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${:[y=-x-11],[y=x^(2)-5x-7]:} Which of the following is a solution to the system of equations? Choose 1 answer: (A) (-2,-9) (B) (0,-7) (c) (2,-13) (D) (3,-13)$

$\begin{array}{l} y=-x-11 \\ y=x^{2}-5 x-7 \end{array}$ $\newline$ Which of the following is a solution to the system of equations? $\newline$ Choose $1$ answer: $\newline$ (A) $(-2,-9)$ $\newline$ (B) $(0,-7)$ $\newline$ (C) $(2,-13)$ $\newline$ (D) $(3,-13)$

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\begin{array}{l} y=-x-11 \\ y=x^{2}-5 x-7 \end{array}

\newline

Which of the following is a solution to the system of equations?

\newline

Choose

1

answer:

\newline

(A)

(-2,-9)

\newline

(B)

(0,-7)

\newline

(C)

(2,-13)

\newline

(D)

(3,-13)

System of Equations: We have the system of equations: $\newline$ $y = -x - 11$ $\newline$ $y = x^2 - 5x - 7$ $\newline$ To find the solution to the system, we need to find a pair of x and y values that satisfy both equations simultaneously.
Substituting and Rearranging: Let's substitute the first equation into the second one to solve for x: $\newline$ $-x - 11 = x^2 - 5x - 7$ $\newline$ Now, we will rearrange the equation to set it to zero and solve for x: $\newline$ $x^2 - 5x + x - 7 + 11 = 0$ $\newline$ $x^2 - 4x + 4 = 0$
Factoring the Quadratic Equation: We can factor the quadratic equation: $\newline$ $(x - 2)^2 = 0$ $\newline$ This gives us one solution for x: $\newline$ $x = 2$
Finding the Value of x: Now that we have the value of x, we can substitute it back into either of the original equations to find the corresponding y value. Let's use the first equation: $\newline$ $y = -x - 11$ $\newline$ $y = -2 - 11$ $\newline$ $y = -13$
Substituting $x$ into the Original Equation: We have found a pair $(x, y) = (2, -13)$ that satisfies both equations. Let's check if this pair is one of the given choices: $\newline$ (A) $(-2,-9)$ $\newline$ (B) $(0,-7)$ $\newline$ (C) $(2,-13)$ $\newline$ (D) $(3,-13)$ $\newline$ The correct answer is (C) $(2, -13)$ .