Solve the system of equations. y = x^2 + 18x + 36 y = 4x + 12 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 = x^2 + 18x + 36 y = 4x + 12 So, x^2 + 18x + 36 = 4x + 12.Subtract and Simplify: Subtract 4x + 12 from both sides to set the equation to zero. x^2 + 18x + 36 - 4x - 12 = 0 This simplifies to x^2 + 14x + 24 = 0.Factor Quadratic Equation: Factor the quadratic equation. (x + 2)(x + 12) = 0Solve for x: Solve for x by setting each factor equal to zero. x + 2 = 0 or x + 12 = 0 This gives us x = -2 or x = -12.Substitute x = -2: Substitute x = -2 into one of the original equations to find the corresponding y value. Using y = 4x + 12: y = 4(-2) + 12 y = -8 + 12 y = 4 So one intersection point is (-2, 4).Substitute x = -12: Substitute x = -12 into one of the original equations to find the corresponding y value. Using y = 4x + 12: y = 4(-12) + 12 y = -48 + 12 y = -36 So the other intersection point is (-12, -36).

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Solve the system of equations. $\newline$ $y = x^2 + 18x + 36$ $\newline$ $y = 4x + 12$ $\newline$ 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 = x^2 + 18x + 36

\newline

y = 4x + 12

\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 $y$ . $y = x^2 + 18x + 36$ $y = 4x + 12$ So, $x^2 + 18x + 36 = 4x + 12$ .
Subtract and Simplify: Subtract $4x + 12$ from both sides to set the equation to zero. $\newline$ $x^2 + 18x + 36 - 4x - 12 = 0$ $\newline$ This simplifies to $x^2 + 14x + 24 = 0$ .
Factor Quadratic Equation: Factor the quadratic equation. $\newline$ $(x + 2)(x + 12) = 0$
Solve for x: Solve for x by setting each factor equal to zero. $\newline$ $x + 2 = 0$ or $x + 12 = 0$ $\newline$ This gives us $x = -2$ or $x = -12$ .
Substitute $x = -2$ : Substitute $x = -2$ into one of the original equations to find the corresponding $y$ value. $\newline$ Using $y = 4x + 12$ : $\newline$ $y = 4(-2) + 12$ $\newline$ $y = -8 + 12$ $\newline$ $y = 4$ $\newline$ So one intersection point is $(-2, 4)$ .
Substitute $x = -12$ : Substitute $x = -12$ into one of the original equations to find the corresponding $y$ value. $\newline$ Using $y = 4x + 12$ : $\newline$ $y = 4(-12) + 12$ $\newline$ $y = -48 + 12$ $\newline$ $y = -36$ $\newline$ So the other intersection point is $(-12, -36)$ .