Solve the system of equations. y = -31x + 73 y = x^2 - 31x - 48 Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Set Equations Equal: We have the system of equations: y = -31x + 73 y = x^2 - 31x - 48 Set the two equations equal to each other to find the x-values where they intersect. -31x + 73 = x^2 - 31x - 48Simplify Quadratic Equation: Simplify the equation by moving all terms to one side to form a quadratic equation. 0 = x^2 - 31x - 48 + 31x - 73 0 = x^2 - 121Solve for x: Solve the quadratic equation for x. x^2 - 121 = 0 Factor the difference of squares. (x - 11)(x + 11) = 0Factor and Solve: Set each factor equal to zero and solve for x. x - 11 = 0 or x + 11 = 0 x = 11 or x = -11Find Y-Values: Find the corresponding y-values for each x-value by substituting back into one of the original equations. We'll use y = -31x + 73. For x = 11: y = -31(11) + 73 y = -341 + 73 y = -268 For x = -11: y = -31(-11) + 73 y = 341 + 73 y = 414

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Solve the system of equations. $\newline$ $y = -31x + 73$ $\newline$ $y = x^2 - 31x - 48$ $\newline$ Write the coordinates in exact form. Simplify all fractions and radicals. $\newline$ (,) $\newline$ (,)

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Algebra 2

Solve a system of linear and quadratic equations: parabolas

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Q. Solve the system of equations.

\newline

y = -31x + 73

\newline

y = x^2 - 31x - 48

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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: $\newline$ $y = -31x + 73$ $\newline$ $y = x^2 - 31x - 48$ $\newline$ Set the two equations equal to each other to find the $x$ -values where they intersect. $\newline$ $-31x + 73 = x^2 - 31x - 48$
Simplify Quadratic Equation: Simplify the equation by moving all terms to one side to form a quadratic equation. $\newline$ $0 = x^2 - 31x - 48 + 31x - 73$ $\newline$ $0 = x^2 - 121$
Solve for x: Solve the quadratic equation for x. $\newline$ $x^2 - 121 = 0$ $\newline$ Factor the difference of squares. $\newline$ $(x - 11)(x + 11) = 0$
Factor and Solve: Set each factor equal to zero and solve for $x$ . $x - 11 = 0$ or $x + 11 = 0$ $x = 11$ or $x = -11$
Find Y-Values: Find the corresponding y-values for each x-value by substituting back into one of the original equations. We'll use $y = -31x + 73$ . $\newline$ For $x = 11$ : $\newline$ $y = -31(11) + 73$ $\newline$ $y = -341 + 73$ $\newline$ $y = -268$ $\newline$ For $x = -11$ : $\newline$ $y = -31(-11) + 73$ $\newline$ $y = 341 + 73$ $\newline$ $y = 414$