Linear Equations • Topic 4 of 4

Literal Equations

A literal equation contains several letters, and you are asked to solve for one of them in terms of the others — rearranging a formula rather than finding a number. The technique is identical to solving a numeric equation: treat the target variable as the unknown and every other letter as if it were a constant, then isolate it with inverse operations. To solve d = rt for t, divide both sides by r to get t = d/r. When the target appears in two terms, factor it out first: from P = 2L + 2W, solving for L gives L = (P − 2W)/2. This is exactly what the GRE tests when it hands you a physics-style or finance-style formula and asks you to express one variable in terms of the rest, and it is the setup skill behind many word problems.

✅ Solved examples

1. Solve d = rt for r.
Divide both sides by t: r = d/t.
2. Solve P = 2L + 2W for W.
Subtract 2L: P − 2L = 2W. Divide by 2: W = (P − 2L)/2.
3. Solve A = (1/2)bh for h.
Multiply both sides by 2: 2A = bh. Divide by b: h = 2A/b.
4. Solve y = mx + c for x.
Subtract c: y − c = mx. Divide by m: x = (y − c)/m.

✏️ Practice — try these, take hints as needed

1. Solve V = lwh for h.
h is multiplied by lw.
Divide both sides by lw.
h = V/(lw).
h = V/(lw)
2. Solve C = 2πr for r.
r is multiplied by 2π.
Divide by 2π.
r = C/(2π).
r = C/(2π)
3. Solve ax + b = c for x.
Subtract b.
ax = c − b.
Divide by a.
x = (c − b)/a
4. Solve F = (9/5)C + 32 for C.
Subtract 32.
F − 32 = (9/5)C.
Multiply by 5/9.
C = (5/9)(F − 32)
5. Solve S = n(a + b) for b.
Divide both sides by n.
S/n = a + b.
Subtract a.
b = S/n − a

📝 Topic test — 8 questions

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