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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} \\](https://content.bartleby.com/qna-images/question/3a7c8b7c-6421-40e7-b33a-5c89a0da3579/09b599a3-ef65-4f1e-800b-2894194bc48e/jst2g_thumbnail.png)
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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