Solve by eliminating variables:

Question

System of equations to solve by eliminating variables in college algebra course.

Accepted Answer

Hello. Everyone in this video. We're going to be looking at this practice problem where we want to find the values of X, Y and Z. Using the elimination method for the set of equations X minus two Y is equal to negative 73 X plus five Y plus Z is equal to six and two X plus three Y plus Z is equal to five. So because we're using the elimination method, I'm going to be looking for equations that have similar terms and are easy for me to eliminate. So the second and third equation you can see both have X. Y and Z variables and They only have one Z. variable and then three X and two X. So if I subtracted the third equation from the second I would get rid of my Z. And I would have only one X left. So having only one x left would help me to find the value of X. So I'm going to go ahead and do that. So I'll do equation too minus equation three. So that is three X plus five, Y plus Z is equal to six minus two. X plus three. Y plus Z is equal to five. And if I distribute this negative I'll have negative two x minus three, Y minus Z and minus five. So I'm gonna go ahead and do this subtraction three x -2 x is x five, Y -3, Y is two, Y Z -Z is zero and 6 -5 is one. So I'm left with X plus two Y is equal to one and I'll call that equation four. And now I want to look at equation for with equation one since I haven't used equation one and from here you can see that I can eliminate the Y or the wise by themselves because I have this negative two, I and I have this positive two I here so I'm going to do one plus or equation one plus equation four. So equation one is x minus two. Y is equal to negative seven and equation four is X plus two, Y is equal to positive one. So if I do this addition X plus X is two, X negative two, Y plus two, Y is zero and negative seven plus one is negative six. So I have two. X is equal to negative six. And from here I can solve for X by dividing by two. So that leaves me with X is equal to negative three. And now you can see I have a value for one of my variables X. And I'm going to go ahead and plug this value back into equation four to solve for why? So equation four is X plus two, Y is equal to one. And if I plug in the value of X which is negative three, I have negative three plus two. Y. Is equal to one. I'm going to add three to both sides. So it leaves me with two, Y is equal to four and I'm going to divide by two to isolate Y. And that gives me why is equal to positive two. So now I have the value for a second variable and I can go ahead and plug into another equation to find the value of Z. So I'm going to go ahead and plug in. X. is equal to negative three and Y is equal to two Into Equation # three. And notice how you can do equation number three or equation number two. I just chose to do equation three. So I have equation three years two X plus three Y plus Z is equal to five. If I plug in the values I have two times negative three plus three times two Plus Z is equal to five. If I multiply this I have negative six plus positive six plus Z is equal to five. The six cancel the sixes cancel out because negative six plus six is zero. I'm left with Z is equal to five. And now you can see I have the values for all of my variables. So my solution set for X. Y. And Z would be the value of X. Which we said was -3. The value of why? Which we said or we found to be too. And the value of Z. Which we found to be five. So this would be my final answer. And you can see that this aligns with answer choice C. So hopefully this video helped and see you next time

Solve by eliminating variables:

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