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Solve the system of equations. $\newline$ $y = 32x^2 + 32x - 23$ $\newline$ $y = 32x + 9$ $\newline$ 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 = 32x^2 + 32x - 23

\newline

y = 32x + 9

\newline

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

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: We have the system of equations: $\newline$ $y = 32x^2 + 32x - 23$ $\newline$ $y = 32x + 9$ $\newline$ Set the two equations equal to each other to find the $x$ -values where they intersect. $\newline$ $32x^2 + 32x - 23 = 32x + 9$
Subtract and Simplify: Subtract $32x$ and $9$ from both sides to move all terms to one side and set the equation to zero. $\newline$ $32x^2 + 32x - 23 - 32x - 9 = 32x + 9 - 32x - 9$ $\newline$ $32x^2 - 32 = 0$
Factor Common Term: Factor out the common term $32$ from the left side of the equation. $\newline$ $32(x^2 - 1) = 0$
Factor Difference of Squares: Recognize that $x^2 - 1$ is a difference of squares and can be factored further. $\newline$ $32(x + 1)(x - 1) = 0$
Set Factors Equal: Set each factor that contains an $x$ equal to zero and solve for $x$ . $(x + 1) = 0$ and $(x - 1) = 0$ $x = -1$ and $x = 1$
Substitute and Solve: Substitute the $x$ -values back into either original equation to find the corresponding $y$ -values. We'll use $y = 32x + 9$ . For $x = -1$ : $y = 32(-1) + 9$ $y = -32 + 9$ $y = -23$ For $x = 1$ : $y = 32(1) + 9$ $y = 32 + 9$ $y$ $0$
Write Coordinates: Write the coordinates in exact form. $\newline$ First Coordinate: $(-1, -23)$ $\newline$ Second Coordinate: $(1, 41)$

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