Question

solve the given system - or show that no solution exists:
$x + 2y = 1$
$3x + 2y + 4z = 7$
$-2x + y - 2z = -1$

Explanation:

Step1: Label the equations

Let's label the equations for clarity:

1. \( x + 2y = 1 \) (Equation 1)
2. \( 3x + 2y + 4z = 7 \) (Equation 2)
3. \( -2x + y - 2z = -1 \) (Equation 3)

Step2: Eliminate \( y \) from Equation 1 and Equation 2

Subtract Equation 1 from Equation 2:
\( (3x + 2y + 4z) - (x + 2y) = 7 - 1 \)
Simplify:
\( 3x - x + 2y - 2y + 4z = 6 \)
\( 2x + 4z = 6 \)
Divide by 2:
\( x + 2z = 3 \) (Equation 4)

Step3: Eliminate \( y \) from Equation 1 and Equation 3

From Equation 1, express \( x = 1 - 2y \) (Equation 1a)
Substitute \( x = 1 - 2y \) into Equation 3:
\( -2(1 - 2y) + y - 2z = -1 \)
Expand:
\( -2 + 4y + y - 2z = -1 \)
Simplify:
\( 5y - 2z = 1 \) (Equation 5)

Step4: Now, let's work with Equation 4 and Equation 5. First, from Equation 4, express \( x = 3 - 2z \) (Equation 4a)

From Equation 1: \( x = 1 - 2y \), so \( 3 - 2z = 1 - 2y \)
Simplify: \( 2y = 2z - 2 \) => \( y = z - 1 \) (Equation 6)

Step5: Substitute \( y = z - 1 \) into Equation 5

\( 5(z - 1) - 2z = 1 \)
Expand:
\( 5z - 5 - 2z = 1 \)
Simplify:
\( 3z - 5 = 1 \)
\( 3z = 6 \)
\( z = 2 \)

Step6: Substitute \( z = 2 \) into Equation 4

\( x + 2(2) = 3 \)
\( x + 4 = 3 \)
\( x = -1 \)

Step7: Substitute \( x = -1 \) into Equation 1 to find \( y \)

\( -1 + 2y = 1 \)
\( 2y = 2 \)
\( y = 1 \)

Step8: Verify the solution in all equations

• Equation 1: \( -1 + 2(1) = -1 + 2 = 1 \) (Correct)
• Equation 2: \( 3(-1) + 2(1) + 4(2) = -3 + 2 + 8 = 7 \) (Correct)
• Equation 3: \( -2(-1) + 1 - 2(2) = 2 + 1 - 4 = -1 \) (Correct)

Answer:

\( x = -1 \), \( y = 1 \), \( z = 2 \)