Question

If y varies directly as x and z, and y= 8/3 when x = 1 and 2 = 4, find y when x = 6 and z = 3.

Answer 1

$\begin{gathered} y\propto x,y\propto z \\ So,\text{ } \\ y=k_1x,\Rightarrow\frac{8}{3}=k_1\times1\Rightarrow k_1=\frac{8}{3} \\ y=k_2z\Rightarrow\frac{8}{3}=k_2\times4\Rightarrow k_2=\frac{2}{3} \\ \text{Now, } \\ y=k_1x\Rightarrow\frac{8}{3}\times6=16 \\ y=k_2x\Rightarrow\frac{2}{3}\times3\Rightarrow2 \end{gathered}$