Solve the system of equations. y = -47x + 5 y = 3x^2 - 47x - 43 Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Set Equations Equal: Set the two equations equal to each other since they both equal y. y = -47x + 5 y = 3x^2 - 47x - 43 So, -47x + 5 = 3x^2 - 47x - 43.Simplify Equation: Simplify the equation by moving all terms to one side to set the equation to zero. 0 = 3x^2 - 47x + 47x - 43 - 5 0 = 3x^2 - 48Solve for x: Solve for x by adding 48 to both sides and then dividing by 3. 3x^2 = 48 x^2 = 48 / 3 x^2 = 16Take Square Root: Take the square root of both sides to solve for x. x = ±√16 x = ±4Substitute x: Substitute x = 4 into one of the original equations to solve for y. Using y = -47x + 5, we get: y = -47(4) + 5 y = -188 + 5 y = -183Coordinate Points: Substitute x = -4 into the same equation to solve for y. y = -47(-4) + 5 y = 188 + 5 y = 193Write the solution as coordinate points. The coordinate points are (4, -183) and (-4, 193).

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Solve the system of equations. $\newline$ $y = -47x + 5$ $\newline$ $y = 3x^2 - 47x - 43$ $\newline$ 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 = -47x + 5

\newline

y = 3x^2 - 47x - 43

\newline

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

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: Set the two equations equal to each other since they both equal $y$ . $y = -47x + 5$ $y = 3x^2 - 47x - 43$ So, $-47x + 5 = 3x^2 - 47x - 43$ .
Simplify Equation: Simplify the equation by moving all terms to one side to set the equation to zero. $\newline$ $0 = 3x^2 - 47x + 47x - 43 - 5$ $\newline$ $0 = 3x^2 - 48$
Solve for x: Solve for x by adding $48$ to both sides and then dividing by $3$ . $\newline$ $3x^2 = 48$ $\newline$ $x^2 = \frac{48}{3}$ $\newline$ $x^2 = 16$
Take Square Root: Take the square root of both sides to solve for $x$ . $x = \pm\sqrt{16}$ $x = \pm4$
Substitute $x$ : Substitute $x = 4$ into one of the original equations to solve for $y$ . Using $y = -47x + 5$ , we get: $y = -47(4) + 5$ $y = -188 + 5$ $y = -183$
Coordinate Points: Substitute $x = -4$ into the same equation to solve for $y$ . $y = -47(-4) + 5$ $y = 188 + 5$ $y = 193$
Coordinate Points: Substitute $x = -4$ into the same equation to solve for $y$ . $y = -47(-4) + 5$ $y = 188 + 5$ $y = 193$ Write the solution as coordinate points.The coordinate points are $(4, -183)$ and $(-4, 193)$ .