In Exercises 61–66, find the solution set for each system by grap...

Question

In Exercises 61–66, find the solution set for each system by graphing both of the system's equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations.

Accepted Answer

Hello, today we're going to be determining a solution to the system of equations and verifying that solution afterward. So we're given a set of two equations. We're going to label the first equation as one and the second equation as two. Let's go ahead and take a look at the first equation. The first equation is given to us as X squared plus 16, Y squared is equal to 16. Now, normally, when we have X squared and Y squared terms, we want both of these terms to have a coefficient of war of one. But currently, Y squared doesn't have a coefficient of one. It has a coefficient of 16. So let's go ahead and divide both sides of this equation by 16. In order to give Y a coefficient of one, this is going to give us the first equation as X squared over 16 plus Y squared over one is equal to 16 divided by 16, which is one. Now, if we take a look at this equation, this is given to us as an equation of an ellipse, but we need to determine the major axis for this ellipse. Well, notice that the value underneath the X term is greater than the value underneath the Y term. That means that the major axis of this ellipse is going to be the X axis. And that means that the form of the equation of the ellipse is going to be X squared over A squared plus Y squared over B squared is equal to one. And the rough shape of this ellipse is going to have the major vertices along the X axis. And the minor vertices along the Y axis, we can represent the values of the major vertices with the A variable and the values of the minor vertices with the B variable. So let's go ahead and solve for the A and B values. Now, in the given equation, we know that the bigger denominator is going to be 16. So that means that A squared is equal to 16. And that also means that the B squared term is going to equal to the smaller denominator. So B squared is equal to one. If we solve for the A equation, we can take the square root of both sides. And that is going to give us A s equal to plus or minus four. And if we take the square root of both sides of the B equation, we get B is equal to plus or minus one. That means that the equation of the ellipse is gonna look roughly like this where we have the major vertices at four comma zero. Negative four comma zero. And we have the minor vertices at zero, comma one and zero comma negative one. And it's gonna have a very thin shape roughly like this. So that's a rough shape of the ellipse. The first equation. Now we want to determine the shape of the second equation. Now, the second equation is given to us as X squared plus Y squared is equal to 16. This is going to be the equation of a circle. And the equation of a circle is normally given to us as X squared plus Y squared is equal to R squared. So we need to find the value of the radius in order to graph this circle. Well, we know that in this case R squared is equal to 16. And if we take the square root of both sides, in order to solve for R, we get R is equal to four. So that means that the equation of the or the second equation is going to have the shape of a circle with the radius of four. And that means that we're going to have the vertices at four, comma zero, negative four, comma 00, comma four and zero comma negative four. And this is going to be the rough shape of the equation of the circle. So now we want to look at the intercepting points for the system of the equations and looking at the rough graph that we drew, we have two intercepting points for the circle and the ellipse, the two intercepting points are going to be at four comma zero and negative four comma zero. That means that both of these coordinates should be the solutions to the equations. But we need to verify these solutions in order to verify it. We're going to be plugging in both of these points into the original equations. Now recall that the first equation given to us is X squared plus 16, Y squared is equal to 16. We're going to start by plugging in the first coordinate which is four comma zero. In order to verify the solution plugging in this value is going to give us four squared which is 16 plus zero, which is equal to 16. This is going to simplify to 16 is equal to 16, which is a true statement. Then we're gonna verify the second coordinate which is negative four comma zero. This is going to give us negative force squared which is 16 plus zero is equal to 16. This is also going to simplify to 16 is equal to 16, which is a true statement. So that means that the system of solutions verifies the first equation. And now we're gonna verify this system of solutions with the second equation. Now the second equation is given to us as X squared plus Y squared is equal to 16. We're plug, we're going to plug in the first point which is four comma zero. This is going to give us the value of 16 plus zero is equal to 16, which simplifies to 16 is equal to 16, which is a true statement. Then we're gonna plug in the next point which is negative four comma zero. This is going to give us negative four squared which is positive 16 plus zero is equal to 16. This also simplifies to 16 is equal to 16, which is a true statement. So the system of coordinate solutions verifies both of the equations, which means that the solution to the system of equations is going to be four comma zero and negative four comma zero. And we also have the rough shape of the graph. And with that being said, the answer to this problem is going to be a. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

Struggling with College Algebra?

In Exercises 61–66, find the solution set for each system by graphing both of the system's equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations.

Watch next

Ellipses: Standard Form practice set