Solve the system of equations. y = 32x + 97 y = x^2 + 32x - 47 Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Substitute y into second equation: Substitute y from the first equation into the second equation: y = x^2 + 32x - 47 becomes 32x + 97 = x^2 + 32x - 47.Solve for x: Subtract 32x from both sides to get 97 = x^2 - 47.Calculate y for x=12: Add 47 to both sides to get x^2 = 144.Calculate y for x=-12: Take the square root of both sides to find x. This gives us x = ±12.Write solution as coordinate points: Substitute x = 12 into the first equation to find y. y = 32(12) + 97.Calculate y for x = 12. y = 384 + 97 = 481.Now substitute x = -12 into the first equation to find y. y = 32(-12) + 97.Calculate y for x = -12. y = -384 + 97 = -287.Write the solution as coordinate points. The coordinate points are (12, 481) and (-12, -287).

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Solve the system of equations. $\newline$ $y = 32x + 97$ $\newline$ $y = x^2 + 32x - 47$ $\newline$ 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 = 32x + 97

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y = x^2 + 32x - 47

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

\newline

(______,______)

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(______,______)

Substitute $y$ into second equation: Substitute $y$ from the first equation into the second equation: $y = x^2 + 32x - 47$ becomes $32x + 97 = x^2 + 32x - 47$ .
Solve for x: Subtract $32x$ from both sides to get $97 = x^2 - 47$ .
Calculate $y$ for $x=12$ : Add $47$ to both sides to get $x^2 = 144$ .
Calculate $y$ for $x=-12$ : Take the square root of both sides to find $x$ . This gives us $x = \pm 12$ .
Write solution as coordinate points: Substitute $x = 12$ into the first equation to find $y$ . $y = 32(12) + 97$ .
Write solution as coordinate points: Substitute $x = 12$ into the first equation to find $y$ . $y = 32(12) + 97$ .Calculate $y$ for $x = 12$ . $y = 384 + 97 = 481$ .
Write solution as coordinate points: Substitute $x = 12$ into the first equation to find $y$ . $y = 32(12) + 97$ .Calculate $y$ for $x = 12$ . $y = 384 + 97 = 481$ .Now substitute $x = -12$ into the first equation to find $y$ . $y = 32(-12) + 97$ .
Write solution as coordinate points: Substitute $x = 12$ into the first equation to find $y$ . $y = 32(12) + 97$ .Calculate $y$ for $x = 12$ . $y = 384 + 97 = 481$ .Now substitute $x = -12$ into the first equation to find $y$ . $y = 32(-12) + 97$ .Calculate $y$ for $x = -12$ . $y$ $1$ .
Write solution as coordinate points: Substitute $x = 12$ into the first equation to find $y$ . $y = 32(12) + 97$ .Calculate $y$ for $x = 12$ . $y = 384 + 97 = 481$ .Now substitute $x = -12$ into the first equation to find $y$ . $y = 32(-12) + 97$ .Calculate $y$ for $x = -12$ . $y$ $1$ .Write the solution as coordinate points. The coordinate points are $y$ $2$ and $y$ $3$ .