Solve the system of equations. y = x^2 + 26x - 47 y = 26x + 97 Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Set Equations Equal: Set the two equations equal to each other since they both equal y. y = x^2 + 26x - 47 y = 26x + 97 So, x^2 + 26x - 47 = 26x + 97.Subtract to Zero: Subtract 26x + 97 from both sides to set the equation to zero. x^2 + 26x - 47 - 26x - 97 = 0 This simplifies to x^2 - 144 = 0.Factor Quadratic Equation: Factor the quadratic equation. x^2 - 144 = (x - 12)(x + 12) = 0Solve for x: Solve for x by setting each factor equal to zero. x - 12 = 0 or x + 12 = 0 This gives us x = 12 or x = -12.Substitute x Values: Substitute x = 12 into one of the original equations to find the corresponding y value. Using y = 26x + 97, we get y = 26(12) + 97 = 312 + 97 = 409.Write Coordinate Points: Substitute x = -12 into the same equation to find the corresponding y value. Using y = 26x + 97, we get y = 26(-12) + 97 = -312 + 97 = -215.Write the solution as coordinate points. The coordinate points are (12, 409) and (-12, -215).

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Solve the system of equations. $\newline$ $y = x^2 + 26x - 47$ $\newline$ $y = 26x + 97$ $\newline$ $\newline$ 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.

\newline

y = x^2 + 26x - 47

\newline

y = 26x + 97

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

\newline

(______,______)

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(______,______)

Set Equations Equal: Set the two equations equal to each other since they both equal $y$ . $y = x^2 + 26x - 47$ $y = 26x + 97$ So, $x^2 + 26x - 47 = 26x + 97$ .
Subtract to Zero: Subtract $26x + 97$ from both sides to set the equation to zero. $\newline$ $x^2 + 26x - 47 - 26x - 97 = 0$ $\newline$ This simplifies to $x^2 - 144 = 0$ .
Factor Quadratic Equation: Factor the quadratic equation. $x^2 - 144 = (x - 12)(x + 12) = 0$
Solve for x: Solve for x by setting each factor equal to zero. $\newline$ $x - 12 = 0$ or $x + 12 = 0$ $\newline$ This gives us $x = 12$ or $x = -12$ .
Substitute x Values: Substitute $x = 12$ into one of the original equations to find the corresponding $y$ value. $\newline$ Using $y = 26x + 97$ , we get $y = 26(12) + 97 = 312 + 97 = 409$ .
Write Coordinate Points: Substitute $x = -12$ into the same equation to find the corresponding $y$ value. $\newline$ Using $y = 26x + 97$ , we get $y = 26(-12) + 97 = -312 + 97 = -215$ .
Write Coordinate Points: Substitute $x = -12$ into the same equation to find the corresponding $y$ value. $\newline$ Using $y = 26x + 97$ , we get $y = 26(-12) + 97 = -312 + 97 = -215$ .Write the solution as coordinate points. $\newline$ The coordinate points are $(12, 409)$ and $(-12, -215)$ .