Literal Equations
A free Algebra I lesson from the “Algebra Foundations” unit, with a worked example and practice problems including step-by-step solutions.
A literal equation has two or more variables. Solving for one variable means isolating that variable so the formula is ready to use in a new situation. In Algebra Foundations, students need more than a memorized rule: they need to recognize the structure, select a method, carry out the algebra cleanly, and interpret the answer in a graph, table, equation, or real context. The expanded practice now mixes skill fluency, transfer questions, and cumulative review so the lesson builds durable Algebra I readiness.
What you'll learn
- Solve formulas for a chosen variable
- Undo operations while keeping equations balanced
- Interpret rearranged formulas
Worked example
Problem. Solve A = lw for w.
- The target variable is w.
- w is multiplied by l, so divide both sides by l.
- w = A/l.
- Connect the result back to Literal Equations so the method and meaning are both clear.
Answer: w = A/l
Lesson preview: this page shows 5 practice items. The interactive activity contains 28 practice items and 15 quiz items (multiple choice and typed response). Counts include repeated prompts and variants.
Practice problems
1. Solve A = lw for w.
Choices: w = A/l · w = Al · w = l/A · w = A - l
Show solution
- w is multiplied by l.
- Divide both sides by l.
Answer: w = A/l
2. Solve y = x + b for b. Enter the expression for b.
Show solution
- Subtract x from both sides.
- b = y - x.
Answer: y - x
3. Solve p = 2l + 2w for p.
Choices: p = 2l + 2w · p = l + w · p = 4lw · p = 2(l - w)
Show solution
- p is already isolated.
- No rearranging is needed.
Answer: p = 2l + 2w
4. Solve d = rt for t.
Choices: t = d/r · t = dr · t = r/d · t = d - r
Show solution
- t is multiplied by r.
- Divide both sides by r.
Answer: t = d/r
5. Solve C = 2pi r for r.
Choices: r = C/(2pi) · r = 2pi/C · r = C - 2pi · r = 2Cpi
Show solution
- r is multiplied by 2pi.
- Divide both sides by 2pi.
Answer: r = C/(2pi)
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