Overview
A literal equation expresses a relationship among several quantities. Solve for the requested letter using the same equality-preserving steps as a numerical equation, treating the other letters as fixed inputs. When the requested letter appears in multiple terms, collect and factor it before division.
Consider this example: If p = 2a + 2w, solve for w. Treat a as fixed. Subtract 2a to get 2w = p − 2a. Divide by 2: w = p/2 − a. The entire numerator p − 2a must be divided by 2; p − 2a itself is twice w.
The method also handles a different case: For h ≠ 0, solve a = bh/2 for b. Multiply by 2: 2a = bh. Divide by nonzero h to get b = 2a/h. The choice a/(2h) divides by 2 instead of undoing the original division.
For a further application, consider: Solve p = ax + b(x − p) for x, assuming a + b ≠ 0. Expand the right side to (a + b)x − bp. Add bp, obtaining p(1 + b) = (a + b)x. Divide by the nonzero coefficient a + b: x = p(1 + b)/(a + b). Both the repeated target x and the repeated parameter p must be gathered before division.
Dividing by a sum term by term does not isolate a variable. A formula containing a denominator also needs the denominator’s nonzero condition. Name the target letter and group every term containing it. After solving, substitute the formula back into the original relationship and state the exceptional inputs where division is unavailable.
A target in both numerator and denominator needs collection before division, and its original denominator still controls the answer. Solve p = (x − a)/(x + b), assuming p ≠ 1 and a + b ≠ 0. First record x ≠ −b. Multiply by x + b to obtain px + pb = x − a, then collect x terms: (p − 1)x = −a − pb. Thus x = (a + bp)/(1 − p). The first assumption makes this new denominator nonzero. To check the old restriction, add b to the solution: x + b = (a + b)/(1 − p), which is nonzero by both assumptions. A rearrangement that gives the formula without this last check has not established that it solves the original rational equation.