In Exercises 25–35, solve each system by the method of your choic...

Question

In Exercises 25–35, solve each system by the method of your choice.  This  is a piecewise function, refer to textbook problem.

Accepted Answer

Hello today we're going to solve a system of equations using any of our preferred methods. So the equations give it to us is x squared plus y squared equal to nine. And our second equation is x minus y equal to three. So I'm gonna go ahead and just label this equations just to keep ourselves organized. What we can do is go ahead and use the substitution method and in order to do that we can use our second equation and solve for one of the variables to come up with a substitution for our first equation. So let's go ahead and solve for X in our second equation. In order to solve for X. All we need to do is go ahead and add Y to both sides of this equation. And what we end up with is X is equal to Y plus three. So this is going to act as a substitution for X. For our first equation, plugging this into our first equation is going to give us y plus three squared and keep in mind that X is squared in this first equation Plus y squared is equal to nine. Now, what we're going to do is expand this value and simplify the left hand side in order to solve for y. Well, in order to open up this value here, we're going to need to foil because why Plus three squared is the same thing as saying why? Plus three times Y plus three. And we're going to need to foil this in order to open up the parentheses while foiling y plus three squared is going to give us why squared Plus 6? Y plus nine Plus y squared equal to nine. Now, in order to in order to simplify, we're just gonna go ahead and combine like terms this y squared can combine with this white square to give us two Y squared and that gives us two Y squared plus six, Y plus nine equal to nine. Now, in order to simplify this even further, we can go ahead and subtract nine to both sides of this equation. Doing so is going to leave us with two Y squared plus six Y equal to zero. And now if you notice this, y squared has two and this Y has six which are multiples of two. So we can go ahead and divide everything by two and that'll give us y squared plus three Y equal to zero. Now, finally we can go ahead and factor out a Y from the left hand side of this equation. And that leaves us with Y times Y plus three equal to zero. Now, if we apply the zero property we can go ahead and let both of these factors equal to zero. And that lets us get why equal to zero and y plus three equal to zero. So in order to sulfur Y on the right hand side we just need to subtract three to both sides of the equation and that gives us Y is equal to negative three. So if you notice we have two values for y, which means that this system of equations is going to have two solutions. So how do we solve for X one? In order to solve for X? We can go ahead and take our values of y and plug it into our second equation. Doing so when why is equal to zero? What we are left with Is X zero equal to three. And that means that X is going to equal to three and when y is equal to negative three, what we are left with is x minus negative three is equal to three. Now minus negative three is gonna produce a positive value. So what we get is X plus three equal to three. And in order to solve for X, we just need to subtract three from both sides of this equation and we are left with X is equal to zero. So just to put our solutions together, we know that X is equal to three when y is equal to zero and X is equal to zero when y is equal to negative three, that's gonna be the solutions to this system of equations. Which also means that the answer to this problem is going to be d. So I hope this video helps you in understanding how to use the substitution method to solve a system of equations and I'll go ahead and see you all in the next video

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In Exercises 25–35, solve each system by the method of your choice. This is a piecewise function, refer to textbook problem.