Question

The variable $x$ varies directly as $y$ : when $y$ is $10$, $x$ is $\frac{3}{2}$. find $y$ when $x$ is $9$.

Answer 1

$x$ varies directly $y$

$y=kx$

$x=3/2;y=10$

$10=k\ast \frac{3}{2}$

${}^{\prime }{k}^{\prime }$ is constant.

$k=\frac{20}{3}$

when $x=9$ & $k=\frac{20}{3}$

$y=?$

$y=\frac{20}{3}\left(9\right)$

$y=60$