Variation is a concept that is based more on the concept of ratio. In general, variation implies change in any condition or level or amount or limits for any given quantities. For example, we say that as labor increases, the wages increases. So there is difference in wages with a minor change in labor even. For if labor increases wages to be paid will increase and if labor will decrease so will be wages. Also if labor increases, time taken will be less and if it decreases, time taken will be more. There are examples of direct and inverse proportion respectively. There are many types of variation depending upon various factors like direct and inverse variation or combined variation, joint variation.

A combined variation, as the name implies, consists of both inversely and directly related variables. For example: speed varies directly with distance traveled and inversely with time. And together they form the formula for speed.

In general, we say that if a quantity is dependent on two or more quantities as directly with some and inversely with some, then we obtain an equality equation of that quantity by combining all other quantities with a constant of proportionality. 
If y varies directly with x and inversely with z then we write a combined equation as y = k ($\frac{x}{z}$) where k is a constant of proportionality.

This equation has an inverse relation and also a direct relation and since both are combined in one, that is why the name it combined variation.
In order to find the combined variation between given quantities as required we follow the following steps.
1) The first and most important step is to write an equation based on the information given in problem. Forming right equation is most important. The variables can be named in accordance to the data items given in question. Also, the problem should be read carefully in order to take in account any changes for like cubes, roots, squares etc.

2) Now use the information given in the problem and solve equation in step 1 to find the value of the constant of proportionality used in step 1.

3) Now we substitute the value of constant obtained in step 2 in the equation of step 1 in order to obtain a required general combined variation equation as given in question. Only this equation can lead us to correct answers as required.

4) Use the data from the question to obtain other values from the equation in step 3 as given and required.
This way we can obtain any general combined variation equation depending upon the given problem and thus use it to find numerous other values. Also, always use units in final answer as given and implies. It is important to understand the question and variations given properly to determine an accurate answer. And also correct equation. If any misinterpretation is done, answer will not be accurate.
Let us look at some examples on combined variation for a better understanding. We will take up both numeric and word problems for better understanding and also to learn real life application of the concept of combined variation.

Example 1:  

It is given that z varies directly with y and inversely with x. Also, when x = 24 and y = 48, z = 4. Find the value of z when x = 4 and y = 12
Solution: 

Z varies directly with y and inversely with x, this implies equation is

Z = k ($\frac{y}{x}$) where k is constant

It is given that y = 48, x = 24 implies z = 4

4 = k ($\frac{48}{24}$)

$\Rightarrow$ 4 = 2k

$\Rightarrow$ K = 2

Therefore, equation is z = 2 $\frac{y}{x}$

Now at y = 12 and x = 4 we have

Z = 2 $\frac{(12)}{4}$ = 6


Example 2:

The stopping distance of a car varies inversely with the friction of the road but directly with the speed of the car. At the speed of 20, car takes 
50 feet to stop on the road with friction value of 2. Find the stopping distance at a speed of 40 with road friction value to be 4.
Solution: 

Let d be the stopping distance, f be the friction value, s be the speed.

Then according to the question we have

D = $\frac{k s}{f}$ where k is constant

We are given that s = 20 and f = 2 implies d = 50

50 = 20 $\frac{k}{2}$

$\Rightarrow$ 20k = 100

$\Rightarrow$ k = 5

Required equation therefore is d = 5 $\frac{s}{f}$

At s = 40 and f = 4, we have

D = 5 $\frac{(40)}{4}$ = 50

Therefore, the required stopping distance is 50 feet.