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Solve the system of equations. $\newline$ $y = 21x + 7$ $\newline$ $y = 2x^2 + 21x - 43$ $\newline$ 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 = 21x + 7

\newline

y = 2x^2 + 21x - 43

\newline

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

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: We have the system of equations: $\newline$ $y = 21x + 7$ $\newline$ $y = 2x^2 + 21x - 43$ $\newline$ To find the intersection points, we set the two equations equal to each other. $\newline$ $21x + 7 = 2x^2 + 21x - 43$
Subtract and Simplify: Subtract $21x + 7$ from both sides to set the equation to zero. $\newline$ $21x + 7 - 21x - 7 = 2x^2 + 21x - 43 - 21x - 7$ $\newline$ $0 = 2x^2 - 50$
Divide and Factor: Divide the equation by $2$ to simplify. $\newline$ $0 = x^2 - 25$
Solve for $x$ : Factor the quadratic equation. $x^2 - 25 = (x - 5)(x + 5)$
Find y-Values: Solve for $x$ by setting each factor equal to zero. $\newline$ $(x - 5) = 0$ or $(x + 5) = 0$ $\newline$ $x = 5$ or $x = -5$
Write Coordinates: Find the corresponding $y$ -values for each $x$ by substituting back into one of the original equations. We'll use $y = 21x + 7$ .
For $x = 5$ :
$y = 21(5) + 7$
$y = 105 + 7$
$y = 112$
For $x = -5$ :
$y = 21(-5) + 7$
$y = -105 + 7$
$x$ $0$
Write Coordinates: Find the corresponding $y$ -values for each $x$ by substituting back into one of the original equations. We'll use $y = 21x + 7$ . For $x = 5$ : $y = 21(5) + 7$ $y = 105 + 7$ $y = 112$ For $x = -5$ : $y = 21(-5) + 7$ $y = -105 + 7$ $x$ $0$ Write the coordinates in exact form. First Coordinate: $x$ $1$ Second Coordinate: $x$ $2$

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