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

Question

Solve each system in Exercises 5–18. 3x+2y−3z=−2, 2x−5y+2z=−2, 4x−3y+4z=  10

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Hello. Today we're going to be solving a system of equations in three variables. Now, before we begin this process I always like to go ahead and label my equations. So we're gonna let the first one being one, the second one B two and the last one B three. Now the idea here is that we want to go ahead and take a combination of any two of the given equations and make another equation of two variables by adding those equations together. Now that does sound a little confusing but this is going to make more sense once we go ahead and jump into the problem. So I want to go ahead and pick two equations to add to each other. Let's go ahead and add equation one with the equation to. So the first equation is five x minus y minus five, Z is equal to nine and the second equation is x minus two, Y plus Z is equal to negative two. Now I want to go ahead and add these together in a way that cancels out one of the variables. And I want to go ahead and cancel out the z variable. Now, in order to cancel the z variable, I would first need to multiply. Our second equation by five and doing so is going to give me five, x minus y minus five, Z is equal to nine and we're going to add five x minus 10, Y plus five. Z is equal to negative 10. Now if you notice once we add these equations together, the z variable is going to cancel out adding everything else together is going to give us five, X plus five X. Which is 10 X negative y minus 10 Y. Which is going to give us negative 11, Y is equal to negative 10 plus nine which is negative one. And this is a new equation that we've come up with in two variables for now, I'm 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 equations now. So let's go ahead and add equation to with the equation three. Again, equation two is x minus two, Y plus Z is equal to negative two and equation three is two, X plus four, Y plus two. Z is equal to four. Now again we want to go ahead and cancel out one of the variables to come up with an equation of two variables. However, we want to cancel the same variable that we did with the first group. So since we canceled out Z in the first group, we want to cancel out Z in this group. So that way we can come up with this equation in two variables X and Y. Now, in order to cancel out the Z, we can go ahead and multiply equation two by -2. And doing this is going to give us negative two, X plus four, y minus two. Z is equal to positive four. And we're going to add equation three to this which is two X plus four, Y plus two, Z is equal to four. Now, if you notice here, once we add these equations together, we have negative two X plus two X. Which is going to cancel X. And we have negative two Z plus two Z. Which is going to cancel it. Z. So this already gives us an equation that's just gonna leave us with why? Which is what we want. Now we're adding four Y with four, Y is going to give us eight Y and four plus four is going to give us eight and we can go ahead and use this equation to solve for Y. And in order to do that we just divide both sides of this equation by eight. Doing this is going to give us why is equal to one and that's gonna be one of the values that we are looking for. Now we have a substitution for why we can take this value of Y and plug it into this four equation that we came up with earlier. So plugging this in is going to give us 10 x minus 11 times Y which is one Equal to -1 and negative 11 times one is just going to give us negative 11. So we have 10 X minus 11 is equal to negative one. Now we need to go ahead and sulfur X. In order to do this. I'm going to first add 11 to both sides of this equation and that's going to leave me with 10, X is equal to 10. And dividing both sides by 10 is going to give me X is equal to one and that's gonna be another value that we're looking for. Now we have a substitution for X and we have a substitution for Y. So we can take these substitution and plug it into any one of the three original equations and we can go ahead and use that to solve for Z. I'm gonna go ahead and take this equation. I mean these solutions and plug it into equation two. Now, equation two is x minus two, Y plus Z is equal to negative two. So X is going to be one minus two times Y, which is one plus Z, which is equal to negative two two times one is just going to give us too. So we have one minus two plus Z is equal to negative two and one minus two is just going to give us negative one. So we have negative one plus Z is equal to negative two. And finally, in order to solve for Z, we're just going to add one to both sides of the equation and that's going to give us Z is equal to negative one. So now we have the values for X for Y and for Z. And we want to go ahead and put this into an ordered pair of three variables. An ordered pair of three variables is of the form X comma Y comma Z. And plugging in the values that we found for X. Y and Z is going to give us one comma one comma negative one. 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 solving 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. 3x+2y−3z=−2, 2x−5y+2z=−2, 4x−3y+4z= 10

Key Concepts

Systems of Linear Equations

Gaussian Elimination

Matrix Representation

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