Solve the system of equations. y = x^2 - 35x + 12 y = -35x + 37 Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Set Equations Equal: We have the system of equations: y = x^2 - 35x + 12 y = -35x + 37 To find the intersection points, we need to set the two equations equal to each other. x^2 - 35x + 12 = -35x + 37Simplify Equation: Now, we simplify the equation by adding 35x to both sides and subtracting 37 from both sides: x^2 - 35x + 12 + 35x - 37 = -35x + 37 + 35x - 37 x^2 - 25 = 0Isolate x^2 Term: Next, we add 25 to both sides to isolate the x^2 term: x^2 - 25 + 25 = 0 + 25 x^2 = 25Find x Values: To find the values of x, we take the square root of both sides: sqrt(x^2) = sqrt(25) x = 5 or x = -5Substitute x into Equation: Now we have two values for x, we need to find the corresponding y values by substituting x back into one of the original equations. We'll use y = -35x + 37. First, for x = 5: y = -35(5) + 37 y = -175 + 37 y = -138Calculate y Values: Next, for x = -5: y = -35(-5) + 37 y = 175 + 37 y = 212Final Coordinates: We now have the coordinates of the intersection points in exact form: First Coordinate: (5, -138) Second Coordinate: (-5, 212)

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Solve the system of equations. $\newline$ $y = x^2 - 35x + 12$ $\newline$ $y = -35x + 37$ $\newline$ 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 = x^2 - 35x + 12

\newline

y = -35x + 37

\newline

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

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: We have the system of equations: $\newline$ $y = x^2 - 35x + 12$ $\newline$ $y = -35x + 37$ $\newline$ To find the intersection points, we need to set the two equations equal to each other. $\newline$ $x^2 - 35x + 12 = -35x + 37$
Simplify Equation: Now, we simplify the equation by adding $35x$ to both sides and subtracting $37$ from both sides: $x^2 - 35x + 12 + 35x - 37 = -35x + 37 + 35x - 37$ $x^2 - 25 = 0$
Isolate $x^2$ Term: Next, we add $25$ to both sides to isolate the $x^2$ term: $\newline$ $x^2 - 25 + 25 = 0 + 25$ $\newline$ $x^2 = 25$
Find x Values: To find the values of x, we take the square root of both sides: $\sqrt{x^2} = \sqrt{25}$ $x = 5$ or $x = -5$
Substitute $x$ into Equation: Now we have two values for $x$ , we need to find the corresponding $y$ values by substituting $x$ back into one of the original equations. We'll use $y = -35x + 37$ .
First, for $x = 5$ :
$y = -35(5) + 37$
$y = -175 + 37$
$y = -138$
Calculate $y$ Values: Next, for $x = -5$ : $y = -35(-5) + 37$ $y = 175 + 37$ $y = 212$
Final Coordinates: We now have the coordinates of the intersection points in exact form: $\newline$ First Coordinate: $(5, -138)$ $\newline$ Second Coordinate: $(-5, 212)$