Solve each system in Exercises 25–26. (x+2)/6 − (y+4)/3 + z/2 = 0...

Question

Solve each system in Exercises 25–26. (x+2)/6 − (y+4)/3 + z/2 = 0, (x+1)/2 + (y−1)/2 − z/4 = 9/2, (x−5)/4 + (y+1)/3 + (z−2)/2 = 19/4

Accepted Answer

Hello. Today we're gonna be solving the system of equations of three variables. Now if you notice each of our equations have fractions inside of it, we don't want to work with this involving fractions because the fractions are going to make the process of solving this very difficult. So let's go ahead and get rid of the denominators by multiplying each equation by its respective lowest common denominator in equation one the denominators are four and two. So the lowest common denominator is going to be four. So multiplying everything by four in equation one is going to give me x minus one plus Y plus two - times Z is equal to -1. Now I need to multiply negative two N. Z. Together and that's going to give me negative two. Z. And two minus one is going to give me positive one. So I have X plus y minus two, Z plus one is equal to negative one. And finally I can simplify this by subtracting one to both sides of the equation. And so our new equation one is going to be X plus y minus two, Z is equal to negative two. Now I want to go ahead and repeat this process for equation to an equation three for equation two we have the denominators six and two. So the lowest common denominator here is going to be six and multiplying everything by six is going to give me three times x plus 2 -3, Y minus Z plus six is equal to five. Now in order to simplify this, I need to distribute three into X plus two and negative one into Z plus six and this is going to give me three, X plus six minus three, y minus z minus six is going to equal to five and six minus six is going to give me zero. So for equation to what I'm left with is three x minus three, Y minus Z is equal to five. Now finally we're going to repeat this process for equation three. For equation three we have the denominators 6, 12, 18 and 36. So the lowest common denominator for equation three is going to be and multiplying everything by 36 in equation three is going to give me x plus four minus two times Y plus two plus three times Z is equal to six. Now in order to simplify this, I need to go ahead and distribute negative two into y plus two. And doing this is going to give me x plus four minus two, y minus four plus three, Z is equal to six and four minus four is just going to give me zero. So my new equation three is going to be x minus two, Y plus three, Z is equal to six. Now let's just go ahead and group up everything together. So my new system of equations, equation one is going to be x plus y minus two, Z is equal to negative two, Equation two is going to be three, X minus three, Y minus Z is equal to five. And equation three is going to be x minus two, Y plus three, Z is equal to six. Now we can go ahead and solve for the system of equations and we're going to do this by picking two equations, adding them together to get rid of one variable and then we're gonna come up with another equation with two variables. So let's go ahead and add equations one and three together again, equation one is x plus y minus two. Z is equal to negative two and equation three is x minus two, Y plus three, Z is equal to six. Since I want to make an equation with two variables, we can go ahead and cancel out the X variable by multiplying equation three by negative one. Doing this is going to give me x plus y minus two, Z is equal to negative two. And we're going to be adding negative X plus two Y minus three. Z is equal to negative six. If we add X and negative X is going to cancel out the X variable and that's going to leave us with two, Y plus y which is three Y native three Z minus two, Z. Which is going to give us negative five, Z is equal to negative six minus two which is negative eight. And since this is an equation with two variables, I'm gonna go ahead and call this equation four. Now we want to go ahead and repeat this process but with a different combination of variables I mean of equations. So let's just go ahead and pick equation one and add equation to to it. So again, equation one is X plus y minus two, Z is equal to negative two. And equation two is three, X minus three, Y minus Z is equal to five. Since we cancel out the X variable in the first pair of equations, we wanna go ahead and cancel out the X variable in this pair. And we can do that by multiplying the first equation by negative three. Doing this is going to give me negative three X minus three, Y plus six. Z is equal to positive six. And we're going to be adding three x minus three, Y minus Z is equal to five. So again negative three, X plus three X is just going to cancel each other out, -3 Y -3. Y is going to give me -6, Y and six Z minus Z is going to give me five Z. And finally 6-plus 5 is going to give me 11. Now I have essentially made a second equation with two variables Y and Z. And I'm gonna go ahead and call this equation five. Now what we want to go ahead and do is add equation four and five together and cancel out one of the variables. So that way we can sell for either y or Z. So let's go ahead and add equation four and 5 together. Doing this is going to give me three Y minus five, Z is equal to negative eight. And we're gonna be adding negative six, Y plus five, Z is equal to 11. Now notice that if we add these two equations together, Z is going to cancel out because negative five, Z plus five, Z gives us zero negative six, Y plus three Y is going to give us negative three. Y and 11 minus eight is going to give me three in order to solve for Y. I divide both sides of this equation by negative three and why is going to equal to negative one? Now we have a value for y. We can go ahead and take this value of Y and plug it in for either equation four or equation five. So that way we can solve for Z. Let's go ahead and plug this in to equation four. So plugging in Y into equation four is going to give us three times negative one minus five. Z is equal to negative eight, three times negative one is going to give me negative three -5. Z is equal to -8. In order to solve for Z. I'm going to add three to both sides of the equation and this is going to give me negative five. Z is equal to negative five. And finally to solve for Z. I'm going to divide both sides of this equation by negative five and Z is going to equal to one and that's another variable that we are solving for. Now. We have a value for Y and a value for Z. So we can go ahead and plug this into any one of the three original equations. That way we can solve for X. Let's go ahead and plug in these two values into equation one. So that's going to give us X plus Y, which is minus one minus two times E. Which is going to be negative two times one is equal to negative two, negative two times one is just going to give me negative two and negative two minus one is going to give me negative three. So we have X minus three is equal to negative two. And in order to sulfur X, I'm going to add three to both sides of the equation and X is going to equal to one which is the final variable that we are solving for. Now we just need to go ahead and put this into an order triplet. Now recall that an order triplet is of the form X comma Y, comma Z. So putting these values into an order triplet is going to give us one, comma negative one, comma 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 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 25–26. (x+2)/6 − (y+4)/3 + z/2 = 0, (x+1)/2 + (y−1)/2 − z/4 = 9/2, (x−5)/4 + (y+1)/3 + (z−2)/2 = 19/4

Key Concepts

Systems of Equations

Linear Equations

Fractional Coefficients

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