y = 4 x y = x 2 - 12 A solution to the given system

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