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Advanced Math / Nonlinear equations in one variable and systems of equations in two variables Difficulty: Hard

$8 x + y = - 11$

$2 x^{2} = y + 341$

The graphs of the equations in the given system of equations intersect at the point $(x, y)$ in the $x y$ -plane. What is a possible value of $x$ ?

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Explanation

Choice A is correct. It's given that the graphs of the equations in the given system of equations intersect at the point $(x, y)$ . Therefore, this intersection point is a solution to the given system. The solution can be found by isolating $y$ in each equation. The given equation $8 x + y = - 11$ can be rewritten to isolate $y$ by subtracting $8 x$ from both sides of the equation, which gives $y = - 8 x - 11$ . The given equation $2 x^{2} = y + 341$ can be rewritten to isolate $y$ by subtracting $341$ from both sides of the equation, which gives $2 x^{2} - 341 = y$ . With each equation solved for $y$ , the value of $y$ from one equation can be substituted into the other, which gives $2 x^{2} - 341 = - 8 x - 11$ . Adding $8 x$ and $11$ to both sides of this equation results in $2 x^{2} + 8 x - 330 = 0$ . Dividing both sides of this equation by $2$ results in $x^{2} + 4 x - 165 = 0$ . This equation can be rewritten by factoring the left-hand side, which yields $(x + 15) (x - 11) = 0$ . By the zero-product property, if $(x + 15) (x - 11) = 0$ , then $(x + 15) = 0$ , or $(x - 11) = 0$ . It follows that $x = - 15$ , or $x equals 11$ . Since only $negative 15$ is given as a choice, a possible value of $x$ is $negative 15$ .

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.