Solve the system of equations. $\newline$ $y = -27x - 23$ $\newline$ $y = x^2 - 13x + 26$ $\newline$ Write the coordinates in exact form. Simplify all fractions and radicals. $\newline$ (,)

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Algebra 2

Solve a nonlinear system of equations

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Q. Solve the system of equations.

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y = -27x - 23

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y = x^2 - 13x + 26

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

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(______,______)

Set Equations Equal: Since both equations are equal to $y$ , we can set them equal to each other to find $x$ . This gives us the equation $-27x - 23 = x^2 - 13x + 26$ .
Rearrange and Solve for x: Rearrange the equation to set it to zero and solve for $x$ . This means we will move all terms to one side to get $x^2 - 13x + 26 = 27x + 23$ . Simplifying this, we get $x^2 - 40x + 3 = 0$ .
Quadratic Formula: This is a quadratic equation, and we can solve for $x$ by factoring, completing the square, or using the quadratic formula. The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . For our equation, $a = 1$ , $b = -40$ , and $c = 3$ .
Calculate Discriminant: Calculate the discriminant $\Delta = b^2 - 4ac$ which is $\Delta = (-40)^2 - 4(1)(3)$ . This gives us $\Delta = 1600 - 12$ , so $\Delta = 1588$ .
Find Real Solutions: Since the discriminant is positive, we have two real solutions. Now we use the quadratic formula to find the values of $x$ : $x = \frac{40 \pm \sqrt{1588}}{2}$ . Simplifying under the radical, $\sqrt{1588} = \sqrt{(4\cdot397)} = 2\sqrt{397}$ . So the solutions for $x$ are $x = \frac{40 \pm 2\sqrt{397}}{2}$ .
Substitute $x$ into Original Equation: Simplify the solutions for $x$ by dividing by $2$ : $x = 20 \pm \sqrt{397}$ .
Find $y$ for $x = 20 + \sqrt{397}$ : Now we substitute the values of $x$ back into one of the original equations to find $y$ . Let's use $y = -27x - 23$ . First, we'll find $y$ for $x = 20 + \sqrt{397}$ . This gives us $y = -27(20 + \sqrt{397}) - 23$ .
Find $y$ for $x = 20 - \sqrt{397}$ : Simplify the expression for $y$ : $y = -540 - 27\sqrt{397} - 23$ . Combining like terms, we get $y = -563 - 27\sqrt{397}$ .
Final Pairs of Solutions: Now we find $y$ for $x = 20 - \sqrt{397}$ . This gives us $y = -27(20 - \sqrt{397}) - 23$ .
Final Pairs of Solutions: Now we find $y$ for $x = 20 - \sqrt{397}$ . This gives us $y = -27(20 - \sqrt{397}) - 23$ .Simplify the expression for $y$ : $y = -540 + 27\sqrt{397} - 23$ . Combining like terms, we get $y = -563 + 27\sqrt{397}$ .
Final Pairs of Solutions: Now we find $y$ for $x = 20 - \sqrt{397}$ . This gives us $y = -27(20 - \sqrt{397}) - 23$ .Simplify the expression for $y$ : $y = -540 + 27\sqrt{397} - 23$ . Combining like terms, we get $y = -563 + 27\sqrt{397}$ .We now have two pairs of solutions for the system of equations: $(20 + \sqrt{397}, -563 - 27\sqrt{397})$ and $(20 - \sqrt{397}, -563 + 27\sqrt{397})$ .