To find the solution(s) of the system of equations algebraically, we need to solve for the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy both equations simultaneously:
1. [tex]\( y = -x^2 + 6x + 16 \)[/tex]
2. [tex]\( y = -4x + 37 \)[/tex]
Since both equations equal [tex]\(y\)[/tex], we can set them equal to each other to solve for [tex]\(x\)[/tex]:
[tex]\[
-x^2 + 6x + 16 = -4x + 37
\][/tex]
Rearrange the equation so that all terms are on one side, resulting in a standard quadratic equation:
[tex]\[
-x^2 + 6x + 16 + 4x - 37 = 0
\][/tex]
Combine like terms:
[tex]\[
-x^2 + 10x - 21 = 0
\][/tex]
Multiply the entire equation by [tex]\(-1\)[/tex] to make the coefficient of [tex]\(x^2\)[/tex] positive:
[tex]\[
x^2 - 10x + 21 = 0
\][/tex]
To solve this quadratic equation, we can use the quadratic formula, [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -10\)[/tex], and [tex]\(c = 21\)[/tex].
First, calculate the discriminant:
[tex]\[
b^2 - 4ac = (-10)^2 - 4(1)(21) = 100 - 84 = 16
\][/tex]
Since the discriminant is positive, there are two real solutions. Calculate the solutions using the quadratic formula:
[tex]\[
x = \frac{-(-10) \pm \sqrt{16}}{2(1)} = \frac{10 \pm 4}{2}
\][/tex]
This gives us two values for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{10 + 4}{2} = \frac{14}{2} = 7
\][/tex]
[tex]\[
x = \frac{10 - 4}{2} = \frac{6}{2} = 3
\][/tex]
Now, we need to find the corresponding [tex]\(y\)[/tex] values for these [tex]\(x\)[/tex] values using either of the original equations. Let's use [tex]\( y = -4x + 37 \)[/tex]:
For [tex]\( x = 7 \)[/tex]:
[tex]\[
y = -4(7) + 37 = -28 + 37 = 9
\][/tex]
So, one solution is [tex]\( (7, 9) \)[/tex].
For [tex]\( x = 3 \)[/tex]:
[tex]\[
y = -4(3) + 37 = -12 + 37 = 25
\][/tex]
So, another solution is [tex]\( (3, 25) \)[/tex].
Thus, the solution set to the system of equations [tex]\( y = -x^2 + 6x + 16 \)[/tex] and [tex]\( y = -4x + 37 \)[/tex] is:
[tex]\[
(3, 25) \text{ and } (7, 9)
\][/tex]
Therefore, the correct answer is:
[tex]\[
(3, 25) \text{ and } (7, 9)
\][/tex]
