Solve the system of equations. y = 42x + 81 y = x^2 + 42x - 40 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 = 42x + 81 y = x^2 + 42x - 40 So, x^2 + 42x - 40 = 42x + 81Subtract 42x: Subtract 42x from both sides to start isolating the x^2 term. x^2 + 42x - 40 - 42x = 42x + 81 - 42x This simplifies to: x^2 - 40 = 81Add 40: Add 40 to both sides to isolate the x^2 term completely. x^2 - 40 + 40 = 81 + 40 This simplifies to: x^2 = 121Take Square Root: Take the square root of both sides to solve for x. √(x^2) = ±√(121) This gives us: x = ±11Substitute x = 11: Substitute x = 11 into one of the original equations to find the corresponding y value. Using y = 42x + 81, we get: y = 42(11) + 81 y = 462 + 81 y = 543 So one of the points of intersection is (11, 543).Substitute x = -11: Substitute x = -11 into the same equation to find the other corresponding y value. Using y = 42x + 81, we get: y = 42(-11) + 81 y = -462 + 81 y = -381 So the other point of intersection is (-11, -381).

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Solve the system of equations. $\newline$ $y = 42x + 81$ $\newline$ $y = x^2 + 42x - 40$ $\newline$ 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 = 42x + 81

\newline

y = x^2 + 42x - 40

\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 $y$ . $y = 42x + 81$ $y = x^2 + 42x - 40$ So, $x^2 + 42x - 40 = 42x + 81$
Subtract $42x$ : Subtract $42x$ from both sides to start isolating the $x^2$ term. $\newline$ $x^2 + 42x - 40 - 42x = 42x + 81 - 42x$ $\newline$ This simplifies to: $\newline$ $x^2 - 40 = 81$
Add $40$ : Add $40$ to both sides to isolate the $x^2$ term completely. $\newline$ $x^2 - 40 + 40 = 81 + 40$ $\newline$ This simplifies to: $\newline$ $x^2 = 121$
Take Square Root: Take the square root of both sides to solve for $x$ . $\sqrt{x^2} = \pm\sqrt{121}$ This gives us: $x = \pm11$
Substitute $x = 11$ : Substitute $x = 11$ into one of the original equations to find the corresponding $y$ value. $\newline$ Using $y = 42x + 81$ , we get: $\newline$ $y = 42(11) + 81$ $\newline$ $y = 462 + 81$ $\newline$ $y = 543$ $\newline$ So one of the points of intersection is $(11, 543)$ .
Substitute $x = -11$ : Substitute $x = -11$ into the same equation to find the other corresponding $y$ value. $\newline$ Using $y = 42x + 81$ , we get: $\newline$ $y = 42(-11) + 81$ $\newline$ $y = -462 + 81$ $\newline$ $y = -381$ $\newline$ So the other point of intersection is $(-11, -381)$ .