Questions: Use the elimination method to find all solutions of the system: x^2+y^2=5 x^2-y^2=1 The four solutions of the system are: (the one with x<0, y<0 is) x= y= (the one with x<0, y>0 is) x= y= (the one with x>0, y<0 is) x= y= (the one with x>0, y>0 is) x= y=

Solution Steps

To solve the system of equations using the elimination method, we can add and subtract the equations to eliminate one of the variables. This will allow us to solve for the remaining variable and then back-substitute to find the other variable.

1. Add the two equations to eliminate \( y^2 \).
2. Subtract the second equation from the first to eliminate \( x^2 \).
3. Solve the resulting equations for \( x \) and \( y \).
4. Determine the signs of \( x \) and \( y \) to find all four solutions.

We start with the system of equations: \[ \begin{cases} x^2 + y^2 = 5 \\ x^2 - y^2 = 1 \end{cases} \]

By adding the two equations, we eliminate \( y^2 \): \[ (x^2 + y^2) + (x^2 - y^2) = 5 + 1 \implies 2x^2 = 6 \implies x^2 = 3 \] Thus, we find: \[ x = \sqrt{3} \quad \text{or} \quad x = -\sqrt{3} \]

Next, we subtract the second equation from the first to eliminate \( x^2 \): \[ (x^2 + y^2) - (x^2 - y^2) = 5 - 1 \implies 2y^2 = 4 \implies y^2 = 2 \] Thus, we find: \[ y = \sqrt{2} \quad \text{or} \quad y = -\sqrt{2} \]

Now we combine the values of \( x \) and \( y \) to find all possible solutions:

1. \( x = \sqrt{3}, y = \sqrt{2} \)
2. \( x = \sqrt{3}, y = -\sqrt{2} \)
3. \( x = -\sqrt{3}, y = \sqrt{2} \)
4. \( x = -\sqrt{3}, y = -\sqrt{2} \)

Final Answer