Math101 · Independent Ontario learning library
Math101learn.math101.caLiteral Equations
A literal equation contains two or more variables and is often a formula. Solving for one variable rewrites the same relationship with that variable isolated.
Literal rearrangement lets one model answer different questions and supports science formulas, geometry, rates, and later algebraic manipulation.
Intuition and core definition
A literal equation contains two or more variables and is often a formula. Solving for one variable rewrites the same relationship with that variable isolated. Other variables are treated as quantities, not numbers to guess. Valid algebraic operations must be applied to both sides and any division introduces a nonzero condition.
Notation, language, and conditions
In $A=\frac12bh$, solving for $h$ gives $h=\frac{2A}{b}$ with $b\ne0$. The requested variable is the subject. Equivalent formulas have the same valid tuples of values within their stated domains. Factoring may be needed before dividing when the target appears in multiple terms.
Why this idea matters
Rearranging a literal equation isolates one quantity while preserving a relationship that can later accept many different data sets.
A dependable method
- Circle the target variable and note any restrictions.
- Clear fractions or distribute only when doing so simplifies access to the target.
- Move all terms containing the target to one side and all other terms to the other.
- Factor the target if it occurs in more than one term.
- Divide by its remaining coefficient, state restrictions, and verify by substitution.
Worked example
Representations and interpretation
A formula triangle or labelled diagram can show which quantities are connected, but algebra records the reversible transformations. A dependency diagram changes its chosen output when the formula is rearranged; the underlying relation remains the same.
Reasoning about variations
For $y=ax+bx$, the target $x$ appears twice. Factoring gives $y=x(a+b)$, then $x=y/(a+b)$ provided $a+b\ne0$. Dividing each term by a different symbol before factoring can obscure this necessary condition.
Common mistakes
How to check your work
- Substitute the rearranged expression back into the original and simplify to an identity.
- Test with a convenient numerical case satisfying restrictions.
- Use units: both sides of the isolated formula must have the target variable’s units.
Practice
- Solve $d=rt$ for $t$.
- Solve $A=\frac12bh$ for $b$.
- Solve $y=mx+b$ for $x$.
Answers and brief solutions
Show answers
- $t=\frac dr$ $t=d/r$ after dividing by $r$, with $r\ne0$.
- $b=\frac{2A}{h}$ Multiply by $2$, then divide by $h\ne0$.
- $x=\frac{y-b}{m}$ Subtract $b$ and divide by $m\ne0$.
Synthesis and transfer
Solving a distance formula for time reveals the nonzero-rate condition; units then confirm that distance divided by rate produces a duration.
From $d=rt$, isolating time gives $t=d/r$ only when $r\ne0$. Dimensional cancellation—distance divided by distance per time—leaves time and supports the rearrangement. The zero-rate case deserves separate interpretation: if $r=0$ and $d\ne0$, the model is inconsistent, while $r=0$ and $d=0$ does not determine a unique elapsed time. Such cases are easy to lose when division is treated as a purely visual move. Rearranging symbolically before inserting numbers also exposes how one quantity depends on the others; doubling distance at fixed rate doubles time, whereas doubling rate at fixed distance halves it.
Related topics
Teaching and accessibility note
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
Solve $d=rt$ for $t$.
- $t=d/r$ after dividing by $r$, with $r\ne0$.
End of lesson
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