Question

$x^2 + y^2 = 13$
$x^2 - y^2 = 5$
check all of the following that are solutions to the system shown.
(-3,-2)
(3, 2)
(3,-2)
(-3, 2)

Explanation:

Step1: Add the two equations

$$x^2 + y^2 + x^2 - y^2 = 13 + 5$$
Simplify to solve for $x^2$:
$$2x^2 = 18$$
$$x^2 = 9$$
$$x = \pm 3$$

Step2: Substitute $x^2=9$ to find $y^2$

Substitute $x^2=9$ into $x^2 + y^2 = 13$:
$$9 + y^2 = 13$$
$$y^2 = 13 - 9 = 4$$
$$y = \pm 2$$

Step3: Verify all candidate pairs

• For $(-3,-2)$: $(-3)^2+(-2)^2=9+4=13$, $(-3)^2-(-2)^2=9-4=5$ ✔️
• For $(3,2)$: $3^2+2^2=9+4=13$, $3^2-2^2=9-4=5$ ✔️
• For $(3,-2)$: $3^2+(-2)^2=9+4=13$, $3^2-(-2)^2=9-4=5$ ✔️
• For $(-3,2)$: $(-3)^2+2^2=9+4=13$, $(-3)^2-2^2=9-4=5$ ✔️

Answer:

(-3,-2), (3, 2), (3,-2), (-3, 2)