Solve the system of equations. y = x^2 + 7x - 43 y = 7x - 7 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 + 7x - 43 y = 7x - 7 So, x^2 + 7x - 43 = 7x - 7Subtract and Simplify: Subtract 7x and add 7 to both sides to set the equation to zero. x^2 + 7x - 43 - 7x + 7 = 0 This simplifies to x^2 - 36 = 0Factor Quadratic Equation: Factor the quadratic equation. x^2 - 36 = (x + 6)(x - 6) = 0Solve for x: Solve for x by setting each factor equal to zero. x + 6 = 0 or x - 6 = 0 This gives us x = -6 or x = 6Substitute x = -6: Substitute x = -6 into one of the original equations to find the corresponding y value. Using y = 7x - 7, we get y = 7(-6) - 7 = -42 - 7 = -49 So one point of intersection is (-6, -49).Substitute x = 6: Substitute x = 6 into one of the original equations to find the corresponding y value. Using y = 7x - 7, we get y = 7(6) - 7 = 42 - 7 = 35 So the other point of intersection is (6, 35).

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Solve the system of equations. $\newline$ $y = x^2 + 7x - 43$ $\newline$ $y = 7x - 7$ $\newline$ 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 + 7x - 43

\newline

y = 7x - 7

\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 $y$ . $\newline$ $y = x^2 + 7x - 43$ $\newline$ $y = 7x - 7$ $\newline$ So, $x^2 + 7x - 43 = 7x - 7$
Subtract and Simplify: Subtract $7x$ and add $7$ to both sides to set the equation to zero. $\newline$ $x^2 + 7x - 43 - 7x + 7 = 0$ $\newline$ This simplifies to $x^2 - 36 = 0$
Factor Quadratic Equation: Factor the quadratic equation. $x^2 - 36 = (x + 6)(x - 6) = 0$
Solve for x: Solve for x by setting each factor equal to zero. $\newline$ $x + 6 = 0$ or $x - 6 = 0$ $\newline$ This gives us $x = -6$ or $x = 6$
Substitute $x = -6$ : Substitute $x = -6$ into one of the original equations to find the corresponding $y$ value. $\newline$ Using $y = 7x - 7$ , we get $y = 7(-6) - 7 = -42 - 7 = -49$ $\newline$ So one point of intersection is $(-6, -49)$ .
Substitute $x = 6$ : Substitute $x = 6$ into one of the original equations to find the corresponding $y$ value. $\newline$ Using $y = 7x - 7$ , we get $y = 7(6) - 7 = 42 - 7 = 35$ $\newline$ So the other point of intersection is $(6, 35)$ .