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Solve the system of equations. $\newline$ $y = 12x^2 + 39x - 45$ $\newline$ $y = 39x + 3$ $\newline$ 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

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Q. Solve the system of equations.

\newline

y = 12x^2 + 39x - 45

\newline

y = 39x + 3

\newline

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

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: We have the following system of equations: $\newline$ $y = 12x^2 + 39x - 45$ $\newline$ $y = 39x + 3$ $\newline$ To find the solution, we will set the two equations equal to each other because they both equal $y$ . $\newline$ $12x^2 + 39x - 45 = 39x + 3$
Simplify by Subtracting: Subtract $39x$ from both sides to start simplifying the equation. $\newline$ $12x^2 + 39x - 45 - 39x = 39x + 3 - 39x$ $\newline$ $12x^2 - 45 = 3$
Isolate Quadratic Term: Add $45$ to both sides to isolate the quadratic term. $\newline$ $12x^2 - 45 + 45 = 3 + 45$ $\newline$ $12x^2 = 48$
Solve for $x^2$ : Divide both sides by $12$ to solve for $x^2$ . $\newline$ $\frac{12x^2}{12} = \frac{48}{12}$ $\newline$ $x^2 = 4$
Solve for x: Take the square root of both sides to solve for x. $\newline$ $\sqrt{x^2} = \sqrt{4}$ $\newline$ $x = 2$ or $x = -2$
Substitute $x$ into $y$ : Now that we have the values for $x$ , we can substitute them back into either of the original equations to find the corresponding $y$ values. Let's use $y = 39x + 3$ . For $x = 2$ : $y = 39(2) + 3$ $y = 78 + 3$ $y = 81$ For $x = -2$ : $y$ $0$ $y$ $1$ $y$ $2$
Final Coordinates: We now have the two sets of coordinates where the two equations intersect. $\newline$ First Coordinate: $(2, 81)$ $\newline$ Second Coordinate: $(-2, -75)$

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