Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN: 9780134463216
Author: Robert F. Blitzer
Publisher: PEARSON
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use the elimination method to find all solutions 

 

(for this equation i kept getting an answer with roots however the program I am using doesnt give me the option to input a root symbol. )

### Solving Systems of Equations Using the Elimination Method

To find all solutions of the given system of equations using the elimination method, consider the following system:

\[
\begin{cases}
x^2 - y^2 + 3 = 0 \\
2x^2 + y^2 - 4 = 0 
\end{cases}
\]

#### Step-by-Step Solution:
1. **Write down the equations:**
   \[
   \begin{aligned}
   & 1. \quad x^2 - y^2 + 3 = 0 \\
   & 2. \quad 2x^2 + y^2 - 4 = 0
   \end{aligned}
   \]

2. **Eliminate \( y^2 \) by adding the two equations:**
   \[
   \begin{aligned}
   & (x^2 - y^2 + 3) + (2x^2 + y^2 - 4) = 0 \\
   & x^2 + 2x^2 + 3 - 4 = 0 \\
   & 3x^2 - 1 = 0 \\
   & 3x^2 = 1 \\
   & x^2 = \frac{1}{3} \\
   & x = \pm \sqrt{\frac{1}{3}}
   \end{aligned}
   \]

3. **Substitute \( x = \sqrt{\frac{1}{3}} \) and \( x = -\sqrt{\frac{1}{3}} \) back into one of the original equations to solve for \( y \).**

4. **For \( x = \sqrt{\frac{1}{3}} \):**
   \[
   \begin{aligned}
   & (\sqrt{\frac{1}{3}})^2 - y^2 + 3 = 0 \\
   & \frac{1}{3} - y^2 + 3 = 0 \\
   & -y^2 + 3 + \frac{1}{3} = 0 \\
   & -y^2 + \frac{10}{3} = 0 \\
   & -y^2 = -\frac{10}{3} \\
   & y^2 = \frac{10}{3} \\
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Transcribed Image Text:### Solving Systems of Equations Using the Elimination Method To find all solutions of the given system of equations using the elimination method, consider the following system: \[ \begin{cases} x^2 - y^2 + 3 = 0 \\ 2x^2 + y^2 - 4 = 0 \end{cases} \] #### Step-by-Step Solution: 1. **Write down the equations:** \[ \begin{aligned} & 1. \quad x^2 - y^2 + 3 = 0 \\ & 2. \quad 2x^2 + y^2 - 4 = 0 \end{aligned} \] 2. **Eliminate \( y^2 \) by adding the two equations:** \[ \begin{aligned} & (x^2 - y^2 + 3) + (2x^2 + y^2 - 4) = 0 \\ & x^2 + 2x^2 + 3 - 4 = 0 \\ & 3x^2 - 1 = 0 \\ & 3x^2 = 1 \\ & x^2 = \frac{1}{3} \\ & x = \pm \sqrt{\frac{1}{3}} \end{aligned} \] 3. **Substitute \( x = \sqrt{\frac{1}{3}} \) and \( x = -\sqrt{\frac{1}{3}} \) back into one of the original equations to solve for \( y \).** 4. **For \( x = \sqrt{\frac{1}{3}} \):** \[ \begin{aligned} & (\sqrt{\frac{1}{3}})^2 - y^2 + 3 = 0 \\ & \frac{1}{3} - y^2 + 3 = 0 \\ & -y^2 + 3 + \frac{1}{3} = 0 \\ & -y^2 + \frac{10}{3} = 0 \\ & -y^2 = -\frac{10}{3} \\ & y^2 = \frac{10}{3} \\
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