Solve the system of equations. y = 12x^2 - x - 19 y = -x + 29 Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Set Equations Equal: We have the following system of equations: y = 12x^2 - x - 19 y = -x + 29 To find the solution, we will set the two equations equal to each other since they both equal y. 12x^2 - x - 19 = -x + 29Simplify and Rearrange: Simplify the equation by moving all terms to one side to set the equation to zero. 12x^2 - x - 19 + x - 29 = 0 12x^2 - 48 = 0Factor and Solve for x: Solve for x by factoring or using the quadratic formula. In this case, we can factor the equation. 12x^2 - 48 = 0 Divide by 12 to simplify: x^2 - 4 = 0 Factor the difference of squares: (x - 2)(x + 2) = 0Substitute x into Equation: Solve for x by setting each factor equal to zero. x - 2 = 0 or x + 2 = 0 x = 2 or x = -2Write Coordinates: Find the corresponding y-values by substituting x back into one of the original equations. We can use y = -x + 29. For x = 2: y = -(2) + 29 y = 27 For x = -2: y = -(-2) + 29 y = 31Write the coordinates in exact form. The solutions to the system of equations are the points where the two graphs intersect, which are the x-values we found and their corresponding y-values. First Coordinate: (2, 27) Second Coordinate: (-2, 31)

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Solve the system of equations. $\newline$ $y = 12x^2 - x - 19$ $\newline$ $y = -x + 29$ $\newline$ Write the coordinates in exact form. Simplify all fractions and radicals. $\newline$ (,) $\newline$ (,)

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Algebra 1

Solve a system of linear and quadratic equations

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Q. Solve the system of equations.

\newline

y = 12x^2 - x - 19

\newline

y = -x + 29

\newline

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

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: We have the following system of equations: $\newline$ $y = 12x^2 - x - 19$ $\newline$ $y = -x + 29$ $\newline$ To find the solution, we will set the two equations equal to each other since they both equal $y$ . $\newline$ $12x^2 - x - 19 = -x + 29$
Simplify and Rearrange: Simplify the equation by moving all terms to one side to set the equation to zero. $\newline$ $12x^2 - x - 19 + x - 29 = 0$ $\newline$ $12x^2 - 48 = 0$
Factor and Solve for $x$ : Solve for $x$ by factoring or using the quadratic formula. In this case, we can factor the equation. $12x^2 - 48 = 0$ Divide by $12$ to simplify: $x^2 - 4 = 0$ Factor the difference of squares: $(x - 2)(x + 2) = 0$
Substitute $x$ into Equation: Solve for $x$ by setting each factor equal to zero. $\newline$ $x - 2 = 0$ or $x + 2 = 0$ $\newline$ $x = 2$ or $x = -2$
Write Coordinates: Find the corresponding $y$ -values by substituting $x$ back into one of the original equations. We can use $y = -x + 29$ .
For $x = 2$ :
$y = -(2) + 29$
$y = 27$
For $x = -2$ :
$y = -(-2) + 29$
$y = 31$
Write Coordinates: Find the corresponding $y$ -values by substituting $x$ back into one of the original equations. We can use $y = -x + 29$ . For $x = 2$ : $y = -(2) + 29$ $y = 27$ For $x = -2$ : $y = -(-2) + 29$ $y = 31$ Write the coordinates in exact form. The solutions to the system of equations are the points where the two graphs intersect, which are the $x$ -values we found and their corresponding $y$ -values. First Coordinate: $x$ $1$ Second Coordinate: $x$ $2$