Solve the system of equations. y = -27x + 49 y = x^2 - 15x + 4 Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Substitute y Equation: Substitute y from the first equation into the second equation: y = x^2 - 15x + 4 becomes -27x + 49 = x^2 - 15x + 4.Rearrange to Set to 0: Rearrange the equation to set it to 0: x^2 - 15x + 4 + 27x - 49 = 0, which simplifies to x^2 + 12x - 45 = 0.Factor Quadratic Equation: Factor the quadratic equation: (x + 15)(x - 3) = 0.Solve for x: Solve for x: x + 15 = 0 or x - 3 = 0, so x = -15 or x = 3.Calculate y for -15: Substitute x = -15 into the first equation to find y: y = -27(-15) + 49.Calculate y for 3: Calculate y for x = -15: y = 405 + 49, which is y = 454.Substitute x to Find y: Substitute x = 3 into the first equation to find y: y = -27(3) + 49.Calculate y for 3: Calculate y for x = 3: y = -81 + 49, which is y = -32.Write Solution as Points: Write the solution as coordinate points: The coordinate points are (-15, 454) and (3, -32).

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Solve the system of equations. $\newline$ $y = -27x + 49$ $\newline$ $y = x^2 - 15x + 4$ $\newline$ 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.

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y = -27x + 49

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y = x^2 - 15x + 4

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

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(______,______)

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(______,______)

Substitute $y$ Equation: Substitute $y$ from the first equation into the second equation: $y = x^2 - 15x + 4$ becomes $-27x + 49 = x^2 - 15x + 4$ .
Rearrange to Set to $0$ : Rearrange the equation to set it to $0$ : $x^2 - 15x + 4 + 27x - 49 = 0$ , which simplifies to $x^2 + 12x - 45 = 0$ .
Factor Quadratic Equation: Factor the quadratic equation: $(x + 15)(x - 3) = 0$ .
Solve for x: Solve for x: $x + 15 = 0$ or $x - 3 = 0$ , so $x = -15$ or $x = 3$ .
Calculate $y$ for $-15$ : Substitute $x = -15$ into the first equation to find $y$ : $y = -27(-15) + 49$ .
Calculate $y$ for $3$ : Calculate $y$ for $x = -15$ : $y = 405 + 49$ , which is $y = 454$ .
Substitute $x$ to Find $y$ : Substitute $x = 3$ into the first equation to find $y$ : $y = -27(3) + 49$ .
Calculate $y$ for $3$ : Calculate $y$ for $x = 3$ : $y = -81 + 49$ , which is $y = -32$ .
Write Solution as Points: Write the solution as coordinate points: The coordinate points are $(-15, 454)$ and $(3, -32)$ .