. Okay so just the combination of those X. And Y values that we found and so that's going to correspond with answer A I hope this video helped. Thanks everyone for watching. See you in the next one

In Exercises 19–28, solve each system by the addition method. x^2+y^2=13,x^2−y^2=5

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