## Precalculus (6th Edition)

When $x=40$, the value of $y$ is $12$.
$y$ varies directly so its equation is of the form $y=kx$. When $x=30$, $y=9$. Substitute these values into $y=kx$ to obtain: $y=kx \\9=k(30) \\\frac{9}{30}=\frac{k(30)}{30} \\\frac{3}{10}=k$ Thus, the direct variation is $y=\frac{3}{10}x$. To find the value of $y$ when $x=40$, substitute $40$ to $x$ in the equation above to obtain: $y=\frac{3}{10}x \\y=\frac{3}{10}(40) \\y=\frac{120}{10} \\y=12$