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Solve the system of equations. $\newline$ $y = 5x^2 + 28x - 15$ $\newline$ $y = 28x + 30$ $\newline$ 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 = 5x^2 + 28x - 15

\newline

y = 28x + 30

\newline

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

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: We have the system of equations: $\newline$ $y = 5x^2 + 28x - 15$ $\newline$ $y = 28x + 30$ $\newline$ Set the two equations equal to each other to find the $x$ -values where they intersect. $\newline$ $5x^2 + 28x - 15 = 28x + 30$
Simplify and Isolate: Subtract $28x + 30$ from both sides to set the equation to zero. $\newline$ $5x^2 + 28x - 15 - 28x - 30 = 0$ $\newline$ Simplify the equation. $\newline$ $5x^2 - 45 = 0$
Solve for x: Add $45$ to both sides to isolate the quadratic term. $\newline$ $5x^2 = 45$ $\newline$ Divide both sides by $5$ to solve for $x^2$ . $\newline$ $x^2 = 9$
Find y-values: Take the square root of both sides to solve for $x$ . $x = \pm\sqrt{9}$ $x = \pm3$ We have two possible $x$ -values where the graphs of the equations intersect.
Intersection Points: Find the corresponding $y$ -values for each $x$ -value by substituting back into one of the original equations. We'll use $y = 28x + 30$ .
First, for $x = 3$ :
$y = 28(3) + 30$
$y = 84 + 30$
$y = 114$
So one intersection point is $(3, 114)$ .
Intersection Points: Find the corresponding $y$ -values for each $x$ -value by substituting back into one of the original equations. We'll use $y = 28x + 30$ . First, for $x = 3$ : $y = 28(3) + 30$ $y = 84 + 30$ $y = 114$ So one intersection point is $(3, 114)$ .Now, for $x = -3$ : $y = 28(-3) + 30$ $x$ $0$ $x$ $1$ So the second intersection point is $x$ $2$ .

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