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^2−4y^2=−7, 3x^2+y^2=31

Accepted Answer

Hey everyone welcome back in this problem. We are asked to solve the system of equations given below using the addition method. Okay, in the system of equations were given is negative X squared plus three, Y squared equals 24 3 X squared plus four. Y squared equals 136. Okay, so we have a system of two equations and these are nonlinear equations. So first up we're just gonna go ahead and label these equations so that we can refer to them. So equation one is going to be negative X squared plus three, Y squared equals 24. An equation two is going to be three, X squared plus four, Y squared equals 136. Okay now we're asked to use the addition method with the addition method. What we want to do is we want to add these equations together and eliminate one of the variables. So we can choose variable X. Or the variable Y. In order to eliminate them by adding we need the coefficient in front of them to be the same in each equation, just with a different sign, the opposite sign. Okay, so in this case let's choose to eliminate X. The X squared term. Okay. And we can have do that by multiplying equation one by three. Okay, so we multiply equation one by three and add it to equation to Okay, we're going to be able to eliminate sorry that X squared term. Let's go ahead and do that. So equation one times three is gonna give us negative three X squared plus nine. Where I squared equals 72. Okay and then equation two is three X squared plus four where I squared equals adding these together we get zero negative three X squared plus three X squared zero nine Y squared plus four Y squared is gonna give us 13 Y squared In 72 plus 136 gives us 208. Okay? So dividing by 13 we get y squared is equal to 16. And if I just move so we have a little teeny bit more room to write. Okay we're gonna have Y. Is equal to plus or minus four. Okay? And don't forget when you're taking square roots that you get the positive and negative route. Okay So we have Y. Is equal to plus or minus four. So we found two Y values. Now we need to find the corresponding X values. Okay remember for a solution we need the X and the Y value. So let's go ahead and substitute one of our y values into our equation. Okay and we can choose either equation. This point is going to satisfy both of them. Um so either equation will work. So let's choose to substitute why equals four into equation one. Now if we look at our equations minus X squared plus three Y squared. Okay so we substitute for squared here But we noticed this that because we have this y squared whether we substitute the positive or negative, y value, we're going to get the same X values because when we square it it's going to be 16. Okay, so we can actually do this in one step and save ourselves a little bit of writing and a little bit of redoing the same steps. Okay, So we can look at the plus or minus four at the same time. So we have minus X squared plus three and again this is time 16. Whether it's plus four or minus four, you square you get 16 Okay. 24 on the right hand side, so we get minus X squared plus 48 is equal to 24. So we have x squared as we go to 24. Taking the square root X is equal to again plus or minus the square root of 24. Okay, now when we're looking at the square of 24 we can simplify this a little bit. So we can take we know that 24 can be written as four times six. Okay? And we know that the square root of forest to when we have things multiplied under the root, we can separate them in case we get the square root of two or sorry, the square root of four times the square root of six. Which is going to give us to route six. Okay, So these are X values. Okay. And so we found two X values the plus and the negative. So we have two X values two Y values. That's gonna give us four points. So our solution set it's gonna be four points. It's going to be negative to route six case with a negative X. Value with the negative Y. Value negative four. We have negative to Route six with the positive y. Value four. Then we have the positive X. Value to six with the negative Y. Value and finally the positive X. Value with the positive Y. Value. Okay so there's our solution set containing those four points that we found that's going to correspond with answer B. Okay and that's it for this one. Thanks everyone for watching. See you in the next video.

In Exercises 19–28, solve each system by the addition method. x^2−4y^2=−7, 3x^2+y^2=31

Key Concepts

Systems of Equations

Addition Method (Elimination Method)

Quadratic Equations

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