Solve each system in Exercises 5–18.

Question

Solve each system in Exercises 5–18. $\begin{cases}2x - 4y + 3z = 17 \x + 2y - z = 0 \4x - y - z = 6\end{cases}$

Accepted Answer

Hello today we're going to be solving a system of equations with three variables. Now before we start this I'm just gonna go ahead and label each equation with its own number. So I'm gonna consider the first equation as number one. The second one is two and the third one is three. And the idea that we want to do here is we want to pick a combination of any two of the three equations and get rid of one of the variables so that we can come up with another equation of two variables. That sounds a little bit confusing at first but will make more sense once we jump into this. So I'm just gonna go ahead and pick equation one which is X plus four, Y plus Z is equal to 11. And I'm gonna go ahead and add equation to to this which is three X plus two, Y minus Z is equal to one. Now again I want to go ahead and pick a variable to get rid of from these two equations. And I'm gonna go ahead and get rid of the Z variable because if we add these two equations together, the Z variable is going to cancel out. So adding equation one with equation two is gonna give us three X plus X which is four X plus four, Y plus two Y. Which is going to give us six Y. The z's are going to cancel out because Z minus 00 and that's going to be equal to 11 plus one which is 12. And so we get this new equation, the two variables four X plus six, Y. Is equal to 12. And I'm just gonna go ahead and label this as equation four. Now I want to go ahead and repeat this process but with a different combination of variables from our system of equations and I'm gonna go ahead and pick equation too which is three, X plus two, Y minus Z is equal to one and add equation three which is five X minus two, Y minus Z is equal to seven. And I want to go ahead and do the same thing where I want to add these in a way that's going to cancel out one of the variables. Now keep in mind we cancel out dizzy in the first combination of equations. So we're going to want to cancel out Z in the second combination of equations so that we can come up with another equation of two variables that involve X and Y. Now in order to do this, I'm gonna go ahead and multiply equation three by negative one and doing this is going to give me three, X plus two, Y minus Z is equal to one and we're going to be adding negative five, X plus two, Y plus Z is equal to negative seven. Now if you notice once we add a negative Z with positive z Z's are going to cancel out and now we just need to go ahead and add the other terms three X minus five X is going to give us negative two X and two Y plus two. Y. Is going to give us four Y. And that's going to be equal to negative seven plus one which is negative six. Now we've come up with another equation of two variables and I'm gonna go ahead and label that as equation number five. Now what we can go ahead and do is add equation four and equation five together. So that way we can go ahead and sell for X and Y. So equation four is four, X plus six, Y. Is equal to 12. And equation five is negative two X plus four Y. Is equal to negative six. I'm gonna go ahead and cancel out the x terms and I can go ahead and do that by multiplying equation five by two. And doing this is going to give me four, X plus six Y. Is equal to 12. And we're going to be adding negative four, X plus eight Y. Is equal to negative 12. Now, if you notice here, once we add four X. With negative four X. We're going to cancel out the X terms And we're just going to be left with y. Now eight plus six is going to give us 14. So we have 14 Y. Is equal to 12 minus 12 which is zero. And now we can just go ahead and sulfur. Why In order to solve for why we're just going to go ahead and divide both sides of this equation by 14 And 0/14 is just going to be zero. So what we are left with is why is equal to zero and that's going to be one of the three solutions that we are looking for. Now what we can go ahead and do is take this value of Y and plug it into either equation four or equation five. In order to solve for X. I'm gonna go ahead and plug this into equation four. So I'm gonna have equation four which is four X plus six, Y equal to 12. And this is where Y is equal to zero. So plugging in Y is going to give me four X plus six times zero equal to 12, 6 times zero is just zero and four X plus zero is just four X. So what we have is four X is equal to 12. And in order to solve for X, I'm just gonna divide both sides of this equation by four And X is going to equal to three. And that's our second solution. Now we have a substitution for X. And we have a substitution for Y. We can take these values of X and Y and plug it into any one of the original three equations in order to solve for Z. And I'm gonna go ahead and plug those TV values into equation one. So recall that equation one is X plus four, Y plus Z is equal to 11 and this is where X is equal to three and y is equal to zero. So plugging in the substitution is going to be X, which is three plus four times Y, which is four times zero plus Z is equal to 11. Now four times zero is just going to give me zero and three plus zero is just three. So we have three plus Z is equal to 11. And in order to solve rosie, I'm just gonna go ahead and subtract three to both sides of the equation and 11 minus three is going to give me eight. So I have Z is equal to eight and that's going to be the final solution for a system of equations. Now we want to go ahead and write our solution as an ordered pair and an order pair in three variables is going to be of the form X comma Y, comma Z. And plugging in the values of X, y and Z into this ordered pair is going to give us three comma zero, comma eight. And this is going to be the solution to our system of equations. And that means that the answer to this problem is going to be a so I hope this video helps you in understanding how to solve a system of equations with 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 - 4y + 3z = 17 \\(\x\) + 2y - z = 0 \\4x - y - z = 6\(\end{cases}$

Key Concepts

Systems of Linear Equations

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

Three-Variable Systems

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