Solve each system in Exercises 5–18. x+0y+2z=11, x+0y+3z=14, x+2y...

Question

Solve each system in Exercises 5–18. x+0y+2z=11, x+0y+3z=14, x+2y−0z=5

Accepted Answer

Hello Today we're going to be solving the given system of equations of three variables. Now, before I start this I always like to go ahead and label my equations. The first one is going to be one, the second one is two and the last one is three. Now what we want to go ahead and do is add some of the equations together and solve for one of the given variables. Because even though the system of equations involves three different variables, if you notice equation one and equation to only have X and Z. So let's go ahead and add these two equations together to sulfur X and sulfur Z. So we're going to add equation one which is two, X plus Z is equal to six and add equation to to it which is X plus Z is equal to two. Now we want to go ahead and add this in a way where we can cancel out one of the variables. And in order to do that, let's just go ahead and multiply the bottom equation by negative one. That way we can go ahead and cancel out the z variable doing this is going to give us two. X plus Z is equal to six and we're going to add negative X minus Z is equal to negative two. Now adding Z with negative Z is just going to cancel it out, leaving us with just the X term two X minus X is just going to give us X and six minus two is going to give us four and thus X is going to equal four and that is one of the solutions that we are looking for. Now. We can go ahead and take this value of X and plug it into one of the either equation one or equation to to solve for Z. Let's go ahead and plug it into equation two, equation two is X plus Z is equal to two, but X is equal to four. So we have four plus Z is equal to two. In order to solve for Z. We're just gonna subtract four to both sides of the equation. And doing this is going to give us Z is equal to negative two. And that is the other solution that we are looking for. Now finally we have a substitution for Z. So we can go ahead and plug this in to our third equation to sulfur Y. Our third equation is y minus four times E. Or in this case we're going to have y minus four times negative two is equal to 16, negative four times negative two is going to give us positive eight. So we have y plus eight is equal to 16. And in order to sulfur Y. Which is gonna subtract eight to both sides of this equation. And that's going to give us Y. Is equal to positive eight. And that's going to be the final solution that we are looking for. So we have our values for X, Y and Z. But now we want to go ahead and put our solution in the form of an ordered pair. Now an ordered pair of three variables is gonna be of the form X comma y comma Z. So we have 48 and negative two. So our solution is going to be four comma eight comma negative two. And this is going to be the answer To our system of equations of three variables. And that means that the answer to this problem is going to be D. So I hope this video helps you in understanding how to solve a system of equations of three variables. And I'll go ahead and see you all in the next video.

Solve each system in Exercises 5–18. x+0y+2z=11, x+0y+3z=14, x+2y−0z=5

Key Concepts

Systems of Linear Equations

Matrix Representation

Gaussian Elimination

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