15. Solve the following system of equations.

x + y – z = -1
x + y + z = 3
3x – 2y – z = -4:

A. (-1, 3, 3)

B. (0, 1, 2)

C. (1, 0, 2)

D. (3, -1, 2)

Since this is multiple choice, why not just try which of the three sets of numbers satisfy all equations? Obviously, answer A does not satisfy x+y+z=3

Alternatively, use standard methods of algebra two reduce the number of vaiables to one, or use determinants.

I will be happy to critique your work.

To solve the system of equations, we will use the method of elimination.

Step 1: We will add the second equation to the first equation and eliminate the variable y.
(x + y – z) + (x + y + z) = -1 + 3
2x + 2y = 2

Step 2: We will multiply the third equation by 2 and add it to the second equation to eliminate the variable y.
3x – 2y – z + 2(3x – 2y – z) = -4 + 2(-4)
3x – 2y – z + 6x – 4y – 2z = -4 – 8
9x – 6y – 3z = -12

Step 3: We will multiply the first equation by -3 and add it to the third equation to eliminate the variable z.
-3(x + y – z) + (9x – 6y – 3z) = -3(-1) + (-12)
-3x – 3y + 3z + 9x – 6y – 3z = 3 + (-12)
6x – 9y = -9

Step 4: We can simplify the second equation, which is 2x + 2y = 2, by dividing it by 2.
x + y = 1

Now we have a simplified system of equations:

2x + 2y = 2
6x – 9y = -9
x + y = 1

Step 5: We can solve the simplified system of equations using any method, such as substitution or elimination. In this case, we will use substitution.

From the third equation, we get x = 1 - y. We can substitute this value of x into the other two equations.

Substituting x = 1 - y into the first equation:
2(1 - y) + 2y = 2
2 - 2y + 2y = 2
2 = 2

This equation does not give us any new information. It implies that the first equation is dependent on the second equation and does not provide an additional constraint on the variables.

Substituting x = 1 - y into the second equation:
6(1 - y) - 9y = -9
6 - 6y - 9y = -9
-15y = -15
y = 1

Now substituting y = 1 into the third equation:
x + 1 = 1
x = 0

So the solution to the system of equations is x = 0, y = 1.

Checking the solutions in the original third equation:
3(0) – 2(1) – z = -4
0 – 2 – z = -4
-2 – z = -4
z = 2

Therefore, the solution to the system of equations is x = 0, y = 1, z = 2.

Looking at the answer choices, we can see that option B. (0, 1, 2) matches the solution we obtained.

Therefore, the correct answer is B. (0, 1, 2).