Solve each system in Exercises 25–26. (x+3)/2 − (y−1)/2 + (z+2)/4...

Question

Solve each system in Exercises 25–26. (x+3)/2 − (y−1)/2 + (z+2)/4 = 3/2, (x−5)/2 + (y+1)/3 − z/4 = − 25/6, (x−3)/4 − (y+1)/2 + (z−3)/2= − 5/2

Accepted Answer

Hello. Today we're gonna be solving the system of equations involving three variables. Now notice that each equation in our system of equations are of the form of fractions. We don't want to go ahead and deal with fractions because that's gonna make solving this equations very tedious and time consuming. So let's go ahead and get rid of the fractions by multiplying each equation by its respective lowest common denominator. In equation one we have six and two as the denominators. That means that the lowest common denominator for equation one is going to be six. So multiplying everything in Equation one by six is going to give us x plus one minus three Times y plus 3 0 - is equal to seven. Now in order to simplify this, I'm going to distribute negative three into y plus three and negative one into z minus one. And that's going to give me x plus one minus three, Y minus nine minus z plus one is equal to seven. Now negative nine plus one plus one is going to give me negative seven. So I have x minus three, Y minus z minus seven is equal to seven. And in order to simplify this further, I'm going to go ahead and add seven to both sides of the equation and that's going to give me x minus three y minus Z is equal to 14 and that's gonna be our new equation one. Now let's go ahead and repeat this process for equation too. In equation two we have 30 2060 and 15 as the denominators. So the lowest common denominator is going to be 60. Multiplying everything by 60 is going to give us - times x minus two plus three times y minus two -Z -2 Is equal to two times 4. Now, in order to simplify this, I'm going to distribute to into or negative two into x minus two, I'm going to distribute three into y minus two and I'm going to distribute negative one into z minus two and four times two is going to give me eight. So doing this is going to give me negative two, X plus four plus three, Y minus six minus Z plus two is equal to eight and negative six plus four plus two is going to give me zero. So my new equation two is going to be negative two, X plus three, y minus Z is equal to eight. And finally we need to go ahead and repeat this process for equation three. Now, equation three has the denominators of 24 8, 12 and six. So the lowest common denominator here is going to be 24. So for equation three, multiplying everything by 24 is going to give me x plus three -3 times y plus three minus two Time Z -3 Is equal to four. Now, in order to simplify this further, I'm gonna distribute negative three into y plus three and negative two into z minus three. And this is going to give me x plus three -3, Y -9 minus two. Z plus six is equal to four and three minus nine plus six is going to give me zero. So my new equation three is going to be x minus three y minus two, Z is equal to four. So let's just go ahead and put all of our new equations into a system of equations. So our new system of equations is going to be equation one which is X -3, Y -Z is equal to 14. Our second equation is going to be negative two, X plus three, Y minus Z is equal to eight. And our third equation is going to be x minus three, y minus two, Z is equal to four. And now we can go ahead and solve this system of equations. And in order to do that, I'm gonna go ahead and pick a combination of any two of our given equations and solve them to make another equation of two variables. So let's just go ahead and pick equations one and two. Again, equation one is x minus three, Y minus Z is equal to 14. And equation two is negative two, X plus three, Y minus Z is equal to eight. I'm gonna go ahead and add these equations together to cancel out why? Because negative three, Y plus three, Y is just gonna cancel each other out and that's going to give me negative two, X plus X which is negative X. And negative Z minus Z. Is going to give me negative two. Z and 14 plus eight is going to give me 22. And for now now I have an equation with two variables. I'm gonna call this equation four and now I want to pick another combination of equations to make another equation of two variables. So let's go ahead and pick equation to an equation three, Equation two is negative two, X plus three, Y minus Z is equal to eight and equation three is x minus three, Y minus two, Z is equal to four. Now once again I want to go ahead and cancel out the same variable that I did with the first pair of equations. So we want to go ahead and cancel out y 3, -3. Y is just going to cancel each other out and negative two, X plus X is going to give me negative X and negative two Z minus Z is going to give me minus three. Z. And that is going to equal to eight plus four which is 12. Now we've made an entirely new equation of two variables and I'm gonna go ahead and call this equation five. Now, what we can go ahead and do is add equation four and five together and cancel out one of the variables so that we can solve for either X or Z. So adding equation four with equation five is going to give me negative x minus two, Z is equal to 22 plus negative X minus three, Z is equal to 12. Let's go ahead and start this by canceling out the X variable. And we can do this by multiplying equation five by negative one. This is going to give me negative x minus two, Z is equal to 22 and we're going to be adding X plus three, Z is equal to negative 12. Now negative X plus X is just going to cancel each other out and three Z minus two. Z is going to give me a single Z and this is going to be equal to 22 minus 12, which is 10. So we've already solved for one of our variables which is Z. Z is equal to 10. Now what I can go ahead and do is take this variable Z and plug it into either equation four or five. And that way I can solve for X and I'm gonna go ahead and plug this in into equation four, doing this is going to give me negative X -2 times e, which is 10 is equal to 22. Now negative two times 10 is going to give me negative 20 so I have negative X minus 20 is equal to 22. I want to get X by itself. So let's go ahead and add both sides of the equation by 20 and that's going to give me negative X is equal to 42. And finally I need to divide both sides of the equation by negative one and that's going to give me X is equal to negative that's gonna be the value of X that we're looking for. Now, we have a value of X and a value of Z. We can go ahead and plug this into any one of the equations right here. And that way we can go ahead and solve for y. Well let's go ahead and plug this in to equation one, doing this is going to give me X which is negative 42 minus three, Y minus Z, which is minus 10 and that's going to equal to 14, negative 42 minus 10 is going to give me negative 52. So I have negative 52 minus three, Y is equal to 14. Now, what I'm gonna go ahead and do is add 52 to both sides of the equation and this is going to give me negative three, Y is equal to 66. And finally, I'm gonna divide both sides of this equation by negative three and y is going to equal to negative 22 that's going to be the final variable that we are looking for. Finally, we want to take our values of X, y and z and put it into an order triplet. Now recall that an order triplet is of the form X comma Y comma Z. So putting this into an order triplet is going to give me negative 42 -22 and 10, and this is going to be the solution to the system of equations that was given to us. 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 involving three variables, and I'll go ahead and see you all in the next video.

My Courses

Chemistry

Biology

Math

Physics

Business

Social Sciences

Programming

Product & Marketing

Solve each system in Exercises 25–26. (x+3)/2 − (y−1)/2 + (z+2)/4 = 3/2, (x−5)/2 + (y+1)/3 − z/4 = − 25/6, (x−3)/4 − (y+1)/2 + (z−3)/2= − 5/2