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Solve the system of equations. $\newline$ $y = 3x^2 - 6x - 33$ $\newline$ $y = -6x + 15$ $\newline$ Write the coordinates in exact form. Simplify all fractions and radicals. $\newline$ (,) $\newline$ (,)

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Solve a system of linear and quadratic equations

Full solution

Q. Solve the system of equations.

\newline

y = 3x^2 - 6x - 33

\newline

y = -6x + 15

\newline

Write the coordinates in exact form. Simplify all fractions and radicals.

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: We have the system of equations: $\newline$ $y = 3x^2 - 6x - 33$ $\newline$ $y = -6x + 15$ $\newline$ Set the two equations equal to each other to find the $x$ -coordinates of the intersection points. $\newline$ $3x^2 - 6x - 33 = -6x + 15$
Simplify Equation: Simplify the equation by moving all terms to one side. $\newline$ $3x^2 - 6x - 33 + 6x - 15 = 0$ $\newline$ $3x^2 - 48 = 0$
Divide and Simplify: Divide the equation by $3$ to simplify further. $\newline$ $x^2 - 16 = 0$
Factor Quadratic Equation: Factor the quadratic equation. $x^2 - 16 = (x - 4)(x + 4)$
Solve for x: Solve for x by setting each factor equal to zero. $\newline$ $(x - 4) = 0$ or $(x + 4) = 0$ $\newline$ $x = 4$ or $x = -4$
Find $y$ for $x=4$ : Find the corresponding $y$ -values by substituting $x$ back into one of the original equations. Let's use $y = -6x + 15$ . For $x = 4$ : $y = -6(4) + 15$ $y = -24 + 15$ $y = -9$
Find $y$ for $x=-4$ : Find the $y$ -value for $x = -4$ :
$y = -6(-4) + 15$
$y = 24 + 15$
$y = 39$
Write Coordinates: Write the coordinates in exact form. $\newline$ First Coordinate: $(4, -9)$ $\newline$ Second Coordinate: $(-4, 39)$

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