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Solve the system of equations. $\newline$ $y = x^2 + 26x - 50$ $\newline$ $y = 26x + 50$ $\newline$ Write the coordinates in exact form. Simplify all fractions and radicals. $\newline$ (,) $\newline$ (,)

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Solve a system of linear and quadratic equations

Full solution

Q. Solve the system of equations.

\newline

y = x^2 + 26x - 50

\newline

y = 26x + 50

\newline

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

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: Set the two equations equal to each other since they both equal $y$ . $y = x^2 + 26x - 50$ $y = 26x + 50$ $x^2 + 26x - 50 = 26x + 50$
Subtract to Zero: Subtract $26x + 50$ from both sides to set the equation to zero. $\newline$ $x^2 + 26x - 50 - 26x - 50 = 0$ $\newline$ $x^2 - 100 = 0$
Factor Quadratic Equation: Factor the quadratic equation. $x^2 - 100 = (x + 10)(x - 10)$
Solve for x: Set each factor equal to zero and solve for x. $\newline$ $(x + 10) = 0$ or $(x - 10) = 0$ $\newline$ $x = -10$ or $x = 10$
Substitute for $y (-10)$ : Substitute $x = -10$ into $y = 26x + 50$ to find the corresponding $y$ -value. $\newline$ $y = 26(-10) + 50$ $\newline$ $y = -260 + 50$ $\newline$ $y = -210$
Substitute for $y$ ( $10$ ): Substitute $x = 10$ into $y = 26x + 50$ to find the corresponding $y$ -value. $\newline$ $y = 26(10) + 50$ $\newline$ $y = 260 + 50$ $\newline$ $y = 310$
Write Coordinates: Write the coordinates in exact form. $\newline$ First Coordinate: $(-10, -210)$ $\newline$ Second Coordinate: $(10, 310)$

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