Question

use a system of equations to solve $-x^{2}+3x + 14=-\frac{1}{4}x^{2}+5$. part a drag the solution(s) of the system into the bin labeled system solutions. keyboard help system solutions $(-2,6)$ $(6,-2)$ $(-2,4)$ $(6,-4)$ $(2,16)$ $(2,4)$ $(-6,-40)$ $(-6,-4)$ no solution

Explanation:

Step1: Define the system

We can write the system of equations as:

$$
LATEXBLOCK0
$$

Since both equal \(y\), set them equal: \(-x^2 + 3x + 14 = -\frac{1}{4}x^2 + 5\)

Step2: Solve for \(x\)

Multiply all terms by 4 to eliminate the fraction:

$$ -4x^2 + 12x + 56 = -x^2 + 20 $$

Bring all terms to left side:

$$ -4x^2 + 12x + 56 + x^2 - 20 = 0 $$

Simplify:

$$ -3x^2 + 12x + 36 = 0 $$

Divide by \(-3\):

$$ x^2 - 4x - 12 = 0 $$

Factor:

$$ (x - 6)(x + 2) = 0 $$

So \(x = 6\) or \(x = -2\)

Step3: Find \(y\) for \(x = -2\)

Substitute \(x = -2\) into \(y = -\frac{1}{4}x^2 + 5\):

$$ y = -\frac{1}{4}(-2)^2 + 5 = -\frac{1}{4}(4) + 5 = -1 + 5 = 4 $$

So the point is \((-2, 4)\)

Step4: Find \(y\) for \(x = 6\)

Substitute \(x = 6\) into \(y = -\frac{1}{4}x^2 + 5\):

$$ y = -\frac{1}{4}(6)^2 + 5 = -\frac{1}{4}(36) + 5 = -9 + 5 = -4 $$

So the point is \((6, -4)\)

Answer:

\((-2, 4)\), \((6, -4)\)