Solve the system of equations. y = -12x - 9 y = x^2 + 10x + 31 Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Set Equations Equal: Set the two equations equal to each other since they both equal y. y = -12x - 9 y = x^2 + 10x + 31 So, -12x - 9 = x^2 + 10x + 31Rearrange and Solve: Rearrange the equation to set it to zero and solve for x. 0 = x^2 + 10x + 31 + 12x + 9 0 = x^2 + 22x + 40Factor Quadratic Equation: Factor the quadratic equation. 0 = (x + 2)(x + 20)Solve for x: Solve for x by setting each factor equal to zero. x + 2 = 0 or x + 20 = 0 This gives us x = -2 or x = -20Substitute x Values: Substitute x = -2 into one of the original equations to find the corresponding y value. Using y = x^2 + 10x + 31: y = (-2)^2 + 10(-2) + 31 y = 4 - 20 + 31 y = 15 So one intersection point is (-2, 15).Find Intersection Points: Substitute x = -20 into one of the original equations to find the corresponding y value. Using y = x^2 + 10x + 31: y = (-20)^2 + 10(-20) + 31 y = 400 - 200 + 31 y = 231 So the other intersection point is (-20, 231).

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Solve the system of equations. $\newline$ $y = -12x - 9$ $\newline$ $y = x^2 + 10x + 31$ $\newline$ Write the coordinates in exact form. Simplify all fractions and radicals. $\newline$ (,) $\newline$ (,)

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Algebra 2

Solve a nonlinear system of equations

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Q. Solve the system of equations.

\newline

y = -12x - 9

\newline

y = x^2 + 10x + 31

\newline

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

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: Set the two equations equal to each other since they both equal $y$ . $y = -12x - 9$ $y = x^2 + 10x + 31$ So, $-12x - 9 = x^2 + 10x + 31$
Rearrange and Solve: Rearrange the equation to set it to zero and solve for $x$ . $0 = x^2 + 10x + 31 + 12x + 9$ $0 = x^2 + 22x + 40$
Factor Quadratic Equation: Factor the quadratic equation. $\newline$ $0 = (x + 2)(x + 20)$
Solve for x: Solve for x by setting each factor equal to zero. $\newline$ $x + 2 = 0$ or $x + 20 = 0$ $\newline$ This gives us $x = -2$ or $x = -20$
Substitute x Values: Substitute $x = -2$ into one of the original equations to find the corresponding $y$ value. $\newline$ Using $y = x^2 + 10x + 31$ : $\newline$ $y = (-2)^2 + 10(-2) + 31$ $\newline$ $y = 4 - 20 + 31$ $\newline$ $y = 15$ $\newline$ So one intersection point is $(-2, 15)$ .
Find Intersection Points: Substitute $x = -20$ into one of the original equations to find the corresponding $y$ value. $\newline$ Using $y = x^2 + 10x + 31$ : $\newline$ $y = (-20)^2 + 10(-20) + 31$ $\newline$ $y = 400 - 200 + 31$ $\newline$ $y = 231$ $\newline$ So the other intersection point is $(-20, 231)$ .