Question

unit review a: linear system
problems 10-12 find the solution to the system of linear equations. show your work.
10.
\

$$\begin{cases} x + y - 2z = 9 \\\\ 4x - y - 2z = 8 \\\\ 4x + y + 4z = -6 \\end{cases}$$

(written elimination next to it)

(written substitution next to it)
\

$$\begin{cases} x + 6z = 5 \\to x = 5 - 6z \\\\ 3x + y - 4z = -3 \\\\ 2x - y - 2z = -8 \\end{cases}$$

1. (written elimination next to it)

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$$\begin{cases} 5x - y + 7z = 16 \\\\ x + y - 2z = -5 \\\\ 10x - 2y + 14z = 8 \\end{cases}$$

Explanation:

Step1: Solve Problem 10 (Elimination)

First, label the equations:

1. \( x + y - 2z = 9 \)
2. \( 4x - y - 2z = 8 \)
3. \( 4x + y + 4z = -6 \)

Add Equation 1 and Equation 2 to eliminate \( y \):
\( (x + y - 2z) + (4x - y - 2z) = 9 + 8 \)
\( 5x - 4z = 17 \) (Equation 4)

Add Equation 2 and Equation 3 to eliminate \( y \):
\( (4x - y - 2z) + (4x + y + 4z) = 8 + (-6) \)
\( 8x + 2z = 2 \)
Simplify: \( 4x + z = 1 \) (Equation 5)

Solve Equation 5 for \( z \): \( z = 1 - 4x \)

Substitute \( z = 1 - 4x \) into Equation 4:
\( 5x - 4(1 - 4x) = 17 \)
\( 5x - 4 + 16x = 17 \)
\( 21x = 21 \)
\( x = 1 \)

Substitute \( x = 1 \) into Equation 5: \( 4(1) + z = 1 \) → \( z = -3 \)

Substitute \( x = 1 \), \( z = -3 \) into Equation 1: \( 1 + y - 2(-3) = 9 \) → \( 1 + y + 6 = 9 \) → \( y = 2 \)

Step2: Solve the Substitution Problem

Label the equations (assuming the first equation is \( x + 6z = 5 \) → \( x = 5 - 6z \), second: \( 3x + y - 4z = -3 \), third: \( 2x - y - 2z = -8 \))

Add the second and third equations to eliminate \( y \):
\( (3x + y - 4z) + (2x - y - 2z) = -3 + (-8) \)
\( 5x - 6z = -11 \)

Substitute \( x = 5 - 6z \) into \( 5x - 6z = -11 \):
\( 5(5 - 6z) - 6z = -11 \)
\( 25 - 30z - 6z = -11 \)
\( 25 - 36z = -11 \)
\( -36z = -36 \)
\( z = 1 \)

Then \( x = 5 - 6(1) = -1 \)

Substitute \( x = -1 \), \( z = 1 \) into the second equation: \( 3(-1) + y - 4(1) = -3 \)
\( -3 + y - 4 = -3 \)
\( y - 7 = -3 \)
\( y = 4 \)

Step3: Solve Problem 12 (Elimination)

Label the equations:

1. \( 5x - y + 7z = 16 \)
2. \( x + y - 2z = -5 \)
3. \( 10x - 2y + 14z = 8 \)

Add Equation 1 and Equation 2 to eliminate \( y \):
\( (5x - y + 7z) + (x + y - 2z) = 16 + (-5) \)
\( 6x + 5z = 11 \) (Equation 4)

Notice Equation 3 is \( 2(5x - y + 7z) = 2(16) \)? Wait, \( 10x - 2y + 14z = 2(5x - y + 7z) = 8 \), but \( 2(16) = 32 ≠ 8 \), so the system is inconsistent? Wait, check again.

Wait, Equation 1: \( 5x - y + 7z = 16 \), multiply by 2: \( 10x - 2y + 14z = 32 \), but Equation 3 is \( 10x - 2y + 14z = 8 \). Since \( 32 ≠ 8 \), no solution.

Answer:

• Problem 10: \( x = 1 \), \( y = 2 \), \( z = -3 \)
• Substitution Problem: \( x = -1 \), \( y = 4 \), \( z = 1 \)
• Problem 12: No solution (inconsistent system)