Question

solve the following system of equations.
$y=-x^{2}+22x - 99$
$y=x^{2}-18x + 99$
simplify your answer as much as possible.
if there is more than one solution, enter additional solutions with the \or\ button.
if there is no real solution, use the
o solution\ button.
$(x,y)=(square,square)$

Explanation:

Step1: Set equations equal

Since both equal $y$, set them to each other:
$$-x^2 + 22x - 99 = x^2 - 18x + 99$$

Step2: Rearrange into standard form

Bring all terms to one side:
$$-x^2 + 22x - 99 - x^2 + 18x - 99 = 0$$
$$-2x^2 + 40x - 198 = 0$$
Divide by $-2$ to simplify:
$$x^2 - 20x + 99 = 0$$

Step3: Factor the quadratic

Find two numbers that multiply to 99 and add to -20:
$$(x - 9)(x - 11) = 0$$
Solve for $x$:
$x - 9 = 0 \implies x=9$; $x - 11 = 0 \implies x=11$

Step4: Find $y$ for $x=9$

Substitute $x=9$ into $y=x^2 - 18x + 99$:
$$y = 9^2 - 18(9) + 99 = 81 - 162 + 99 = 18$$

Step5: Find $y$ for $x=11$

Substitute $x=11$ into $y=x^2 - 18x + 99$:
$$y = 11^2 - 18(11) + 99 = 121 - 198 + 99 = 22$$

Answer:

$(x,y)=(9, 18)$ or $(x,y)=(11, 22)$