Solve the system of equations. x = 3y + 20 x^2 + y^2 = 290 Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Substitute x: Substitute x from the first equation into the second equation. x = 3y + 20 x^2 + y^2 = 290 (3y + 20)^2 + y^2 = 290Expand and simplify: Expand the squared term and simplify the equation. (3y + 20)^2 + y^2 = 290 9y^2 + 120y + 400 + y^2 = 290 10y^2 + 120y + 400 = 290Subtract to set zero: Subtract 290 from both sides to set the equation to zero. 10y^2 + 120y + 400 - 290 = 0 10y^2 + 120y + 110 = 0Divide and simplify: Divide the entire equation by 10 to simplify. y^2 + 12y + 11 = 0Factor quadratic equation: Factor the quadratic equation. (y + 11)(y + 1) = 0Solve for y: Solve for y by setting each factor equal to zero. y + 11 = 0 or y + 1 = 0 y = -11 or y = -1Substitute for x (-11): Substitute y back into the first equation to find x. For y = -11: x = 3(-11) + 20 x = -33 + 20 x = -13Substitute for x (-1): Substitute y back into the first equation to find x for the second value of y. For y = -1: x = 3(-1) + 20 x = -3 + 20 x = 17Write coordinates: Write the coordinates in exact form. First Coordinate: (-13, -11) Second Coordinate: (17, -1)

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Solve the system of equations. $\newline$ $x = 3y + 20$ $\newline$ $x^2 + y^2 = 290$ $\newline$ 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

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Q. Solve the system of equations.

\newline

x = 3y + 20

\newline

x^2 + y^2 = 290

\newline

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

\newline

(______,______)

\newline

(______,______)

Substitute $x$ : Substitute $x$ from the first equation into the second equation. $\newline$ $x = 3y + 20$ $\newline$ $x^2 + y^2 = 290$ $\newline$ $(3y + 20)^2 + y^2 = 290$
Expand and simplify: Expand the squared term and simplify the equation. $\newline$ $(3y + 20)^2 + y^2 = 290$ $\newline$ $9y^2 + 120y + 400 + y^2 = 290$ $\newline$ $10y^2 + 120y + 400 = 290$
Subtract to set zero: Subtract $290$ from both sides to set the equation to zero. $\newline$ $10y^2 + 120y + 400 - 290 = 0$ $\newline$ $10y^2 + 120y + 110 = 0$
Divide and simplify: Divide the entire equation by $10$ to simplify. $\newline$ $y^2 + 12y + 11 = 0$
Factor quadratic equation: Factor the quadratic equation. $\newline$ $(y + 11)(y + 1) = 0$
Solve for y: Solve for y by setting each factor equal to zero. $\newline$ $y + 11 = 0$ or $y + 1 = 0$ $\newline$ $y = -11$ or $y = -1$
Substitute for $x (-11)$ : Substitute $y$ back into the first equation to find $x$ . For $y = -11$ : $x = 3(-11) + 20$ $x = -33 + 20$ $x = -13$
Substitute for $x (-1)$ : Substitute $y$ back into the first equation to find $x$ for the second value of $y$ . For $y = -1$ : $x = 3(-1) + 20$ $x = -3 + 20$ $x = 17$
Write coordinates: Write the coordinates in exact form. $\newline$ First Coordinate: $(-13, -11)$ $\newline$ Second Coordinate: $(17, -1)$