Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
Solve this system of equations
![### Mathematical System of Equations
Consider the following system of equation that characterizes a set of geometric conditions:
\[
\begin{cases}
x - 1 = \lambda x \\
y - 2 = \lambda y \\
z = -\lambda z \\
x^2 + y^2 - z^2 = 0
\end{cases}
\]
#### Explanation:
1. **Equation Set:**
- The first three equations represent a system where `x`, `y`, and `z` are variables, and `λ` (lambda) is a parameter.
- \(x - 1 = \lambda x\): This is a linear equation in terms of \(x\) and \(λ\).
- \(y - 2 = \lambda y\): Similarly, this is a linear equation in terms of \(y\) and \(λ\).
- \(z = -\lambda z\): This is also a linear equation involving \(z\) and \(λ\).
2. **Geometric Condition:**
- \(x^2 + y^2 - z^2 = 0\): This equation represents the geometric condition of the system, where the sum of the squares of \(x\) and \(y\) equals the square of \(z\).
#### Detailed Breakdown:
- **First Equation:**
\[
x - 1 = \lambda x
\]
Rearranging this, we get:
\[
x - \lambda x = 1 \implies x(1 - \lambda) = 1 \implies x = \frac{1}{1 - \lambda} \quad \text{(assuming } \lambda \neq 1\text{)}
\]
- **Second Equation:**
\[
y - 2 = \lambda y
\]
Rearranging this, we get:
\[
y - \lambda y = 2 \implies y(1 - \lambda) = 2 \implies y = \frac{2}{1 - \lambda} \quad \text{(assuming } \lambda \neq 1\text{)}
\]
- **Third Equation:**
\[
z = -\lambda z
\]
Rearranging this, we have:
\[
z + \lambda z =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feee9c038-f063-46e4-bc36-b32ae2ae3e4d%2F4becea46-5865-4714-a0f3-2226c74fccc1%2Feamnxro_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Mathematical System of Equations
Consider the following system of equation that characterizes a set of geometric conditions:
\[
\begin{cases}
x - 1 = \lambda x \\
y - 2 = \lambda y \\
z = -\lambda z \\
x^2 + y^2 - z^2 = 0
\end{cases}
\]
#### Explanation:
1. **Equation Set:**
- The first three equations represent a system where `x`, `y`, and `z` are variables, and `λ` (lambda) is a parameter.
- \(x - 1 = \lambda x\): This is a linear equation in terms of \(x\) and \(λ\).
- \(y - 2 = \lambda y\): Similarly, this is a linear equation in terms of \(y\) and \(λ\).
- \(z = -\lambda z\): This is also a linear equation involving \(z\) and \(λ\).
2. **Geometric Condition:**
- \(x^2 + y^2 - z^2 = 0\): This equation represents the geometric condition of the system, where the sum of the squares of \(x\) and \(y\) equals the square of \(z\).
#### Detailed Breakdown:
- **First Equation:**
\[
x - 1 = \lambda x
\]
Rearranging this, we get:
\[
x - \lambda x = 1 \implies x(1 - \lambda) = 1 \implies x = \frac{1}{1 - \lambda} \quad \text{(assuming } \lambda \neq 1\text{)}
\]
- **Second Equation:**
\[
y - 2 = \lambda y
\]
Rearranging this, we get:
\[
y - \lambda y = 2 \implies y(1 - \lambda) = 2 \implies y = \frac{2}{1 - \lambda} \quad \text{(assuming } \lambda \neq 1\text{)}
\]
- **Third Equation:**
\[
z = -\lambda z
\]
Rearranging this, we have:
\[
z + \lambda z =
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