Solve the system of equations. y = x^2 + 28x - 41 y = 28x - 16 Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Set Equations Equal: We have the system of equations: y = x^2 + 28x - 41 y = 28x - 16 To find the intersection points, set the two equations equal to each other. x^2 + 28x - 41 = 28x - 16Simplify Equation: Simplify the equation by subtracting 28x from both sides and adding 16 to both sides. x^2 + 28x - 41 - 28x + 16 = 28x - 16 - 28x + 16 x^2 - 25 = 0Solve Quadratic Equation: Solve the simplified quadratic equation for x. x^2 - 25 = (x - 5)(x + 5) Set each factor equal to zero and solve for x. (x - 5) = 0 or (x + 5) = 0 x = 5 or x = -5Find Y-Values: Find the corresponding y-values for each x-value by substituting back into one of the original equations. We'll use y = 28x - 16. For x = 5: y = 28(5) - 16 y = 140 - 16 y = 124Write Coordinates: For x = -5: y = 28(-5) - 16 y = -140 - 16 y = -156Write the coordinates in exact form. The first coordinate is (5, 124). The second coordinate is (-5, -156).

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Solve the system of equations. $\newline$ $y = x^2 + 28x - 41$ $\newline$ $y = 28x - 16$ $\newline$ Write the coordinates in exact form. Simplify all fractions and radicals. $\newline$ (,) $\newline$ (,)

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Algebra 1

Solve a system of linear and quadratic equations

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Q. Solve the system of equations.

\newline

y = x^2 + 28x - 41

\newline

y = 28x - 16

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

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: We have the system of equations: $\newline$ $y = x^2 + 28x - 41$ $\newline$ $y = 28x - 16$ $\newline$ To find the intersection points, set the two equations equal to each other. $\newline$ $x^2 + 28x - 41 = 28x - 16$
Simplify Equation: Simplify the equation by subtracting $28x$ from both sides and adding $16$ to both sides. $\newline$ $x^2 + 28x - 41 - 28x + 16 = 28x - 16 - 28x + 16$ $\newline$ $x^2 - 25 = 0$
Solve Quadratic Equation: Solve the simplified quadratic equation for $x$ . $x^2 - 25 = (x - 5)(x + 5)$ Set each factor equal to zero and solve for $x$ . $(x - 5) = 0 \text{ or } (x + 5) = 0$ $x = 5 \text{ or } x = -5$
Find Y-Values: Find the corresponding y-values for each x-value by substituting back into one of the original equations. We'll use $y = 28x - 16$ . $\newline$ For $x = 5$ : $\newline$ $y = 28(5) - 16$ $\newline$ $y = 140 - 16$ $\newline$ $y = 124$
Write Coordinates: For $x = -5$ : $\newline$ $y = 28(-5) - 16$ $\newline$ $y = -140 - 16$ $\newline$ $y = -156$
Write Coordinates: For $x = -5$ : $y = 28(-5) - 16$ $y = -140 - 16$ $y = -156$ Write the coordinates in exact form. The first coordinate is $(5, 124)$ . The second coordinate is $(-5, -156)$ .