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Solving Literal Equations (page 1 of 2)

Sections: Solving for a given variable, Solving for "y="

Sometimes you have a formula, such as something from geometry, and you need to solve for some variable other than the "standard" one. For instance, the formula for the perimeter P of a square with sides of length s is P = 4s. It may be that you need to solve this equation for s, so you can plug in a perimeter and figure out the side length.

I don't know why, but this process of solving a formula for a given variable is called "solving literal equations". One of the dictionary definitions of "literal" is "related to or being comprised of letters", and variables are sometimes referred to as literals. So "solving literal equations" may just be another way of saying "taking an equation with lots of letters, and solving for one letter in particular."

At first glance, these problems appear to be much worse than your usual equation solving, but they really aren't that bad.  You pretty much do what you've done all along for solving linear equations, except that, due to all the variables, you won't necessarily be able to simplify your answers as much as you're used to doing. Here's how "solving literal equations" works:

  • Solve A = bh for b

      • b = A/h

  • Solve d = rt for r Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

      • d/t = r

  • Solve P = 2l + 2w for w

      • (P - 2L)/2 = w

  • Solve Q = (c + d)/2 for d

      • 2Q - c = d

  • Solve V = 3k/t for t  

      • t = 3k/V

This next one involves a little "trick" to solve it. See if you can see what it is:

  • Solve Q = 3a + 5ac for a

      • Q/(3 + 5c) = a

Looking at the above example, the "trick" came in the second line, where I factored out the a. This technique doesn't come up often, but it's just about guaranteed to come up in your homework once or twice, and maybe even on the test, because so many students don't see the "trick". When you can't isolate the desired variable because it is split between two or more terms, try to factor it out, and then divide off what's left.

  • Solve A = ( 1/2 )ah – ( 1/2 )bh for h

      • 2A/(a - b) = h

This example used the same "trick" as the previous one. In the third line, I factored out the h. You should expect that you will need to know this!

  • The area A of a sector (pie-wedge-shaped section) of a circle is given by:

      • A = (pi)(r^2)(S)/360

    where r is the radius and S is the angle measure of the sector. Solve for S.

      360A/(pi)(r^2) = S

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Cite this article as:

Stapel, Elizabeth. "Solving Literal Equations." Purplemath. Available from
    http://www.purplemath.com/modules/solvelit.htm. Accessed
 

 

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