Solve the system of equations. y = x^2 - 13x - 25 y = -13x + 11 Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Set Equations Equal: We have the system of equations: y = x^2 - 13x - 25 y = -13x + 11 Set the two equations equal to each other to find the x-coordinates of the intersection points. x^2 - 13x - 25 = -13x + 11Simplify and Rearrange: Simplify the equation by adding 13x to both sides and adding 25 to both sides to get the quadratic equation in standard form. x^2 - 13x - 25 + 13x - 11 = -13x + 11 + 13x - 11 x^2 - 36 = 0Isolate x^2 Term: Add 36 to both sides to isolate the x^2 term. x^2 - 36 + 36 = 0 + 36 x^2 = 36Solve for x: Take the square root of both sides to solve for x. sqrt(x^2) = sqrt(36) x = ±6Substitute x into Equation: We have two x-values: x1 = 6 x2 = -6 Now, substitute these x-values into one of the original equations to find the corresponding y-values. Let's use y = -13x + 11. For x1 = 6: y = -13(6) + 11 y = -78 + 11 y = -67For x2 = -6: y = -13(-6) + 11 y = 78 + 11 y = 89We now have the coordinates of the intersection points in exact form: First Coordinate: (6, -67) Second Coordinate: (-6, 89)

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Algebra 2

Solve a system of linear and quadratic equations: parabolas

Full solution

Q. Solve the system of equations.

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y = x^2 - 13x - 25

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y = -13x + 11

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

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(______,______)

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(______,______)

Set Equations Equal: We have the system of equations: $\newline$ $y = x^2 - 13x - 25$ $\newline$ $y = -13x + 11$ $\newline$ Set the two equations equal to each other to find the $x$ -coordinates of the intersection points. $\newline$ $x^2 - 13x - 25 = -13x + 11$
Simplify and Rearrange: Simplify the equation by adding $13x$ to both sides and adding $25$ to both sides to get the quadratic equation in standard form. $\newline$ $x^2 - 13x - 25 + 13x - 11 = -13x + 11 + 13x - 11$ $\newline$ $x^2 - 36 = 0$
Isolate $x^2$ Term: Add $36$ to both sides to isolate the $x^2$ term. $\newline$ $x^2 - 36 + 36 = 0 + 36$ $\newline$ $x^2 = 36$
Solve for x: Take the square root of both sides to solve for x. $\newline$ $\sqrt{x^2} = \sqrt{36}$ $\newline$ $x = \pm6$
Substitute $x$ into Equation: We have two $x$ -values: $\newline$ $x_1 = 6$ $\newline$ $x_2 = -6$ $\newline$ Now, substitute these $x$ -values into one of the original equations to find the corresponding $y$ -values. Let's use $y = -13x + 11$ . $\newline$ For $x_1 = 6$ : $\newline$ $y = -13(6) + 11$ $\newline$ $y = -78 + 11$ $\newline$ $x$ $0$
Substitute $x$ into Equation: We have two $x$ -values: $x_1 = 6$ $x_2 = -6$ Now, substitute these $x$ -values into one of the original equations to find the corresponding $y$ -values. Let's use $y = -13x + 11$ . For $x_1 = 6$ : $y = -13(6) + 11$ $y = -78 + 11$ $x$ $0$ For $x_2 = -6$ : $x$ $2$ $x$ $3$ $x$ $4$
Substitute $x$ into Equation: We have two $x$ -values: $\newline$ $x_1 = 6$ $\newline$ $x_2 = -6$ $\newline$ Now, substitute these $x$ -values into one of the original equations to find the corresponding $y$ -values. Let's use $y = -13x + 11$ . $\newline$ For $x_1 = 6$ : $\newline$ $y = -13(6) + 11$ $\newline$ $y = -78 + 11$ $\newline$ $x$ $0$ For $x_2 = -6$ : $\newline$ $x$ $2$ $\newline$ $x$ $3$ $\newline$ $x$ $4$ We now have the coordinates of the intersection points in exact form: $\newline$ First Coordinate: $x$ $5$ $\newline$ Second Coordinate: $x$ $6$