Question

8-38. solve the system of equations below.
\\(x^2 + y^2 = 16\\)
\\(y = x^2 - 4\\)
hint:
solve the first equation for \\(x^2\\) (not \\(x\\)), then substitute into the second equation and solve for \\(y\\).

Explanation:

Step1: Isolate $x^2$ from first equation

From $x^2 + y^2 = 16$, rearrange to get:
$x^2 = 16 - y^2$

Step2: Substitute $x^2$ into second equation

Replace $x^2$ in $y = x^2 - 4$:
$y = (16 - y^2) - 4$

Step3: Simplify to quadratic equation

Combine like terms and rearrange:
$y = 12 - y^2$
$y^2 + y - 12 = 0$

Step4: Factor the quadratic equation

Find two factors of -12 that sum to 1:
$(y + 4)(y - 3) = 0$

Step5: Solve for $y$ values

Set each factor equal to 0:
$y + 4 = 0 \implies y = -4$
$y - 3 = 0 \implies y = 3$

Step6: Find $x$ for $y=-4$

Substitute $y=-4$ into $x^2 = 16 - y^2$:
$x^2 = 16 - (-4)^2 = 16 - 16 = 0$
$x = 0$

Step7: Find $x$ for $y=3$

Substitute $y=3$ into $x^2 = 16 - y^2$:
$x^2 = 16 - 3^2 = 16 - 9 = 7$
$x = \sqrt{7}$ or $x = -\sqrt{7}$

Answer:

The solutions to the system are $(0, -4)$, $(\sqrt{7}, 3)$, and $(-\sqrt{7}, 3)$