Question

Suppose that y varies directly with x and inversely with z, and y = 50 when x = 15 and z = 3. Write the equation that models the relationship. Then find y when x = 8 and z = 4.

Answer 1

Any variation problems are either in the form of y = kx or y = (k/x) where k is an unknown constant always in the numerator. When you have something that varies directly or jointly by y, this something is placed in the numerator of the equation. So y varies directly with x simply means y = kx. Anything that varies inversely to y is placed in the denominator of the equation. So when z varies inversely to y, we have y = k/z. Combining both of these conditions, we can say that y = kx/z because x varies directly (numerator) and z varies inversely (denominator), with the unknown k always on top. Now that we have the set up, we'll plug in our values and find k. Plugging in the values gives us
50 = 15k/3. Solving for k comes out to k = 10. So our equation that models the relationship is then y = 10x/z. Now all we need to do for part 2 is find y when x = 8 and z = 4. Plugging these 2 values in, I have y = 10(8)/(4), which means y = 20.