To solve the system of equations, we can use the three given equations:
- \( 3x - 2y + z = 18 \) (Equation 1)
- \( x - 2y + z = -2 \) (Equation 2)
- \( 2x - 2y - z = 12 \) (Equation 3)
First, let's eliminate \( z \) by manipulating these equations. We can subtract Equation 2 from Equation 1:
\[ (3x - 2y + z) - (x - 2y + z) = 18 - (-2) \] This simplifies to: \[ 2x = 20 \implies x = 10 \]
Next, let's confirm that this value of \( x \) satisfies the other equations. We can substitute \( x = 10 \) into Equation 2 to find \( y \) and \( z \):
Substituting \( x = 10 \) into Equation 2: \[ 10 - 2y + z = -2 \implies z = -2 - 10 + 2y \implies z = 2y - 12 \quad (Equation 4) \]
Now we substitute \( x = 10 \) into Equation 3 to determine \( y \): \[ 2(10) - 2y - z = 12 \] Substituting for \( z \) from Equation 4: \[ 20 - 2y - (2y - 12) = 12 \] This simplifies to: \[ 20 - 2y - 2y + 12 = 12 \] \[ 32 - 4y = 12 \] \[ 4y = 32 - 12 \implies 4y = 20 \implies y = 5 \]
Now substituting \( y = 5 \) into Equation 4 to find \( z \): \[ z = 2(5) - 12 \implies z = 10 - 12 \implies z = -2 \]
With \( x = 10 \), \( y = 5 \), and \( z = -2 \), we can verify these values in all original equations.
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For Equation 1: \[ 3(10) - 2(5) + (-2) = 30 - 10 - 2 = 18 \quad \text{(True)} \]
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For Equation 2: \[ 10 - 2(5) + (-2) = 10 - 10 - 2 = -2 \quad \text{(True)} \]
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For Equation 3: \[ 2(10) - 2(5) - (-2) = 20 - 10 + 2 = 12 \quad \text{(True)} \]
All three equations are satisfied with \( x = 10 \).
Thus, the value of \( x \) for the solution to the system of equations is:
D. x equals 10