Solve the system of equations. y = -27x + 1 y = x^2 - 26x - 41 Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Set Equations Equal: We have the system of equations: y = -27x + 1 y = x^2 - 26x - 41 To find the intersection points, we set the two equations equal to each other. -27x + 1 = x^2 - 26x - 41Rearrange and Form Quadratic Equation: Rearrange the equation to bring all terms to one side and set it equal to zero to form a standard quadratic equation. x^2 - 26x - 41 + 27x - 1 = 0 x^2 + x - 42 = 0Factor Quadratic Equation: Factor the quadratic equation. We look for two numbers that multiply to -42 and add up to 1. These numbers are 7 and -6. (x + 7)(x - 6) = 0Solve for x: Solve for x by setting each factor equal to zero. x + 7 = 0 or x - 6 = 0 x = -7 or x = 6Find y for x = -7: Find the corresponding y-values for each x-value by substituting back into one of the original equations. We can use y = -27x + 1. For x = -7: y = -27(-7) + 1 y = 189 + 1 y = 190Find y for x = 6: Find the corresponding y-value for x = 6: y = -27(6) + 1 y = -162 + 1 y = -161Write Coordinates: Write the coordinates in exact form. The first coordinate is (-7, 190). The second coordinate is (6, -161).

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Solve the system of equations. $\newline$ $y = -27x + 1$ $\newline$ $y = x^2 - 26x - 41$ $\newline$ 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: parabolas

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Q. Solve the system of equations.

\newline

y = -27x + 1

\newline

y = x^2 - 26x - 41

\newline

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

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: We have the system of equations: $\newline$ $y = -27x + 1$ $\newline$ $y = x^2 - 26x - 41$ $\newline$ To find the intersection points, we set the two equations equal to each other. $\newline$ $-27x + 1 = x^2 - 26x - 41$
Rearrange and Form Quadratic Equation: Rearrange the equation to bring all terms to one side and set it equal to zero to form a standard quadratic equation. $\newline$ $x^2 - 26x - 41 + 27x - 1 = 0$ $\newline$ $x^2 + x - 42 = 0$
Factor Quadratic Equation: Factor the quadratic equation. $\newline$ We look for two numbers that multiply to $-42$ and add up to $1$ . These numbers are $7$ and $-6$ . $\newline$ $(x + 7)(x - 6) = 0$
Solve for x: Solve for x by setting each factor equal to zero. $\newline$ $x + 7 = 0$ or $x - 6 = 0$ $\newline$ $x = -7$ or $x = 6$
Find $y$ for $x = -7$ : Find the corresponding $y$ -values for each $x$ -value by substituting back into one of the original equations. We can use $y = -27x + 1$ . $\newline$ For $x = -7$ : $\newline$ $y = -27(-7) + 1$ $\newline$ $y = 189 + 1$ $\newline$ $y = 190$
Find $y$ for $x = 6$ : Find the corresponding $y$ -value for $x = 6$ :
$y = -27(6) + 1$
$y = -162 + 1$
$y = -161$
Write Coordinates: Write the coordinates in exact form. $\newline$ The first coordinate is $(-7, 190)$ . $\newline$ The second coordinate is $(6, -161)$ .