Variation problems involve formulas which show the relationship between two or more variables. In some formulas, when one value increases, another value decreases. An example of such a formula is h = 20/b. In such cases, we say that h varies inversely with b or h is inversely proportional to b. In general, if y varies inversely with x then y = k/x for some value of k, k ≠ 0. Direct variation is represented as a linear equation with one variable found by multiplying the other by a constant. In general, if y varies directly with x, then y = kx for some constant k, (k ≠ 0). Many problems combine direct and inverse variation. This is known as combined variation. In general, if y varies directly with x and inversely with z then y = kx/z, where k ≠ 0. If the problem stated that y is directly proportional to x, the equation would be y = kx. Notice the inverse variation has z in the denominator. If the problem stated that y varies inversely with z, the equation would be y = k/z. Combining the two gives us y = kx/z.

**Example:** Suppose that y varies directly with x and inversely with z2. If y = 6, when x = 4 and z = 3, what is the value of y when x = 2 and z = 4 ?

The equation for y varies directly with x is y = kx. The equation for y varies inversely with z2 is y = k/z2. Therefore, the equation for the combined variation is y = kx/z2.

Substitute the values for x, y and z in the original equation and solve for k.

6 = k(4)/(32)

6 = 4k/9

54 = 4k

13 ½ = k

Now substitute 13 ½ for k, 2 for x and 4 for z in the original equation and solve for y.

y = (13 1/2)(2)/(4)2

y = 27/16

**Example:** The time it takes to mow a rectangular area of grass varies directly with the length and width of the area to be mowed and inversely with the number of workers. If it takes 10 hours for 5 workers to mow a rectangular area of land 1000 feet long and 450 feet wide, how long will it take 8 workers to mow an area of grass 2000 feet long and 345 feet wide?

Let t = mowing time, l = length, w = width and n = number of workers. The equation is t = klw/n (t = klw is direct variation, t = k/n is the inverse variation). Substitute the values in for t, l, w and n and solve for k.

10 = k(1000)(450)/5

10 = 450000k/5

k = 1/9000

Now substitute 1/9000 for k, 2000 for l, 340 for w, 8 for n and solve for t.

t = (1/9000)(2000)(340)/8

t = 9.4 hours (multiplied and divided on a calculator)

When solving problems involving combined variation, it’s important to first understand direct and inverse variation. Then see how both types of variation are used in a single problem.

This guide with examples should ease any confusion on the topic of combined variation.