Solve each system in Exercises 5–18. 3(2x+y)+5z=−1, 2(x−3y+4z)=−9...

Question

Solve each system in Exercises 5–18. 3(2x+y)+5z=−1, 2(x−3y+4z)=−9, 4(1+x)=−3(z−3y)

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Hello today we're going to be solving the system of equations involving three variables. But before we start that process we need to go ahead and simplify the given equations. So before we start this we're going to go ahead and label each equation with its own number. And let's go ahead and start by simplifying equation one. So the current equation one is two times the quantity X plus two Y plus three Z is equal to five. In order to simplify this, I'm going to distribute two into the quantity X plus two Y. And that's going to give me two X plus four, Y plus three. Z is equal to five. And this is going to be our new equation one and we're going to repeat this process for equation to an equation three, equation two is three times the quantity two x minus y plus two Z is equal to nine. And in order to simplify this I'm going to distribute three into this entire quantity. Doing so is going to give me six x minus three Y plus six Z is equal to nine. And this is going to be our new equation to. Finally we're going to simplify equation three, equation three is seven times the quantity one plus X is equal to negative six times the quantity two Z minus y. And we're going to need to distribute seven into the quantity one plus X and negative six into the quantity two Z minus Y. Doing so is going to give me seven plus seven X is equal to negative 12 Z plus six Y. Now we want the three variables to equal to a constant. And the only thing we have to do is to go ahead and subtract seven X to both sides of the equation and reorder the equation in terms of x, Y and Z. And doing this is going to give me negative seven X plus six Y minus 12. Z is equal to seven. And this is going to be our new equation three. So let's just go ahead and rewrite our system of equations. The first equation is two, X plus four, Y plus three, Z is equal to five. Our second equation is six x minus three, Y plus six, Z is equal to nine. And our last equation is going to be negative seven X plus six, Y minus 12, Z Is equal to seven. Now, what we're going to do is solve our new system of equations for X, Y and Z. So what we want to go ahead and do is pick a combination of two equations and add them together in a way that cancels out one of the variables that way we can make another equation of two variables. That may sound confusing at first but will make more sense once we jump right into this. So let's go ahead and add equation one and equation two together. So equation one is two, X plus four, Y plus three, Z is equal to five and equation two is six, X minus three, Y plus six. Z is equal to nine. Now I want to go ahead and add these in order to cancel out one of the variables and I want to go ahead and cancel out Z. In order to cancel out Z. I can multiply equation one by negative two and that's going to give me negative four X minus eight, Y minus six. Z is equal to negative 10. And we're going to be adding six X minus three, Y plus six, Z is equal to nine. Once we add these equations together, Z is going to cancel itself out. And what we are left with is six X minus four X. Which is going to give us two X negative eight, Y minus three, Y. Which is going to give me negative 11, Y is equal to negative 10 plus nine which is negative one. What I have here is an equation of two variables involving X and y. So I'm gonna go ahead and call this equation for for now now I want to repeat this process but with a different combination of equations, let's go ahead and add equation one and equation three together again, equation one is two, X plus four, Y plus three, Z is equal to five and equation three is negative seven, X plus six, Y minus 12. Z is equal to seven. Now, since I cancel out Z in the first set of equations, I want to go ahead and cancel out Z in this set of equations as well. And in order to do that, I can go ahead and multiply equation one by four. Doing so is going to give me eight X plus 16, Y plus 12 Z is equal to 20. And we're going to add negative seven X plus six, Y minus 12. Z is equal to seven. Adding these equations together, is going to cancel out the Z term. And what we are left with is eight X minus seven X. Which is just X plus 16, Y plus six Y. Which is going to give us 22 Y. And that is going to equal to 20 plus seven which is 27. And we're gonna go ahead and call this equation five. Now that we have two equations involving two variables, we can go ahead and add these equations together and cancel out one of the variables in order to solve for X and Y. So let's go ahead and add equation four, which again is two, X minus 11, Y is equal to negative one. And we're going to add equation five to this, which is X plus 22 Y is equal to 27. I'm gonna go ahead and cancel out the Y term. And in order to do that, I'm going to multiply equation four by two, this is going to give me four, X minus 22 Y is equal to negative two. And we're going to add X plus 22 Y is equal to 27. Adding these equations together is going to cancel out the y term. And what we are left with is four X plus X. Which is five, X. Is equal to 27 minus two, which is 25. And in order to solve for X, I'm gonna divide both sides of this equation by five and X is going to equal to five. Now I have my first variable which is X. I can go ahead and take X and plug it into either equation four or five. In order to solve for Y. And I'm gonna go ahead and plug this in to equation five, plugging in X two, equation five is going to give me five plus 22 Y is equal to 27. And in order to solve for Y, I'm gonna first subtract five to both sides of the equation, this is going to give me 22, Y is equal to 22. And what we need to do now is just divide both sides of this equation by 22. And this is going to give me why is equal to one. Now we have a value for X and we have a value for Y. We can go ahead and plug this into one of the three original equations in order to solve for Z. And I'm gonna go ahead and plug this into equation one. Now recall that equation one was two, X plus four, Y plus three, Z is equal to five, plugging in X and Y is going to give me two times two times five plus four times one plus three Z is equal to 52 times five is going to give me 10 and four times one is going to give me four plus three, Z is equal to 5, 10 plus four is 14 And I'm going to subtract 14 to both sides of this equation. That's going to give me three, Z is equal two negative nine. And in order to solve for Z, I'm just going to divide both sides of this equation by three And this is going to give me Z is equal to negative three. Now that we have our values for X, Y and Z. We want to go ahead and put our solution in the form of an order triplet. Now recall that the order triplet is of the form X comma, y, comma Z. And plugging in X, Y and Z is going to give us five, comma one, comma negative three. And this is going to be the solution to the system of the equations. 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 involving three variables and I'll go ahead and see you all in the next video

Solve each system in Exercises 5–18. 3(2x+y)+5z=−1, 2(x−3y+4z)=−9, 4(1+x)=−3(z−3y)

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