Solve the system of equations. y = x^2 + 24x - 50 y = 24x + 50 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 + 24x - 50 y = 24x + 50 So, x^2 + 24x - 50 = 24x + 50.Subtract to Zero: Subtract 24x + 50 from both sides to set the equation to zero. x^2 + 24x - 50 - 24x - 50 = 0 This simplifies to x^2 - 100 = 0.Factor Quadratic Equation: Factor the quadratic equation. (x - 10)(x + 10) = 0Solve for x: Solve for x by setting each factor equal to zero. x - 10 = 0 or x + 10 = 0 This gives us x = 10 or x = -10.Substitute x = 10: Substitute x = 10 into one of the original equations to find the corresponding y value. Using y = 24x + 50, we get y = 24(10) + 50. y = 240 + 50 y = 290 So one intersection point is (10, 290).Substitute x = -10: Substitute x = -10 into one of the original equations to find the corresponding y value. Using y = 24x + 50, we get y = 24(-10) + 50. y = -240 + 50 y = -190 So the other intersection point is (-10, -190).

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Solve the system of equations. $\newline$ $y = x^2 + 24x - 50$ $\newline$ $y = 24x + 50$ $\newline$ 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 + 24x - 50

\newline

y = 24x + 50

\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 $y$ . $y = x^2 + 24x - 50$ $y = 24x + 50$ So, $x^2 + 24x - 50 = 24x + 50$ .
Subtract to Zero: Subtract $24x + 50$ from both sides to set the equation to zero. $\newline$ $x^2 + 24x - 50 - 24x - 50 = 0$ $\newline$ This simplifies to $x^2 - 100 = 0$ .
Factor Quadratic Equation: Factor the quadratic equation. $\newline$ $(x - 10)(x + 10) = 0$
Solve for x: Solve for x by setting each factor equal to zero. $\newline$ $x - 10 = 0$ or $x + 10 = 0$ $\newline$ This gives us $x = 10$ or $x = -10$ .
Substitute $x = 10$ : Substitute $x = 10$ into one of the original equations to find the corresponding $y$ value. $\newline$ Using $y = 24x + 50$ , we get $y = 24(10) + 50$ . $\newline$ $y = 240 + 50$ $\newline$ $y = 290$ $\newline$ So one intersection point is $(10, 290)$ .
Substitute $x = -10$ : Substitute $x = -10$ into one of the original equations to find the corresponding $y$ value. $\newline$ Using $y = 24x + 50$ , we get $y = 24(-10) + 50$ . $\newline$ $y = -240 + 50$ $\newline$ $y = -190$ $\newline$ So the other intersection point is $(-10, -190)$ .