Let's solve the given system of equations step by step.
The equations are:
- \( x + 2y - z = -4 \) (Equation 1)
- \( x + 3y + z = -16 \) (Equation 2)
- \( x - 7y + 3z = -6 \) (Equation 3)
We can solve these equations using the method of elimination or substitution. Let's proceed with elimination.
Step 1: Eliminate z from the equations
First, we can add Equations 1 and 2 to eliminate \( z \):
\[ (x + 2y - z) + (x + 3y + z) = -4 - 16 \]
This simplifies to:
\[ 2x + 5y = -20 \quad \text{(Equation 4)} \]
Next, let's manipulate Equations 1 and 3 to eliminate \( z \) as well. We can express \( z \) from Equation 1:
\[ z = x + 2y + 4 \]
Now substitute this expression for \( z \) into Equation 3:
\[ x - 7y + 3(x + 2y + 4) = -6 \]
Expanding this gives:
\[ x - 7y + 3x + 6y + 12 = -6 \]
Combining like terms, we get:
\[ 4x - y + 12 = -6 \]
Now, isolate \( y \):
\[ 4x - y = -6 - 12 \] \[ 4x - y = -18 \] \[ y = 4x + 18 \quad \text{(Equation 5)} \]
Step 2: Substitute Equation 5 into Equation 4
Now substitute \( y \) back into Equation 4:
\[ 2x + 5(4x + 18) = -20 \]
Expanding this gives:
\[ 2x + 20x + 90 = -20 \]
Combine like terms:
\[ 22x + 90 = -20 \] \[ 22x = -20 - 90 \] \[ 22x = -110 \]
Now divide by 22:
\[ x = -5 \]
Step 3: Find y and z
Now that we have \( x \), substitute this value back into Equation 5 to find \( y \):
\[ y = 4(-5) + 18 \] \[ y = -20 + 18 \] \[ y = -2 \]
Next, we find \( z \) using \( x \) in Equation 1:
\[ z = -5 + 2(-2) + 4 \] \[ z = -5 - 4 + 4 \] \[ z = -5 \]
Step 4: Final Solution
We have:
\[ x = -5, \quad y = -2, \quad z = -5 \]
Thus, the solution to the system of equations is:
A. The one solution is \( x = -5, y = -2, z = -5 \).
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