Question

Solve the following system of equations by the method of reduction: \[ x + y + z = 6, \quad y + 3z = 11, \quad x + z = 2y. \]

Hint:

When solving linear systems, using substitution and elimination makes the process straightforward and manageable.

Answer 1

Step 1: Convert to Standard Form
Rewriting the third equation: \[ x + z = 2y \quad \Rightarrow \quad x - 2y + z = 0. \] Thus, the system of equations becomes: \[ x + y + z = 6, \quad y + 3z = 11, \quad x - 2y + z = 0. \] Step 2: Solve for the Variables
- From the equation \( y + 3z = 11 \), we get: \[ y = 11 - 3z. \] - Substituting this expression for \( y \) into the first equation: \[ x + (11 - 3z) + z = 6. \] \[ x + 11 - 2z = 6 \quad \Rightarrow \quad x = 2z - 5. \] Step 3: Solve for \( z \)
Now, substituting \( x = 2z - 5 \) and \( y = 11 - 3z \) into the third equation \( x - 2y + z = 0 \): \[ (2z - 5) - 2(11 - 3z) + z = 0. \] \[ 2z - 5 - 22 + 6z + z = 0. \] \[ 9z - 27 = 0 \quad \Rightarrow \quad z = 3. \] Step 4: Calculate \( x \) and \( y \)
Substitute \( z = 3 \) into the equations for \( y \) and \( x \): \[ y = 11 - 3(3) = 2, \quad x = 2(3) - 5 = 1. \] Final Answer: \( x = 1, y = 2, z = 3 \).