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

Substitute and Simplify: Substitute x = -3y into the second equation x^2 + y^2 = 360. (-3y)^2 + y^2 = 360 9y^2 + y^2 = 360Combine Like Terms: Combine like terms. 10y^2 = 360Divide and Solve: Divide both sides by 10 to solve for y^2. y^2 = 36Find y: Take the square root of both sides to find y. y = ±6Substitute and Find x: Substitute y back into x = -3y to find x. For y = 6: x = -3(6) = -18 For y = -6: x = -3(-6) = 18Write Coordinates: Write the coordinates in exact form. First Coordinate: (-18, 6) Second Coordinate: (18, -6)

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

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Math Problems

Algebra 2

Solve a system of linear and quadratic equations: circles

Full solution

Q. Solve the system of equations.

\newline

x = -3y

\newline

x^2 + y^2 = 360

\newline

\newline

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

\newline

(______,______)

\newline

(______,______)

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