Rearranging
Formulas
Learn to isolate any variable in a formula using the same inverse-operation reasoning you already use to solve equations.
Algebra 2
What Does "Rearrange" Mean?
A formula is an equation that relates two or more variables. Rearranging means rewriting that formula so a different variable is alone on one side: you are solving for that variable in general, not for a specific number.
Why Rearrange Formulas?
Efficiency
Solve for a variable once, then plug in different values without re-deriving every time.
Insight
See how one quantity depends on the others: direct variation, inverse variation, or something more complex.
Flexibility
Science, finance, and engineering problems rarely hand you the formula in the form you need.
A Visual Example: Ohm's Law
The Core Principle
The operations you already know: add, subtract, multiply, divide, square, square-root, reciprocal: they all apply. The only new twist is that your answer is an expression, not a single number.
Inverse Operations Toolkit
| Operation on target | Undo with | Example |
|---|---|---|
| Added to something | Subtract it | \( y = x + c \;\Rightarrow\; x = y - c \) |
| Multiplied by something | Divide by it | \( A = lw \;\Rightarrow\; l = \frac{A}{w} \) |
| In a denominator | Multiply to clear denominator, then isolate | \( d = \frac{v}{t} \;\Rightarrow\; t = \frac{v}{d} \) |
| Squared | Take the square root | \( A = \pi r^2 \;\Rightarrow\; r = \sqrt{\frac{A}{\pi}} \) |
| Under a square root | Square both sides | \( v = \sqrt{2gh} \;\Rightarrow\; h = \frac{v^2}{2g} \) |
Solving Equations vs. Rearranging Formulas
| Solving an Equation | Rearranging a Formula | |
|---|---|---|
| Goal | Find a number | Find an expression |
| Given | Constants + one variable | Multiple variables |
| Strategy | Inverse operations | Inverse operations (identical!) |
| Answer form | \( x = 4 \) | \( R = \frac{V}{I} \) |
How to Apply It: Step by Step
1
Identify the target variable: which quantity do you want isolated?
2
Locate the target: is it added? Multiplied? Squared? In a denominator?
3
Undo outer operations first: work from the outside in, applying inverse operations to both sides.
4
Repeat until the target variable stands alone on one side.
5
Simplify the expression on the other side if possible.
Example 1: Solve \( d = rt \) for \( t \)
Example 2: Solve \( P = 2l + 2w \) for \( w \)
1
Start: \( P = 2l + 2w \)
2
Subtract \( 2l \) from both sides: \( P - 2l = 2w \)
3
Divide both sides by 2: \( \frac{P - 2l}{2} = w \)
Example 3: Solve \( E = mc^2 \) for \( c \)
1
Start: \( E = mc^2 \)
2
Divide both sides by \( m \): \( \frac{E}{m} = c^2 \)
3
Take the square root of both sides: \( c = \sqrt{\frac{E}{m}} \)
Example 4: Solve \( I = \frac{V}{R} \) for \( R \)
1
Start: \( I = \frac{V}{R} \). Target variable \( R \) is in the denominator.
2
Multiply both sides by \( R \): \( IR = V \). Now \( R \) is out of the denominator.
3
Divide both sides by \( I \): \( R = \frac{V}{I} \).
When the Target Appears Twice
Sometimes the target variable appears in more than one term. In that case, collect terms and factor it out.
1
Example: Solve \( ax + bx = c \) for \( x \).
2
Factor out \( x \): \( x(a + b) = c \).
3
Divide both sides by \( (a + b) \): \( x = \frac{c}{a + b} \).
Multi-Step: Solve \( S = \frac{n}{2}(a + l) \) for \( a \)
1
Multiply both sides by 2: \( 2S = n(a + l) \)
2
Divide both sides by \( n \): \( \frac{2S}{n} = a + l \)
3
Subtract \( l \) from both sides: \( a = \frac{2S}{n} - l \)
Common Mistakes
✕ Dividing only one term
When solving \( P - 2l = 2w \) for \(w\), you must divide the entire other side by 2. Correct: \( w = \frac{P - 2l}{2} \).
✕ Forgetting to flip operations
If the variable is subtracted, you must add to undo it: not subtract again.
✕ Dropping the square root
After isolating \( x^2 \), many students write \( x = \frac{A}{k} \) instead of \( x = \sqrt{\frac{A}{k}} \).
Summary of Strategies
| Situation | Strategy | Example Result |
|---|---|---|
| Target is multiplied | Divide both sides | \( t = \frac{d}{r} \) |
| Target is added/subtracted | Subtract/add to isolate | \( w = \frac{P-2l}{2} \) |
| Target is in a denominator | Multiply by denominator first | \( R = \frac{V}{I} \) |
| Target is squared | Isolate, then square-root | \( c = \sqrt{\frac{E}{m}} \) |
| Target appears twice | Collect terms, factor out | \( x = \frac{c}{a+b} \) |
Question 1 of 5
Rearrange \( A = \frac{1}{2}bh \) to solve for \( h \).
Multiply both sides by 2 to get \( 2A = bh \), then divide by \( b \) to isolate \( h = \frac{2A}{b} \).
Question 2 of 5
Solve \( F = ma \) for \( m \).
Since \( m \) is multiplied by \( a \), divide both sides by \( a \): \( m = \frac{F}{a} \).
Question 3 of 5
Rearrange \( C = 2\pi r \) to solve for \( r \).
\( r \) is multiplied by \( 2\pi \). Divide both sides by \( 2\pi \): \( r = \frac{C}{2\pi} \).
Question 4 of 5
Solve \( A = \pi r^2 \) for \( r \).
Divide by \( \pi \) first: \( \frac{A}{\pi} = r^2 \). Then take the square root: \( r = \sqrt{\frac{A}{\pi}} \). Remember: square-root undoes squaring.
Question 5 of 5
Rearrange \( y = mx + b \) to solve for \( x \).
Subtract \( b \) from both sides: \( y - b = mx \). Then divide by \( m \): \( x = \frac{y - b}{m} \).