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Question

What is the solution to the following system of equations?

x−y+z=−1
x+y+3z=-3
2x−y+2z=0

Answer 1

To solve this system of equations, we can use the method of substitution or elimination. Let's use the method of substitution.

From the first equation, we can isolate x:
x - y + z = -1
x = y - z - 1

Substitute this value of x into the second equation:
x + y + 3z = -3
(y - z - 1) + y + 3z = -3
2y + 2z - 1 = -3
2y + 2z = -2

Now, let's isolate y in the third equation:
2x - y + 2z = 0
2(y - z - 1) - y + 2z = 0
2y - 2z - 2 - y + 2z = 0
y - 2 = 0
y = 2

Substitute the value of y into the equation 2y + 2z = -2:
2(2) + 2z = -2
4 + 2z = -2
2z = -6
z = -3

Now that we have y = 2 and z = -3, we can substitute these values into the first equation to find x:
x - y + z = -1
x - 2 - 3 = -1
x - 5 = -1
x = 4

Therefore, the solution to the system of equations is x = 4, y = 2, and z = -3.