Solve each system in Exercises 5–18.

Question

Solve each system in Exercises 5–18. $\begin{cases}2x + y = 2 \x + y - z = 4 \3x + 2y + z = 0\end{cases}$

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 label my equations. Let the first one B equation one. The next one B equation two and the last one B. Equation three. Now what we want to go ahead and do is pick any combination of the three equations and add them together in a way that that gets rid of one of the variables in order to make another equation of two variables. That sounds confusing at first but will make more sense once we jump right into the problem. So I want to go ahead and add equation two and 3 together. Equation two is two, X plus y minus two. Z is equal to five. And we're going to be adding equation three to this which is negative X minus three, Y plus Z is equal to zero. Now I can go ahead and cancel out two of the variables here. And in order to do that I can go ahead and multiply equation three by two. Doing this is going to give me two, X plus y minus two. Z is equal to five. And multiplying equation three by two is going to give me negative two X minus six. Y plus two. Z is equal to zero. Now once we add these two equations together two x minus two X is going to cancel each other out and negative two Z plus two. Z is going to cancel each other out and what we are left with is negative six Y plus Y. Which is going to give me negative five Y. And that is going to equal to five plus zero which is five. And in order to solve for y, I'm going to divide both sides of this equation by negative five. And that's going to give me why is equal to negative one. Now, since I have this value for Y, I can go ahead and plug this in into equation one. In order to solve for X. Recall that equation one is X plus y is equal to seven substituting and y is going to give me x minus one is equal to seven. And in order to solve for X, I'm going to add one to both sides of the equation and that's going to give me X is equal to eight. Now that I have the values for X and Y. I can plug this into either equation two or equation three. In order to solve for Z. And I'm gonna go ahead and plug this in two, equation three, plugging into equation three is going to give me negative X, which is negative eight minus three times Y which is going to be negative three times negative one plus Z is equal to zero, negative three times negative one is going to give me positive three. So I have negative eight plus three plus Z is equal to zero and negative eight plus three is going to give me five or negative five. So I have negative five plus Z is equal to zero. And in order to solve for z, I'm just going to add five to both sides of the equation and that's going to give me Z is equal to five. Now that we have the values for X, Y and Z. I need to go ahead and put these values into the form of an order triplet, recall that the order triplet is in the form of X comma y, comma Z and substituting an X. Y and Z is going to give me eight comma negative one, comma five. And this is going to be the solution to the system of equations. And that means that 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 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}\)2x + y = 2 \\(\x\) + y - z = 4 \\3x + 2y + z = 0\(\end{cases}$

Key Concepts

Systems of Linear Equations

Methods for Solving Systems (Substitution, Elimination, and Matrix Methods)

Three-Variable Systems

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