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

Question

In Exercises 19–28, solve each system by the addition method. $\begin{cases}3x^2 + 4y^2 - 16 = 0 \2x^2 - 3y^2 - 5 = 0\end{cases}$

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 equations were given are here five X squared minus seven Y squared minus 31 equals zero and four X squared plus nine, Y squared minus 127 equals zero. Okay. First thing we're gonna do is go ahead and label these equations so we can refer to them. So equation one is five, X squared minus seven, Y squared minus 31 equals zero. An equation two is four, X squared plus nine, Y squared minus 127 equals zero. Okay. Alright. So we're asked to use the addition method. And what we want to do with the addition method is we want one of these variables to have the same coefficient in each equation with a different sign. Okay. What that does is that allows us to add the equations together and cancel out one of those variables. So you can choose either X or y. In this case. Um And we're gonna go ahead and choose X. Now in order to get these to be the same coefficient different sign. We're gonna need to multiply equation one by four. An equation to by negative five. Okay so we're gonna take equation one, we're gonna multiply it by four And then we're gonna add that's our addition method, equation two times negative five. Huh? Alright so equation one times 4, 20 X squared minus Y squared minus 124 is equal to zero. Okay equation two times negative five Negative X squared -45 by squared Plus 635 is equal to zero? And again we want to add these up. We're using our addition method. Okay? And you'll notice that these equations are nonlinear. We have these squared terms. We're using the same method as in the linear case. Okay, there's just a few extra things we need to be cautious of. Alright, so adding together 20 X squared plus negative 20 X squared. That gives us zero negative 28 Y squared plus negative Y squared gives us minus 73 Y squared. Okay negative 124 plus 635 gives us 511. And on the right hand side, zero plus zero is zero. Alright, so just rearranging we get 73 Y squared equals 511 dividing by 73. Let me move down so we have a bit more room to right here dividing by 73. We get y squared is equal to seven so we get y is equal to plus or minus the square root of seven. And don't forget that when you're taking the square root that you get to roots the positive and the negative route. Okay so this means we have two Y values. Okay we're looking for a solution set. So we need x values, we need to find corresponding X values for each of those Y values. Okay so we need to consider both Y values. So let's start with the positive route. Okay, so we're gonna substitute the positive route. Why was route seven into Equation 1? Okay, we could substitute it into either equation one or two were satisfying both equations. So either one is fine, we've chosen one in this case. This is going to give us five X squared minus seven times Y squared. Which is the square root of seven Squared - equals zero. All right. And we're gonna stop here and pause for a minute. Okay? What you'll realize is that because we're squaring this y value with the root, it doesn't matter whether we plug in the positive root or the negative route, we're gonna get the same values. Okay, so let's go ahead and make this easy on ourselves. So we don't have to do this calculation twice. Let's substitute plus or minus that route. Okay, Plus or minus that route When we square it it's gonna be seven positive seven. So we're gonna get the same answers for X. Okay, so we get five X squared minus seven. And again whether it's a positive or negative route, you just get positive seven -31 equals zero. Okay, so five X squared minus 49 minus equals zero. Moving the numbers to the right hand side, we get five X squared is equal to 80 dividing by five X squared is equal to 16. This gives us x values and again plus or minus four. Okay. Alright so now we have two y values and each of them have two X values. Okay? So for each y value we get two X values. That's gonna give us four solutions. Okay? So if we write our solutions out like this we have starting with the X values. We know we can have negative four. Okay and when we have a value of negative four we can have a Y. Value of negative whoops. Not negative square root of y. Negative square root of seven. Okay. We can also have the same X. Value of negative four but a positive square root of seven for the Y value. We also can have positive for for X value. Negative route seven for a Y. Value. And finally positive four. And positive route seven. Okay so that gives us four solutions to this system of equations. That's going to correspond with answer D. That's it for this one. Thanks everyone for watching.

In Exercises 19–28, solve each system by the addition method. $\begin{cases}\)3x^2 + 4y^2 - 16 = 0 \\2x^2 - 3y^2 - 5 = 0\(\end{cases}$

Key Concepts

Systems of Equations

Addition Method (Elimination Method)

Quadratic Equations in Two Variables

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