Question

algebra 2 - test
name: nano tapia
systems of 3 variables
period: 2nd
show all work for credit use a pencil
solve the system of equations using the elimination method.

1. x - 2y + 3z = 12
2. 2x - y - 2z = 5
3. 2x + 2y - z = 4

Explanation:

Step1: Label the equations

Let \(

$$\begin{cases} \text{Equation 1: } x - 2y + 3z = 12 \\ \text{Equation 2: } 2x - y - 2z = 5 \\ \text{Equation 3: } 2x + 2y - z = 4 \end{cases}$$

\)

Step2: Eliminate \( y \) from Equation 1 and Equation 3

Add Equation 1 and Equation 3:
\( (x - 2y + 3z) + (2x + 2y - z) = 12 + 4 \)
Simplify: \( 3x + 2z = 16 \) (let's call this Equation 4)

Step3: Eliminate \( y \) from Equation 2 and Equation 3

Multiply Equation 2 by 2: \( 4x - 2y - 4z = 10 \) (Equation 2a)
Add Equation 2a and Equation 3:
\( (4x - 2y - 4z) + (2x + 2y - z) = 10 + 4 \)
Simplify: \( 6x - 5z = 14 \) (let's call this Equation 5)

Step4: Solve the system of Equation 4 and Equation 5

Equation 4: \( 3x + 2z = 16 \) → Multiply by 2: \( 6x + 4z = 32 \) (Equation 4a)
Subtract Equation 5 from Equation 4a:
\( (6x + 4z) - (6x - 5z) = 32 - 14 \)
Simplify: \( 9z = 18 \) → \( z = 2 \)

Step5: Substitute \( z = 2 \) into Equation 4

\( 3x + 2(2) = 16 \) → \( 3x + 4 = 16 \) → \( 3x = 12 \) → \( x = 4 \)

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

\( 2(4) - y - 2(2) = 5 \) → \( 8 - y - 4 = 5 \) → \( 4 - y = 5 \) → \( -y = 1 \) → \( y = -1 \)

Answer:

The solution is \( x = 4 \), \( y = -1 \), \( z = 2 \) (or as an ordered triple \( (4, -1, 2) \))