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

$y equals 4 x$

$y = x^{2} - 12$

A solution to the given system of equations is $(x, y)$ , where $x greater than 0$ . What is the value of $x$ ?

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Explanation

The correct answer is $6$ . It’s given that $y equals 4 x$ and $y = x^{2} - 12$ . Since $y equals 4 x$ , substituting $4 x$ for $y$ in the second equation of the given system yields $4 x = x^{2} - 12$ . Subtracting $4 x$ from both sides of this equation yields $0 = x^{2} - 4 x - 12$ . This equation can be rewritten as $0 = (x - 6) (x + 2)$ . By the zero product property, $x - 6 = 0$ or $x + 2 = 0$ . Adding $6$ to both sides of the equation $x - 6 = 0$ yields $x equals 6$ . Subtracting $2$ from both sides of the equation $x + 2 = 0$ yields $x = - 2$ . Therefore, solutions to the given system of equations occur when $x equals 6$ and when $x = - 2$ . It’s given that a solution to the given system of equations is $(x, y)$ , where $x greater than 0$ . Since $6$ is greater than $0$ , it follows that the value of $x$ is $6$ .