The graph of an equation representing a direct variation passes through the point (6, 10). Give another point with integral coordinates that is also on the graph of this equation.

[tex]\bf \qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby \begin{array}{llll} k=constant\ of\\ \qquad variation \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ (\stackrel{x}{6},\stackrel{y}{10})\qquad \textit{we know that } \begin{cases} x=6\\ y=10 \end{cases}\implies 10=k6\implies \cfrac{10}{6}=k\implies \cfrac{5}{3}=k \\\\\\ therefore\qquad \boxed{y=\cfrac{5}{3}x}[/tex]to get another point, we simply can pick a random independent variable, namely "x", say hmmmm x = 9, thus[tex]\bf y=\cfrac{5}{3}(\stackrel{\stackrel{x}{\downarrow }}{9})\implies y=5\cdot \cfrac{9}{3}\implies y=10~\hspace{7em}\boxed{(9, 10)}[/tex]