Return to the Purplemath home page


Try a demo lesson Join Login to


Index of lessons | Purplemath's lessons in offline form |
Forums | Print this page (print-friendly version) | Find local tutors


Variation Equations (page 1 of 3)

Variation problems aren't hard once you get the hang of the lingo. The only real difficulty is learning the somewhat specialized vocabulary and the techniques for this classification of problems.

Variation problems involve fairly simple relationships or formulas, involving one variable being equal to one term. That term might be linear (something with just an "x"), quadratic (something in "x2"), more than one variable (such as "r2h"), a square root (something like "sqrt(4 - r^2)"), or something else. But it is always just the one term in the formula, multiplied by some number, usually denoted by "k" if you don't yet know the number's value; this number k is called "the constant of variation".

An example of a variation equation would be the formula for the area of the circle: A = (pi) r^2. In the language of variation, "the area A varies directly with the square of the radius r"; the constant of variation is k = (pi). This formula is an example of "direct" variation. "Direct variation" means that, in the one term of the formula, the variable is "on top".




On the other hand, "inverse variation" means that the variable is underneath, in the bottom of a fraction. Suppose, for instance, that you inherit a money market account containing $100,000, and you wonder how much money your rich uncle initially invested eight years ago. Depending on the average interest rate "r", the formula you would use would be:

    P = 100000 / (1 + r/12)^96

...where P is the principal your uncle invested. (This formula is a variant of the compound-interest formula, by the way.) In the language of variation, this formula reads as "the principal P varies inversely with (1 + r/12)^96", with the constant of variation being k = 100,000.

The other case of variation is "jointly". "Joint variation" means "directly, but with two or more variables". An example would be the formula for the area of a triangle with base "b" and height "h":  A = (1/2)bh. In words, "the area A varies jointly with b and h"; the constant of variation is k = 1/2.

To review:

  • "F varies as x" means F = kx
  • "F varies jointly as x and y" means F = kxy
  • "F varies as x + y" means F = k(x + y)
  • "F varies inversely as x" means F = k/x

Be careful with those middle two. Almost always, when you translate word problems from English into math, "and" means "plus" or "added to". But in joint variation, "and" just means "both of these are together on the same side of the fraction" (usually on top), and you multiply. If you are supposed to add two variables, they'll use the format in that third bulleted example above, or they'll say "varies as the sum of x and y.

Translating variation problems isn't so bad, once you get the hang of it. But then they want you to move on to setting up and solving word problems. These generally fall into two categories: the ones where they want you to find the value of "k", and the ones where they want you to find some other value, but only after you've found "k" first. Here are some examples:

  • If y varies directly as x2, and y = 8 when x = 2, find y when x = 1.

    Since this is direct variation, the formula is "y = kx2". The reason they've given me the data point (x, y) = (2, 8) is that I have to be able to find the value of "k". So I'll plug in the information they've given me, and solve for k:

      y = kx2
      8 = k(22)
      8 = 4k
      2 = k

    Now that I have k, I can rewrite the formula completely:  y = 2x2. With this, I can answer the question they actually asked: "Find y when x = 1."

      y = 2x2
      y = 2(1)2
      y = 2×1
      y = 2

    Then the answer is:  y = 2 Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

  •  If y varies directly as x and z, and y = 5 when x = 3 and z = 4,
    then find
    y when x = 2 and z = 3.

    Translating the formula from English to math, I get:

      y = kxz

    Plugging in the data point they gave me, and solving for the value of k, I get:

      5 = k(3)(4)
      5 = 12k
      5/12 = k

    Now that I have the value of k, I can plug in the new values, and solve for the new value of y:

      y = ( 5/12)xz
      y = ( 5/12)(2)(3)
      y = ( 5/12)(6)
      y = 5/2

    Then the answer is: y = 5/2

Top  |  1 | 2 | 3  |  Return to Index  Next >>

Cite this article as:

Stapel, Elizabeth. "Variation Equations." Purplemath. Available from Accessed



This lesson may be printed out for your personal use.

Content copyright protected by Copyscape website plagiarism search

  Copyright 2000-2014  Elizabeth Stapel   |   About   |   Terms of Use   |   Linking   |   Site Licensing


 Feedback   |   Error?