Understanding the Concept of Variation and Practical Examples using different Types of Variation

Variation problems in algebra involve formulas which show the relationship between two or more variables. The types of variation we will learn about in this article are direct, inverse, joint and combined.

Direct variation is represented as a linear equation with one variable found by multiplying the other by a constant. For example, the formula for circumference of a circle is C = 2∏R. Notice that C is always found by  multiplying R by the constant 2∏. In terms of variation, we say that C varies directly with R or C is directly proportional to R. In general, if y varies directly with x, then y = kx for some constant k, (k ≠ 0). The constant k is known as the constant of variation.

There is a step by step process to solve problems involving variation. First translate the verbal model into an equation. Substitute the values given into the equation and solve for the constant k. Put the value of k back into the equation and all other remaining values. Then solve for the unknown variable.

Example: y varies directly with x. When y is 9, x is 4. What is y when x = 6 ?

Step 1: Write the equation.

Since y varies directly with x, y = kx.

Step 2: Substitute the values in for x and y and solve for k.

9 = k(4)

9 = 4k

9/4 = k

Step 3: Substitute k back into the original equation as well as 6 for x and solve for y.

y = (9/4)(6)

y = 54/4, simplifies to 27/2

There are some practical purposes for direct variation. Suppose you are interested in converting US dollars to Philippine pesos. The number of pesos received is directly proportional to the number of dollars to be exchanged. If 21 US dollars equals 894 pesos,how many pesos will you receive for \$1,400 dollars?

Let y = number of pesos and x = number of US dollars and k = constant of variation.

The equation for this problem is 894 = k(21)

894 = 21k

42.57 = k

To find out how many pesos you will receive for \$1,400, substitute 42.57 for k and \$1,400 for x.

y = 42.57(1,400)

y = 59,598

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.

Example: Suppose y varies inversely with x. If y = 6 when x = 13, what is x when y = 7 ?

Set up the equation for inverse variation and solve for k.

y = k/x

6 = k/13

78 = k

Substitute 78 for k and 7 for y into the original equation and solve for x.

7 = 78/x

7x = 78

x = 78/7

Here’s a practical example using inverse variation. On a dance floor, the amount of floor space per person is inversely proportional to the number of people. If n represents the number of people and s represents the amount of square feet of floor space per person, then s = k/n represents the relationship between s and n.

If there are 15 people on the dance floor and each of them have 10 square feet of floor space, how much floor space will each person have if there are 20 people on the dance floor?

Substitute 15 for n and 10 for s and solve for k.

10 = k/15

150= k

Now substitute 150 for k and 20 for n and solve for s.

s = 150/20

s = 7.5 square feet

Here’s an example involving inverse variation when one of the variables is squared.  If y varies inversely with the square of x and y is 6 when x is 4, what is y when x is 10 ?

The equation for this is y = k/x2 (notice the difference from y = k/x since the variation is with the square of x)

Substitute the value for y and x and solve for k.

6 = k/(42)

6 = k/16

96 = k

Now substitute 16 for k and 10 for x in the original equation and solve for y.

y = 96/(102)

y = 96/100

y= 24/25 or 0.96

A relationship between variables is called joint variation when one variables varies directly with the product of two or more variables. An example of joint variation is the formula for the area of a triangle. Recall that the area of a triangle is . the base times the height. (A = (1/2)bh)In general, if y varies jointly with x and z, then y = kxz, where k is the constant of variation and k > 0.

Example: Suppose y varies jointly with x and z and y is 10 when x is 3 and z is 2. What is the value of y when x is 20 and z is 3 ?

Start with the equation y = kxz and substitute 12 for y, 3 for x and 2 for z and solve for k.

10 = k(3)(2)

10 = 6k

5/3 = k

Now substitute 2 for k, 8 for x and 5 for z in the original equation and solve for y.

y = (5/3)(20)(3)

y = 100

Here’s a real life situation involving joint variation. Suppose the costs incurred by a lumber hauling company varies jointly with the number of trucks used to haul the lumber and the number of hours each truck is used. The cost is \$3,400 when 6 trucks are used for 5 hours each. What are the costs when 8 trucks are used for 7 hours each ?

Let c = costs, n = number of trucks and h = hours each truck is used. Therefore the equation used is

c = khn. Substitute \$3,400 for c, 5 for h and 6 for n. Then solve for k.

3,400 = k(5)(6)

3,400 = 30k

113 1/3 = k

Now substitute 113 1/3 for k, 7 for h and 6 for n in the original equation and solve for c.

c = (113 1/3)(7)(6)

c = (113 1/3)(42)

c = \$4, 760 (multiply on a calculator or change 113 1/3 to 340/3 and multiply by 42 to get (340/3)(42) = 340(14)

= \$4,760)

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.

Suppose that y varies directly with x and inversely with z^2. If y = 8, when x = 4 and z = 2, 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 z^2 is y = k/z^2.

Therefore the equation for the combined variation is y = kx/z^2.

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

8 = k(4)/(2^2)

8 = 4k/4

8 = k

Now substitute 8 for k, 2 for x and 4 for z in the original equation and solve for y.

y = 8(2)/(4^2)

y = 1

Here’s a real life example using combined variation.  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 12 hours for 5 workers to mow a rectangular area of land 1000 feet long and 550 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.

12 = k(1000)(550)/5

10 = 550000k/5

k = 1/11000

Now substitute 1/11000 for k, 2000 for l, 340 for w, 8 for n and solve for t.

t = (1/11000)(2000)(340)/8

t = 7.7 hours

Hope the explanations and examples in this article makes understanding and working with problems involving direct variation much easier.