Solve each system in Exercises 12–13. The is a piecewise functi...

Question

Solve each system in Exercises 12–13.  The  is a piecewise function

Accepted Answer

Hello today we're going to solve a system of equations involving three variables. So in order to approach this, the one thing I like to do to stay organized is go ahead and label all of my equations. Now what we want to do is go ahead and take two combinations of these three equations to make two more equations with respect to two variables. That way we can be able to get a system of equations involving two variables instead of three. So let's go ahead and start that process. What I'm gonna do is go ahead and pick equation one which is X plus y plus Z equal to three. And I'm gonna pick equation too Which is three, X Plus Y -2, Z equal to six. Now what I'm gonna do is go ahead and choose a variable to eliminate from the set of equations. Well I'm gonna go ahead and eliminate Z. So that way I may come up with an equation with respect to X and Y. In order to eliminate Z. I'm gonna go ahead and multiply our first equation by two. Doing so is going to give me two X plus two Y plus two Z equal to six. And just rewriting our second equation. We're going to have three, X plus y minus two, Z equal to six. And I'm gonna go ahead and start adding these equations together While to Z -2. Z is just gonna cancel each other out two X plus three X. Is going to give me five X and two Y plus Y is going to give me three Y. And finally six plus six is going to give me 12. So what I've just done is go ahead and create an equation with respect to variables. And I'm gonna label this equation equation four. Now we're gonna go ahead and repeat this process but with a different combination of variables. So let's go ahead and pick equation one which again is X plus Y plus Z equal to three. And I'm gonna go ahead and pick equation three Which is X -2, Y -Z equal to three. Now, since our fourth equation has X and Y, we want to go ahead and eliminate Z. So that way we we may come up with another equation of X and Y. Well thankfully in this set of equations, we have a positive Z in the first equation and a negative Z in the second equation. So if we were to add these two equations together, Z would cancel itself out, leaving us with X and Y. So let's go ahead and add what we have right now, X plus X is going to give me two, X, y minus two, Y. Is going to give me negative Y. And three plus three is going to give me six. And I'm gonna label this as equation five, not what we're going to do is take our fourth equation and our fifth equation and set it up as a system of equations to solve for either X or either Y. So just gonna go ahead and rewrite it. Our fourth equation is five X plus three Y equal to 12. And our 5th equation Is two X -Y equal to six. And what we want to do again is go ahead and eliminate one of the variables. Well, I'm gonna go ahead and eliminate why in order to do that, I want to go ahead and multiply our fifth equation by three. Doing so is going to give me five, X plus three, Y is equal to 12. And multiplying the second equation by three is gonna give me six X minus three, Y is equal to 18. Now if we add these equations together three y minus three Y is just gonna cancel itself out and five X plus six X. Is going to give me 11 X And 18 plus 12 is going to give me 30 Now in order to solve for X, I'm gonna go ahead and divide both sides by 11 And doing so is going to give me X is equal to 30/11 and that's going to be one of our solutions. Alright, so we just figured out our value for X, what we can do now is go ahead and take this value and plug it into either equation four or five to solve for Y. Well what I'm gonna do is I'm gonna go ahead and take equation five and plug in X. So continuing this work up here. If I plug an X. For five, what I get is two times 30/ -Y. is equal to six. Well 30/11 times two is going to be 60/11 minus Y is equal to six. And since this quantity is being subtracted from Y I'm gonna go ahead and subtract 60/11 from both sides of the equation. Doing so is gonna leave me with negative Y equal to six minus 60/11 which is gonna give me 6/11 and dividing both sides by negative one to cancel out this negative on the left hand side is gonna give me wine equal to negative 6/11. And that's gonna be my second solution. So now we have a value for X. And we have a value for Y. We can go ahead and take both of these values and plug it into any one of our original three equations to solve for Z. Well for me, the easiest one to work with is gonna be equation one. So let's go ahead and do that. So plugging in X. And Y. For equation one is going to give me 30/ -6/11 plus Z is equal to three. Now, 30/11 -6 or 11 is going to give me 20 for over 11 plus Z is equal to three. And since this quantity is being added to Z, I'm gonna go ahead and subtract 20 for over 11 from both sides of the equation. So what I'm left with is 3 -20 for over 11 and that's going to give me 9/11. And that's gonna be our final solution to this system of equations. So just to recap we have X is equal to 30/11, Y is equal to negative 6/11 and Z is equal to 9/11. So that's gonna be the three quick solutions to the system of equations. And the answer to this problem is going to be be so I hope this video helps you in understanding how to solve a system of three equations, and I'll go ahead and see you all in the next video.

Solve each system in Exercises 12–13. The is a piecewise function

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