Question

Use the elimination method to solve the following system of equations for x, y, and z. What is the value of x?

2x+3y−z=−1

3x−8z=−1

5y+7z=−1

(1 point)
Responses

−3
negative 3

4
4

2
2

5
5
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page 13 of 13

Answer 1

To solve the system of equations using the elimination method, we first write down the equations:

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

Let's manipulate these equations to eliminate variables step by step.

### Step 1: Eliminate \( z \)

From **Equation 2**, we can express \( z \) in terms of \( x \):

\[
3x - 8z = -1
\]
\[
8z = 3x + 1
\]
\[
z = \frac{3x + 1}{8}
\]

Now we substitute this expression for \( z \) into **Equations 1 and 3**.

### Substitute into Equation 1:

Using \( z = \frac{3x + 1}{8} \):

\[
2x + 3y - \left(\frac{3x + 1}{8}\right) = -1
\]
To eliminate the fraction, multiply the entire equation by 8:

\[
8(2x) + 8(3y) - (3x + 1) = -8
\]
\[
16x + 24y - 3x - 1 = -8
\]
\[
(16x - 3x) + 24y = -8 + 1
\]
\[
13x + 24y = -7 \quad \text{(Equation 4)}
\]

### Substitute into Equation 3:

Using \( z = \frac{3x + 1}{8} \):

\[
5y + 7\left(\frac{3x + 1}{8}\right) = -1
\]
Again, multiply the entire equation by 8:

\[
8(5y) + 7(3x + 1) = -8
\]
\[
40y + 21x + 7 = -8
\]
\[
40y + 21x = -8 - 7
\]
\[
40y + 21x = -15 \quad \text{(Equation 5)}
\]

### Now we have a system of two equations in \( x \) and \( y \):

1. \( 13x + 24y = -7 \) (Equation 4)
2. \( 21x + 40y = -15 \) (Equation 5)

### Step 2: Eliminate \( y \)

We can manipulate Equations 4 and 5 to eliminate \( y \). We will multiply Equation 4 by 40 and Equation 5 by 24:

1. \( 40(13x + 24y) = 40(-7) \)
- \( 520x + 960y = -280 \) (Equation 6)

2. \( 24(21x + 40y) = 24(-15) \)
- \( 504x + 960y = -360 \) (Equation 7)

Now, we can eliminate \( y \) by subtracting Equation 6 from Equation 7:

\[
(504x + 960y) - (520x + 960y) = -360 + 280
\]
\[
-16x = -80
\]
\[
x = \frac{-80}{-16} = 5
\]

### Conclusion

The value of \( x \) is \( 5 \).

Thus, the answer is \( 5 \).