Full solution

Q. Solve the system of equations.

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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.

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(______,______)

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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 $y$ $3$ .