Question

Margaret is solving the following system of equations for x, y, and z using the elimination method. What is the value of z?

x+y+z=2

6x−4y+5z=31

5x+2y+2z=13

(1 point)
Responses

−2
negative 2

2
2

1
1

3

Answer 1

To solve the system of equations for \( z \), we start with the three equations given:

1. \( x + y + z = 2 \) (Equation 1)
2. \( 6x - 4y + 5z = 31 \) (Equation 2)
3. \( 5x + 2y + 2z = 13 \) (Equation 3)

We can express \( z \) from Equation 1:

\[
z = 2 - x - y
\]

Next, we can substitute this expression for \( z \) into Equations 2 and 3.

**Substituting into Equation 2**:

\[
6x - 4y + 5(2 - x - y) = 31
\]
\[
6x - 4y + 10 - 5x - 5y = 31
\]
\[
(6x - 5x) + (-4y - 5y) + 10 = 31
\]
\[
x - 9y + 10 = 31
\]
\[
x - 9y = 21 \quad \text{(Equation 4)}
\]

**Substituting into Equation 3**:

\[
5x + 2y + 2(2 - x - y) = 13
\]
\[
5x + 2y + 4 - 2x - 2y = 13
\]
\[
(5x - 2x) + (2y - 2y) + 4 = 13
\]
\[
3x + 4 = 13
\]
\[
3x = 9
\]
\[
x = 3 \quad \text{(Equation 5)}
\]

Now we substitute \( x = 3 \) back into Equation 4 to find \( y \):

\[
3 - 9y = 21
\]
\[
-9y = 21 - 3
\]
\[
-9y = 18
\]
\[
y = -2
\]

Now we know \( x \) and \( y \), so we can substitute these values back into Equation 1 to find \( z \):

\[
z = 2 - x - y
\]
\[
z = 2 - 3 - (-2)
\]
\[
z = 2 - 3 + 2
\]
\[
z = 1
\]

Thus, the value of \( z \) is:

\[
\boxed{1}
\]