Solve each system in Exercises 5–18.

Question

Solve each system in Exercises 5–18. $\begin{cases}x + y = -4 \y - z = 1 \2x + y + 3z = -21\end{cases}$

Accepted Answer

Hello. Today we're gonna be solving the system of equations involving three variables. Now, before we start this I always like to number my equations. Let this be equation one equation to an equation three. Now, what we want to go ahead and do is pick a combination of equations to add together in a way that's going to cancel out one of the variables. That way we can come up with another equation involving two variables. That may sound confusing at first but will make more sense once we jump into this. So let's go ahead and add equation three with equation one, equation three is negative X plus y minus Z is equal to negative 10 and equation one is three, X plus two, Y is equal to negative 12. Now you may be thinking to yourself, well wait a minute, how can we add these two equations together If the first equation doesn't have a Z variable in it? Well technically we could add a Z variable since we're not given a Z. Originally, that means that the Z variable here needs to have a coefficient of zero. Now we can go ahead and add the equations together. Now I want to go ahead and do this in a way that's going to cancel out one of the variables and I want to go ahead and cancel out why in order to do that, I can multiply equation three by negative two and that's going to give me two, X minus two, Y plus two, Z is equal to positive 20 and we're going to be adding three X plus two, Y plus zero, Z is equal to negative 12. Now once we add these equations together, the Y variable is going to cancel itself out. And what we are left with is three X plus two X. Which is five X plus two Z plus zero Z. Which is going to give me two Z. And that's going to equal to 20 minus 12 which is going to give me eight. Now, what I've done here is essentially create another equation involving two variables X and Z. I'm gonna go ahead and call this equation four and we're gonna go ahead and repeat this process with a different combination of equations. Let's go ahead and add equation three with the equation too. Again, equation three is negative X plus y minus Z is equal to negative 10. And we're going to add equation to to this Which is going to be three Y minus two, Z is equal to two. Now, just like before we're not given one of the variables in this case it's X. So we're going to add eggs by giving it a coefficient of zero. Now we can go ahead and combine the equations together. Now in the previous combination I went ahead and cancel out the y variable. So I'm gonna go ahead and cancel out the y variable here as well. And in order to do that I can multiply equation three by negative three and doing this is going to give me three, X minus three, Y plus three. Z is equal to positive 30. And we're going to be adding zero X plus three, Y minus two. Z is equal to two. Now once we add these equations together, why is going to cancel itself out? And what we are left with is three X plus zero X. Which is three X plus three, Z minus two. Z. Which is just going to give me one, Z is equal to 30 plus two which is going to give me 32. Now I'm gonna go ahead and label this as equation five. So now what we can go ahead and do is add equations four and five together and cancel out one of the variables. That way we can be able to solve for X and Z. So again we're going to add equation four which is five X plus two, Z is equal to eight. And we're going to add equation five to this which is three, X plus Z Is equal to 32. Now I want to go ahead and cancel out one of the variables and I'm gonna go ahead and cancel out Z. In order to do this, I can multiply equation five by negative two. This is going to give me five X plus two, Z is equal to eight and we're going to be adding negative six, X minus two. Z is equal to negative 64. Now, once we add these together, the Z variable is going to cancel itself out. And what we are left with is negative six X plus five X. Which is negative X is equal to eight minus 64 which is going to give me negative 56. And in order to solve for X, I'm gonna divide both sides of this equation by one. That's going to give me X is equal to 56. So now I have one of the values for X. I can go ahead and take this value for X and plug it into any equation for four and five. That way I can go ahead and solve for Z. Here, I'm gonna go ahead and take X and plug it in to equation four. That's going to give me five times plus two times Z is equal to + times 56 is going to give me And I'm going to subtract 282 both sides of the equation. This is going to give me to Z Is equal to -272. And in order to solve for Z, I'm going to divide both sides of the equation by two. This is going to give me Z is equal to negative 136. Now that we have the values for X and Z. We can plug this into any one of the three original equations. That way we can go ahead and sulfur Y and I'm gonna go ahead and take our X. Value and plug it into equation one. Now recall that equation one is 3 X plus two, Y. Is equal to negative 12 X. Was 56. So we have three times 56 plus two Y. Is equal to negative 12 56 times three is going to give me 168 And I'm going to subtract 168 to both sides of the equation. That's going to give me two Y. Is equal to negative 168 minus 12. Which is going to give me negative 180. And in order to suffer why I'm going to divide both sides of this equation by two And this is going to give me why is equal to negative 90. Now, finally, what we want to go ahead and do is put our values of X, Y and Z. In the form of an order triplet. Again, the order triplet is of the form X comma Y, comma Z. And plugging in our known values is going to give us 56, comma -90, comma -136. And this is going to be the solution to the ordered system of equations. And that means that the answer to this problem is going to be C. So I hope this video helps you in understanding how to solve a system of equations involving three variables and I'll go ahead and see you all in the next video.

Solve each system in Exercises 5–18. $\begin{cases}\)x + y = -4 \\(\y\) - z = 1 \\2x + y + 3z = -21\(\end{cases}$

Key Concepts

Systems of Linear Equations

Substitution and Elimination Methods

Three-Variable Systems

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