Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question
### Solving the System of Equations

Solve the following system of equations:

1. \( x - 5y + z = 12 \)
2. \( -y + 2z = 8 \)
3. \( z = 0 \)

Choose the correct solution from the following options:

- \( (-28, 8, 0) \)
- \( (-28, -8, 0) \)
- \( (52, 8, 0) \)

#### Explanation:

1. From equation (3), we know that \( z = 0 \).
2. Substitute \( z = 0 \) into equation (2):
   \[
   -y + 2(0) = 8 \implies -y = 8 \implies y = -8
   \]
3. Substitute \( y = -8 \) and \( z = 0 \) into equation (1):
   \[
   x - 5(-8) + 0 = 12 \implies x + 40 = 12 \implies x = 12 - 40 \implies x = -28
   \]

Thus, the solution to the system of equations is \( (x, y, z) = (-28, -8, 0) \).

This correct option is:
- \( (-28, -8, 0) \)
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Transcribed Image Text:### Solving the System of Equations Solve the following system of equations: 1. \( x - 5y + z = 12 \) 2. \( -y + 2z = 8 \) 3. \( z = 0 \) Choose the correct solution from the following options: - \( (-28, 8, 0) \) - \( (-28, -8, 0) \) - \( (52, 8, 0) \) #### Explanation: 1. From equation (3), we know that \( z = 0 \). 2. Substitute \( z = 0 \) into equation (2): \[ -y + 2(0) = 8 \implies -y = 8 \implies y = -8 \] 3. Substitute \( y = -8 \) and \( z = 0 \) into equation (1): \[ x - 5(-8) + 0 = 12 \implies x + 40 = 12 \implies x = 12 - 40 \implies x = -28 \] Thus, the solution to the system of equations is \( (x, y, z) = (-28, -8, 0) \). This correct option is: - \( (-28, -8, 0) \)
### Coordinate Selection Exercise

Please select the correct coordinate point from the options below:

- ☐ (-28, 8, 0)
- ☐ (-28, -8, 0)
- ☐ (52, 8, 0)
- ☐ (52, -8, 0)
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Transcribed Image Text:### Coordinate Selection Exercise Please select the correct coordinate point from the options below: - ☐ (-28, 8, 0) - ☐ (-28, -8, 0) - ☐ (52, 8, 0) - ☐ (52, -8, 0)
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