Question

1. \

$$\begin{cases} x + 3y - z = 8 \\\\ 2x - y + 2z = -9 \\\\ 3y = 9 \\end{cases}$$

Explanation:

Step1: Solve for \( y \) from the third equation

The third equation is \( 3y = 9 \). Divide both sides by 3:
\( y = \frac{9}{3} = 3 \)

Step2: Substitute \( y = 3 \) into the first equation

The first equation is \( x + 3y - z = 8 \). Substitute \( y = 3 \):
\( x + 3(3) - z = 8 \)
\( x + 9 - z = 8 \)
Simplify to get: \( x - z = 8 - 9 = -1 \) --- Equation (4)

Step3: Substitute \( y = 3 \) into the second equation

The second equation is \( 2x - y + 2z = -9 \). Substitute \( y = 3 \):
\( 2x - 3 + 2z = -9 \)
Simplify: \( 2x + 2z = -9 + 3 = -6 \)
Divide both sides by 2: \( x + z = -3 \) --- Equation (5)

Step4: Solve the system of Equation (4) and Equation (5)

We have:
\(

$$\begin{cases} x - z = -1 \\ x + z = -3 \end{cases}$$

\)
Add the two equations together:
\( (x - z) + (x + z) = -1 + (-3) \)
\( 2x = -4 \)
Divide by 2: \( x = -2 \)

Step5: Substitute \( x = -2 \) into Equation (5) to find \( z \)

Equation (5): \( -2 + z = -3 \)
Add 2 to both sides: \( z = -3 + 2 = -1 \)

Answer:

The solution to the system is \( x = -2 \), \( y = 3 \), \( z = -1 \), or in ordered triple form \( (-2, 3, -1) \)