Question

y varies jointly as x and z. y = 36 when x= 3 and z=3. Find y when x=5 and z=2.y=

Answer 1

$\begin{gathered} y\propto x\text{ }\times z \\ y\propto xz \\ y\text{ = k}\times\text{ xz} \\ k\text{ = }\frac{y}{xz} \end{gathered}$

Where k is a constant

When y = 36 , x = 3 , z=3

$\begin{gathered} k\text{ = }\frac{y}{xz} \\ k=\text{ }\frac{36}{3\times3}\text{ =}\frac{36}{9}\text{ = 4} \end{gathered}$$\begin{gathered} y\text{ = 4xz} \\ x=\text{ 5 , z = 2} \\ y\text{ = 4 }\times5\times2=40 \end{gathered}$

The value of y is 40