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\(x\) \(=\ ?\)

Solving for a
Variable in Formulas

Learn to isolate any variable in a formula by substituting known values and using inverse operations step by step.

Pre-Algebra

What Is a Formula?

A formula is an equation that shows a relationship between two or more variables. You have already seen many formulas in math and science:

\( A = l \times w \)
Area of a rectangle
\( d = r \times t \)
Distance formula
\( P = 2l + 2w \)
Perimeter of a rectangle

What Does "Solve for a Variable" Mean?

Solving for a variable means you are given numeric values for every variable except one. Your job: substitute the known values and then find the unknown.

\( A = l \times w \)
\( 24 = l \times 4 \)
\( l = 6 \)
Formula Substitute Solve

Why This Skill Matters

Formulas appear everywhere: geometry, science, finance, and everyday life. Being able to solve for any variable in a formula gives you the power to answer questions like:

Geometry
If you know the area and width, find the length.
Science
If you know the distance and rate, find the time.
Finance
If you know the interest and rate, find the principal.

The Core Rule: Inverse Operations

To isolate a variable, undo whatever is being done to it by using the inverse (opposite) operation. Whatever you do to one side, you must do to the other side.

\( x + 5 \)
\( 12 \)
=
\( -5 \)
\( -5 \)
\( x = 7 \)

Inverse Operations Reference

If... Undo it by... Example
A number is added to the variable Subtracting that number \( x + 3 = 10 \;\Rightarrow\; x = 7 \)
A number is subtracted from the variable Adding that number \( x - 4 = 9 \;\Rightarrow\; x = 13 \)
The variable is multiplied by a number Dividing by that number \( 5x = 30 \;\Rightarrow\; x = 6 \)
The variable is divided by a number Multiplying by that number \( \frac{x}{3} = 8 \;\Rightarrow\; x = 24 \)

Evaluating vs. Solving

These two tasks look similar but are different. Understanding the distinction helps you choose the right approach.

Evaluating
Plug in all variable values and simplify to get a single number.

Example: Find \( A \) when \( l = 5 \), \( w = 3 \).
\( A = 5 \times 3 = 15 \)
Solving
Plug in some values, then use inverse operations to find the remaining unknown.

Example: Find \( l \) when \( A = 15 \), \( w = 3 \).
\( 15 = l \times 3 \;\Rightarrow\; l = 5 \)

How to Apply the Process

1
Write down the formula. Identify every variable and what it represents.
2
Identify the unknown. Which variable are you solving for?
3
Substitute. Replace every known variable with its given value.
4
Simplify any arithmetic you can (multiply, add, etc.).
5
Use inverse operations to isolate the unknown variable.
6
Check. Plug your answer back into the original formula to verify.

Example 1: Area of a Rectangle

The area of a rectangle is \( A = l \times w \). If \( A = \) 36 and \( w = \) 4, find \( l \).

\( w = 4 \)
\( l = ? \)
\( A = 36 \)
1
Formula: \( A = l \times w \)
2
Substitute: \( 36 = l \times 4 \)
3
Divide both sides by 4: \( \frac{36}{4} = l \)
4
Result: \( l = 9 \)

Example 2: Perimeter

The perimeter of a rectangle is \( P = 2l + 2w \). If \( P = \) 30 and \( l = \) 9, find \( w \).

1
Formula: \( P = 2l + 2w \)
2
Substitute: \( 30 = 2(9) + 2w \)
3
Simplify: \( 30 = 18 + 2w \)
4
Subtract 18 from both sides: \( 12 = 2w \)
5
Divide both sides by 2: \( w = 6 \)

Example 3: Distance Formula

The distance formula is \( d = r \times t \). A car travels 180 miles at a rate of 60 mph. Find the time \( t \).

Start End
\( d = 180 \text{ mi} \)
\( r = 60 \text{ mph}, t = ? \)
1
Formula: \( d = r \times t \)
2
Substitute: \( 180 = 60 \times t \)
3
Divide both sides by 60: \( \frac{180}{60} = t \)
4
Result: \( t = 3 \text{ hours} \)

Example 4: Temperature Conversion

The formula to convert Celsius to Fahrenheit is \( F = \frac{9}{5}C + 32 \). If \( F = \) 68, find \( C \).

1
Substitute: \( 68 = \frac{9}{5}C + 32 \)
2
Subtract 32 from both sides: \( 36 = \frac{9}{5}C \)
3
Multiply both sides by \( \frac{5}{9} \): \( C = 36 \times \frac{5}{9} \)
4
Simplify: \( C = 20 \)

The Order of Undoing

When a formula has multiple operations on the unknown variable, undo them in reverse order (like peeling off layers):

Layer 2: \(+32\) (undo FIRST: subtract 32)
Layer 1: \(\times \frac{9}{5}\) (undo SECOND: \(\times \frac{5}{9}\))
\( C \)

Common Mistakes to Avoid

✕ Forgetting to substitute
Trying to rearrange the formula without plugging in the given values first. Always substitute known values before solving.
✕ Undoing in the wrong order
Dividing before subtracting when you should subtract first. Undo the outermost operation first.
✕ One-sided operations
Performing an operation on only one side of the equation. Whatever you do to the left, you must do to the right.

Formula Reference Table

Formula Solving for... Steps
\( A = l \times w \) \( l \) Divide: \( l = \frac{A}{w} \)
\( P = 2l + 2w \) \( w \) Subtract \( 2l \), then divide by 2
\( d = r \times t \) \( t \) Divide: \( t = \frac{d}{r} \)
\( F = \frac{9}{5}C + 32 \) \( C \) Subtract 32, then multiply by \( \frac{5}{9} \)
\( I = P \times r \times t \) \( P \) Divide: \( P = \frac{I}{r \times t} \)

Question 1 of 5

The formula for the area of a triangle is \( A = \frac{1}{2}bh \). If \( A = 30 \) and \( b = 10 \), what is \( h \)?

Question 2 of 5

Using \( d = r \times t \), a cyclist rides 48 miles in 4 hours. What is the rate \( r \)?

Question 3 of 5

The perimeter of a rectangle is \( P = 2l + 2w \). If \( P = 40 \) and \( l = 12 \), what is \( w \)?

Question 4 of 5

Using \( F = \frac{9}{5}C + 32 \), if \( F = 212 \), what is \( C \)?

Question 5 of 5

Simple interest is \( I = P \times r \times t \). If \( I = 120 \), \( r = 0.04 \), and \( t = 3 \), what is the principal \( P \)?