Question

suppose y varies directly with the square of x and inversely with z. when x=2 and z=6, then y=2 6/11. find y if x=1 and z=3

Answer 1

Direct variation: y = kx

Inverse variation: y = k/x

y varies directly with x^2 ... : y = kx^2
... and inversely with z: y = kx^2/z

Now we use the given information to find k.
x = 2, z = 6, y = 2 6/11

y = kx^2/z

2 6/11 = k * (2^2)/6

28/11 = 4k/6

28/11 = 2k/3

k = 28/11 * 3/2

k = 84/22 = 42/11

Now we can use our value of k in the function.

y = kx^2/z

y = (42/11)(x^2/z)

Now we use our function to find y when x = 1 and z = 3.

y = (42/11)(x^2/z)

y = (42/11)(1^1/3)

y = 42/33

y = 14/11

y = 1 3/11