Solve each nonlinear system of equations. Give all solutions, ...

Question

Solve each nonlinear system of equations. Give all solutions, including those with nonreal complex components. See Examples 1–5.

$x^{2} + 2 y^{2} = 9$

$x^{2} + y^{2} = 25$

Accepted Answer

Hey, everyone in this problem, we're asked to provide all solutions for the given non-linear system of equations including non-real complex components. The system of equations we're given is two equations. The first X squared plus three, Y squared is equal to three. And the second seven X squared plus Y squared is equal to 49. We're given four answer choices A through D, each of them contain four possible points as solutions. And we're gonna come back to those as we work through this problem. Now, the first thing we're gonna do is label the equations we were given. OK. So this first equation X squared plus Y squared equals three is gonna be equation one. The second equation seven X squared plus Y squared equals 49. We're gonna call equation two. Now, we want to find the solutions of this system of equations and there's a couple of different methods to do this. OK. When we say solutions, you can also think of that as intersection points, we could use the substitution method. We could use the elimination method and whichever one you're more comfortable with or whichever one you think fits this problem the best in this case, we're gonna go ahead and use the elimination method. OK? And what we're gonna do is subtract these two equations. Now, when we do this, what we wanna do is eliminate one of the variables. If we were just to subtract equation one and two as they are right now, we wouldn't be eliminating any of those variables. So what we're gonna do is we're gonna take equation one and we're gonna subtract three, multiplied by equation two. OK? If we do that, then both of those equations will have a three Y squared term. And when we subtract them, we are gonna get the zero, we're gonna have no more Y term. We're gonna have an equation with just X that we can solve. So let's go ahead and do that. Our first equation X squared plus three, Y squared is equal to three. We're gonna subtract that from three times the second equation. OK? So three multiplied by seven X squared gives us 21 X squared. Three multiplied by Y squared, gives us three Y squared and three multiplied by 49 gives us 147. So we have 21 X squared plus three Y squared equals 147. Now we're gonna subtract X squared minus 21 X squared. Well, that's gonna give us negative 20 X squared, three Y squared minus three Y squared that gives us zero. OK. Those Y squared terms are now gone. And on the right hand side, this is equal to three minus 147 which gives us negative 144. And again, just like we wanted, we have an equation with just X, we can go ahead and solve. So we're gonna divide both sides by negative 20. We get that X squared is equal to 144 divided by 20. And the two negative signs will divide it. And then we're gonna take the square root X is equal to the square root of the right hand side. Now, when we take the square root of 144 we know that that's gonna be equal to 12. So in the numerator we're gonna have plus or minus 12. Can you remember that when you take the square root, you get the positive and the negative root and in the denominator, we have the square root of 20. Now, we wanna try to simplify this denominator if we can. Well, let's recall that 20 A is equal to four multiplied by five. So we can write this as the square root of four, multiplied by five, the square root of four, we know is two. So we can write our X value as X is equal to plus or minus 12 divided by two multiplied by the square root of five. And now we can simplify a little bit. OK? We have 12 divided by two. That's gonna give us six. And so we have plus or minus six divided by the square root of five. All right. So these are the X values we have for our. So sure remember that the solution we need X and Y values. So now we need to substitute our X values into one of our original equations to figure out what the corresponding Y values would be. Hm So we could sub this into either equation. We're gonna choose to sub it into equation too. So we're gonna substitute X equals plus or minus six divided by the square root of five into equation two. And why I chose two in this case is because that Y squared term only has a coefficient of one. So when we want to solve for Y squared, it just saves us a step of dividing by that constant. OK. But again, either equation one or two will work, you will get the same result because it is going to be a solution to either equation. So equation two, we have seven multiplied by X squared. OK. Well, we know that X is plus or minus six divided by the square root of five. So we have seven multiplied by plus or minus six, divided by the square root of five, all squared plus Y squared is equal to 49. Let's expand and simplify our brackets. OK. In the numerator we have plus or minus six squared on this first term. That's gonna give us 36 then the denominator, we have the square root of five squared. That's gonna give us five. So our equation becomes seven multiplied by 36 divided by five plus Y squared is equal to 49. We're gonna move our seven multiplied by 36 divided by five to the right hand side by subtracting. And as we do that, we're gonna wanna find this common denominator of five. And so what we have is Y squared is equal to, we have 49 but we wanna have that common denominator of five. OK. In order to have a denominator five, we need to multiply the denominator by five. If we do that to the denominator, we need to multiply the numerator by five as well. And when we do that, we get 245 divided by five minus a seven multiplied by 36 which is 252 divided by five. So Y squared is therefore gonna be equal to negative seven divided by five. And we found that common denominator. So now we can subtract in that numerator, we have Y squared is equal to negative seven divided by five. We need to go ahead and take the square root. So Y is gonna be equal to plus or minus. OK? Don't forget that you're getting two roots, the square root of negative seven divided by five. Hm. Now we have a negative inside of our square, we call that the square root of negative one is defined as I. And so we can take out that factor of negative one and we can write this as plus or minus I multiplied by the square root of seven divided by five. OK. And these are corresponding wide. Hm. And the question specified that they wanted non-real values as well. And in this case, we do have non-real values. We were taking the square root of a negative value and we had to introduce I here. All right. So we have two possible X values which lead to two possible Y values that gives us a total of four solutions. So if we were to write them out, OK, we have that negative X value negative six divided by the square root of five with the positive Y value. OK. I multiplied by the square root of seven divided by five. That's our first point. And we have the negative X value negative six divided by the square root of five with the negative Y value negative I multiplied by the square root of seven divided by five. Then we move to our positive X value six divided by the square of five with the positive Y value I multiplied by the square root of seven divided by five. And finally, the positive X value six divided by the square root of five with the negative Y value negative I multiplied by the square root of seven. Divided by five and those are our four possible solutions to this system of equations. Hm All right. So what we did here, we used the elimination method and we found that there were four solutionss. If we go up to our answer choices, we can see that the only correct answer choice we have is option C OK. Option C matches all of those points we found with the correct X and Y values. That's it for this one. Thanks everyone for watching. I hope this video helped.

Solve each nonlinear system of equations. Give all solutions, including those with nonreal complex components. See Examples 1–5.
$x^{2} + 2 y^{2} = 9$
$x^{2} + y^{2} = 25$

Key Concepts

Nonlinear Systems of Equations

Substitution and Elimination Methods

Complex Solutions in Algebra

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