Use the elimination method to find all solutions of {x^2 + y^2 = 5, x^2 - y^2 = 3

Question:

Use the elimination method to find all solutions of

{eq}\left\{\begin{array}{r} x^{2} + y^{2} = 5 \\ x^{2} - y^{2} = 3 \end{array}\right. {/eq}

Elimination Method

There are many methods to find the solutions of a pair of equations like the substitution method and elimination method. In the elimination method, we eliminate one variable and find its value and then use the obtained value to find the other variable.

Answer and Explanation:

For the problem, we are asked to find all solution for a given system of equations. Specifically, we are asked to use the elimination method. The system of equations is given by

$$x^{2} + y^{2} = 5 \\ x^{2} - y^{2} = 3 $$

For the elimination, we can add the two equations. This would eliminate the variable {eq}y {/eq}, thus give us the value/s of {eq}x {/eq}.

$$\begin{align*} 2x^2&=8\\ x^2&=4\\ x&=\pm2 \end{align*} $$

Now, using the value of {eq}x {/eq}, we can substitute it into either one of the given equations. We can use the first one. Take note that {eq}x {/eq} here is squared, so whichever of the two values of {eq}x {/eq} we use, it will just result to the same value.

$$\begin{align*} 4+y^2&=5\\ y^2&=1\\ y&=\pm1 \end{align*} $$

We now have the values of {eq}x {/eq} and {eq}y {/eq} which are {eq}x = \pm 2 {/eq} and {eq}y = \pm 1 {/eq}. We can use all of these values to make combinations for points, since all terms of {eq}x {/eq} and {eq}y {/eq} in the system are squared.

Therefore, the all the possible solutions for the system are {eq}(-2, -1), (-2, 1), (2, -1), {/eq} and {eq}(2, 1) {/eq}.


Learn more about this topic:

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Elimination Method in Algebra: Definition & Examples

from High School Algebra II: Help and Review

Chapter 7 / Lesson 9
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