Solve the system of equations. x^2 + y^2 = 360 y = 3x Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Substitute y = 3x: Substitute y = 3x into x^2 + y^2 = 360. x^2 + (3x)^2 = 360 x^2 + 9x^2 = 360 10x^2 = 360Divide by 10: Divide both sides by 10 to solve for x^2. x^2 = 36Take square root: Take the square root of both sides to find x. x = ±6Substitute x back: Substitute x back into y = 3x to find y. For x = 6: y = 3(6) = 18 For x = -6: y = 3(-6) = -18Write coordinates: Write the coordinates in exact form. First Coordinate: (6, 18) Second Coordinate: (-6, -18)

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Solve the system of equations. $\newline$ $x^2 + y^2 = 360$ $\newline$ $y = 3x$ $\newline$ Write the coordinates in exact form. Simplify all fractions and radicals. $\newline$ (,) $\newline$ (,)

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Algebra 2

Solve a system of linear and quadratic equations: circles

Full solution

Q. Solve the system of equations.

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x^2 + y^2 = 360

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y = 3x

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

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(______,______)

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(______,______)

Substitute $y = 3x$ : Substitute $y = 3x$ into $x^2 + y^2 = 360$ . $\newline$ $x^2 + (3x)^2 = 360$ $\newline$ $x^2 + 9x^2 = 360$ $\newline$ $10x^2 = 360$
Divide by $10$ : Divide both sides by $10$ to solve for $x^2$ . $\newline$ $x^2 = 36$
Take square root: Take the square root of both sides to find $x$ . $x = \pm 6$
Substitute $x$ back: Substitute $x$ back into $y = 3x$ to find $y$ . $\newline$ For $x = 6$ : $y = 3(6) = 18$ $\newline$ For $x = -6$ : $y = 3(-6) = -18$
Write coordinates: Write the coordinates in exact form. $\newline$ First Coordinate: $(6, 18)$ $\newline$ Second Coordinate: $(-6, -18)$