6.5 Literal Equations & Rearranging Formulas

Literal Equations & Rearranging Formulas

Every formula you have memorized is really three or four formulas, and TABE Level A expects you to produce the one you need. The DRC specification lists two related requirements: "solve linear equations and inequalities in one variable, including equations with coefficients represented by letters" and "rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations."

What Makes an Equation "Literal"

A literal equation contains two or more letters. Solving it does not produce a number — it produces an expression.

The procedure is identical. The only mental adjustment: every letter that is not your target is treated as an ordinary number. In $ax + b = c$, if $a$ were 3 you would divide by 3, so with $a$ you divide by $a$.

The Four-Step Isolation Procedure

1. Name the target. Circle the variable you are solving for.
2. Clear anything added or subtracted from the target's side, using inverse operations on both sides.
3. Clear anything multiplied or divided by the target.
4. Check by substituting easy numbers into both the original and your rearranged version.

Worked example 1: $P = 2l + 2w$, solve for $w$

$P = 2l + 2w$ $P - 2l = 2w \qquad \text{(subtract } 2l \text{ from both sides)}$ $\frac{P - 2l}{2} = w \qquad \text{(divide both sides by 2)}$

$\mathbf{w = \frac{P - 2l}{2}}$

Check with numbers. If $l = 5$ and $w = 3$, then $P = 16$. Feeding $P = 16$, $l = 5$ into the new formula: $(16 - 10)/2 = 3$. ✓

A common error: writing $w = \dfrac{P}{2} - 2l$. Dividing by 2 must apply to the entire left side, not just the first term.

Worked example 2: $A = \frac{1}{2}bh$, solve for $h$

$A = \tfrac{1}{2}bh \;\Rightarrow\; 2A = bh \;\Rightarrow\; \mathbf{h = \frac{2A}{b}}$

Multiply away the fraction first; it keeps the algebra clean.

Worked example 3: $I = Prt$, solve for $r$

$I = Prt \;\Rightarrow\; \mathbf{r = \frac{I}{Pt}}$

All three of $P$, $r$, and $t$ are multiplied together, so isolating any one of them is a single division by the product of the other two.

Worked example 4: $C = \frac{5}{9}(F - 32)$, solve for $F$

$C = \tfrac{5}{9}(F - 32)$ $\tfrac{9}{5}C = F - 32 \qquad \text{(multiply both sides by } \tfrac{9}{5}\text{)}$ $\mathbf{F = \tfrac{9}{5}C + 32}$

This derivation is worth doing once by hand, because it turns two memorized temperature formulas into one.

When the Target Appears Twice

If the variable you want shows up in more than one term, you cannot divide it away directly. Collect, factor, divide.

Solve $ax + 3 = bx - 7$ for $x$:

$ax - bx = -7 - 3 \qquad \text{(collect } x \text{ terms left, constants right)}$ $x(a - b) = -10 \qquad \text{(factor out } x\text{)}$ $\mathbf{x = \frac{-10}{a - b}}$

Solve $S = \dfrac{n}{2}(a + L)$ for $L$:

$2S = n(a + L) \;\Rightarrow\; \frac{2S}{n} = a + L \;\Rightarrow\; \mathbf{L = \frac{2S}{n} - a}$

The Formulas TABE Rearranges Most

Why This Matters on Test Day

Word problems frequently hand you the output of a formula and ask for an input:

A cylindrical tank holds 1,570 cubic feet and has a radius of 10 feet. How tall is it? (Use $\pi \approx 3.14$.)

Rearranging first is cleaner than substituting first:

$h = \frac{V}{\pi r^2} = \frac{1{,}570}{3.14 \times 100} = \frac{1{,}570}{314} = \mathbf{5 \text{ feet}}$

One rearrangement, one division. Substituting first and then unwinding the equation takes twice as long and doubles the chances of a sign or division error.