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Advanced Math Difficulty: Hard

$y = x + 9$

$y = x^{2} + 16 x + 63$

A solution to the given system of equations is $(x, y)$ . What is the greatest possible value of $x$ ?

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Explanation

Choice A is correct. It's given that $y = x + 9$ and $y = x^{2} + 16 x + 63$ ; therefore, it follows that $x + 9 = x^{2} + 16 x + 63$ . This equation can be rewritten as $x + 9 = (x + 9) (x + 7)$ . Subtracting $left parenthesis x plus 9 right parenthesis$ from both sides of this equation yields $0 = (x + 9) (x + 7) - (x + 9)$ . This equation can be rewritten as $0 = (x + 9) ((x + 7) - 1)$ , or $0 = (x + 9) (x + 6)$ . By the zero product property, $x + 9 = 0$ or $x + 6 = 0$ . Subtracting $9$ from both sides of the equation $x + 9 = 0$ yields $x = - 9$ . Subtracting $6$ from both sides of the equation $x + 6 = 0$ yields $x = - 6$ . Therefore, the given system of equations has solutions, $(x, y)$ , that occur when $x = - 9$ and $x = - 6$ . Since $negative 6$ is greater than $negative 9$ , the greatest possible value of $x$ is $negative 6$ .

Choice B is incorrect. This is the negative of the greatest possible value of $x$ when $y equals 0$ for the second equation in the given system of equations.

Choice C is incorrect. This is the value of $y$ when $x equals 0$ for the first equation in the given system of equations.

Choice D is incorrect. This is the value of $y$ when $x equals 0$ for the second equation in the given system of equations.