Solve each system in Exercises 5–18. 4x−0y+2z=11, x+2y−z=−1, 2x+2...

Question

Solve each system in Exercises 5–18. 4x−0y+2z=11, x+2y−z=−1, 2x+2y−3z=−1

Accepted Answer

Hello today we're going to be solving the system of equations of three variables. Now before we jump into this equation I always like to label my given equations. So I'm gonna let the first one the equation one, the second one B. Equation two and the third one B. Equation three. Now what we want to go ahead and do is add any of the two equations together. So that way we can come up with another equation of two variables. I know that sounds a little confusing but this is going to make more sense once we jump into the problem now notice that our first equation is already in the form of an equation of two variables and it involves X and Z. So let's go ahead and add equations two and three together. But add it in a way where we can go ahead and cancel out the y variable. So equation two is two, X plus two, y minus Z is equal to eight. And our third equation is three, X plus two, y minus two, Z is equal to eight. Now again we want to go ahead and cancel out the y term. So let's go ahead and multiply. Our third equation by negative one. And doing this is going to give us two, X plus two, y minus Z is equal to eight and we're going to be adding through negative three x minus two, Y plus two. Z is equal to negative eight. Now if you notice once we add these equations together, the y term is going to cancel out because two y minus two Y is going to give us zero. So let's just go ahead and add the rest of the terms together, negative three X plus two X is going to give us negative X and two Z minus Z is going to give us positive Z. And that is going to equal to eight minus eight which is zero. And so what we've essentially done is come up with another equation but of two variables X and Z. And we're gonna go ahead and label this as equation four. Now we have equation four of two variables X and Z. And we have the first equation of two variables which is also X and Z. So let's go ahead and add the first equation with our new equation, so the first equation is negative x minus seven, Z is equal to 16 and our new equation is negative X plus Z is equal to zero. Now again we want to go ahead and add these in a way that's going to go ahead and cancel out one of the variables. And in order to do that, let's just go ahead and multiply Our new equation by -1. Because doing this is going to give us negative x minus seven, Z equal to 16 and we're gonna add negative one times negative X, which is positive x -Z is equal to zero. Now, once we add these values together, the exes are going to cancel out and now we have a system of equations of one variable and we can use this to solve for z. Now negative seven Z minus Z is going to give us negative eight Z and this is going to equal to 16. Now in order to solve for z, we can just go ahead and divide both sides of this equation by negative eight and Z is going to equal to negative two and this is one of the values that we are looking for Now. We can go ahead and take this value of Z and plug it into our new equation in order to solve for X. So our new equation was negative X plus Z, which is going to be minus two is equal to zero. Now, in order to solve for X, we're going to add two to both sides of this equation and this is going to give us negative X is equal to two. Now, finally, in order to solve for X, which is gonna divide both sides of this equation by negative one to get rid of the negative sign and this is going to give us X is equal to negative two and that's another value that we are looking for. Now we have a substitution for X and we have a substitution for Z. We can go ahead and plug this in into either equation two or equation three. In order to solve for Y let's go ahead and plug the substitution into equation two. Now, equation two is two, X or two times negative two plus two Y -Z, which is -2. So plus two is equal to eight now, two times negative two is going to give us negative four. So we have negative four plus two, Y plus two is equal to eight, negative four. Plus two is going to give us negative two. So we have negative two plus two, Y is equal to eight. We can go ahead and add two to both sides of this equation to give us two, Y is equal to 10. And once we divide both sides of this equation by two, what we are left with is Y is equal to five and this is the final value that we are looking for. Now we have the values for x, Y and z. But we want to go ahead and put this in the form of an ordered pair, an ordered pair in three variables is going to be of the form X comma, y comma C. And again X is equal to negative two, Y is equal to five and z is equal to negative two. So our answer is going to be negative two comma five comma negative two. And this is going to be the solution to the equation of our the system of equations of three variables. 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 and three variables, and I'll go ahead and see you all in the next video.

Solve each system in Exercises 5–18. 4x−0y+2z=11, x+2y−z=−1, 2x+2y−3z=−1

Key Concepts

Systems of Linear Equations

Gaussian Elimination

Matrix Representation

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