Question

problem 7 write a system of linear inequalities to represent the graph.
a.
problems 8-9 find the solution to the system of linear equations. show your work.
8.
\

$$\begin{cases} 4x - y - z = 27 \\\\ x - y + 2z = -3 \\\\ 2x + y - 2z = 21 \\end{cases}$$

9.
\

$$\begin{cases} x + y + z = 8 \\\\ 3x - y + 4z = 25 \\\\ 6x - 2y + 8z = 50 \\end{cases}$$

Explanation:

Step1: Eliminate \( y \) from equations 2 and 3

Equation 2: \( x - y + 2z = -3 \)
Equation 3: \( 2x + y - 2z = 21 \)
Add them: \( (x + 2x) + (-y + y) + (2z - 2z) = -3 + 21 \)
Simplify: \( 3x = 18 \)
Solve: \( x = \frac{18}{3} = 6 \)

Step2: Substitute \( x = 6 \) into equations 2 and 1

Equation 2: \( 6 - y + 2z = -3 \) → \( -y + 2z = -9 \) (Equation 2a)
Equation 1: \( 4(6) - y - z = 27 \) → \( 24 - y - z = 27 \) → \( -y - z = 3 \) (Equation 1a)

Step3: Eliminate \( y \) from Equation 2a and 1a

Equation 2a: \( -y + 2z = -9 \)
Equation 1a: \( -y - z = 3 \)
Subtract 1a from 2a: \( (-y + 2z) - (-y - z) = -9 - 3 \)
Simplify: \( 3z = -12 \) → \( z = -4 \)

Step4: Substitute \( x = 6 \), \( z = -4 \) into Equation 2

\( 6 - y + 2(-4) = -3 \) → \( 6 - y - 8 = -3 \) → \( -y - 2 = -3 \) → \( -y = -1 \) → \( y = 1 \)

Answer:

\( x = 6 \), \( y = 1 \), \( z = -4 \) (or the ordered triple \( (6, 1, -4) \))