2.5 Solving Literal Equations (Formulas)

Scott McDaniel

2.5 Solving Literal Equations (Formulas)

Learning Objectives

After completing this section, you will be able to:

Solve formulas and literal equations for a specified variable.
Apply inverse operations and properties of equality with symbolic expressions.
Rearrange common geometry and science formulas.
Recognize when to isolate a variable that appears in multiple terms.

Launch – What Is a Literal Equation?

A literal equation (or formula) contains two or more variables. You are often asked to solve for one variable in terms of the others.

Example: Area of a rectangle: $A = LW$

If you know $A$ and $W$ , you can solve for $L$ :

$L = \frac{A}{W}$

These problems use the same skills you have already practiced:

Apply inverse operations
Keep both sides balanced
Use algebra to undo operations symbolically

Solving for a Variable in Geometry Formulas

Example 2.5.1: Area of a Rectangle

$A = LW$ , solve for $L$

Rectangle with length L and width W labeled. — **Figure 2.5A**

Solution

Divide both sides by $W$ :

$L = \frac{A}{W}$

✅ Length written in terms of area and width.

Example 2.5.2: Simple Interest Formula

$I = Prt$ , solve for $t$

Solution

Divide both sides by $Pr$ :

$t = \frac{I}{Pr}$

✅ Time written in terms of interest, principal, and rate.

Example 2.5.3: Perimeter of a Triangle

$P = a + b + c$ , solve for $b$

Triangle with sides labeled a, b, and c. — **Figure 2.5B**

Solution

Subtract $a$ and $c$ from both sides:

$b = P - a - c$

✅ Side $b$ in terms of the perimeter and other sides.

Try It

Working with Coefficients and Fractions

Example 2.5.4: Volume of a Pyramid

$V = \frac{1}{3}Bh$ , solve for $h$

Pyramid with base area B and height h labeled. — **Figure 2.5C**

Solution

Multiply both sides by 3:

$3V = Bh$

Divide both sides by $B$ :

$h = \frac{3V}{B}$

✅ Height written in terms of volume and base area.

Example 2.5.5: Cylinder Surface Area

$A = 2\pi rh$ , solve for $r$

Cylinder with radius r and height h labeled. — **Figure 2.5D**

Solution

Divide both sides by $2\pi h$ :

$r = \frac{A}{2\pi h}$

✅ Radius written in terms of surface area and height.

Try It

Solving for a Variable in a Linear Equation

Literal equations don’t always come from geometry—sometimes they’re just algebraic relationships between variables.

Example 2.5.6: Slope-Intercept Equation

$y = mx + b$ , solve for $x$

Solution

Subtract $b$ :

$y - b = mx$

Divide by $m$ (assuming $m \ne 0$ ):

$x = \frac{y - b}{m}$

Example 2.5.7: Slope Formula Rearranged

$y - b = mx$ , solve for $m$

Solution

Divide both sides by $x$ (assuming $x \ne 0$ ):

$m = \frac{y - b}{x}$

Example 2.5.8: Two-Variable Equation

$4x - 3y = 12$ , solve for $y$

Solution

Subtract $4x$ from both sides:

$-3y = -4x + 12$

Divide by –3:

$y = \frac{4}{3}x - 4$

Try It

Solving for Other Variables in Science Formulas

Literal equations are everywhere in science. The process is identical—undo operations to isolate the desired variable.

Example 2.5.9: Newton’s Second Law

$F = ma$ , solve for $m$

Solution

Divide both sides by $a$ (assuming $a \ne 0$ ):

$m = \frac{F}{a}$

Example 2.5.10: Distance Formula

$d = rt$ , solve for $r$

Solution

Divide both sides by $t$ (assuming $t \ne 0$ ):

$r = \frac{d}{t}$

Example 2.5.11: Ohm’s Law (Rearranged)

$E = IR$ , solve for $R$

Solution

Divide both sides by $I$ (assuming $I \ne 0$ ):

$R = \frac{E}{I}$

Example 2.5.12: Volume of a Rectangular Solid

$V = LWH$ , solve for $H$

Rectangular solid labeled length L, width W, and height H. — **Figure 2.5E**

Solution

Divide both sides by $LW$ (assuming $L \ne 0$ and $W \ne 0$ ):

$H = \frac{V}{LW}$

Example 2.5.13: Average Temperature Formula

$Q = mc\Delta T$ , solve for $\Delta T$

Solution

Divide both sides by $mc$ (assuming $m \ne 0$ and $c \ne 0$ ):

$\Delta T = \frac{Q}{mc}$

Example 2.5.14: Velocity Formula

$a = \frac{v - u}{t}$ , solve for $v$

Solution

Multiply both sides by $t$ (assuming $t \ne 0$ ):

$at = v - u$

Add $u$ to both sides:

$v = at + u$

Try It

More Complex Formulas

Some formulas require more than one step to isolate the desired variable. Apply the same equation-solving principles you have already learned, performing inverse operations one step at a time.

Example 2.5.15: Rectangle Perimeter

$P = 2l + 2w$ , solve for $w$

Solution

Subtract $2l$ :

$P - 2l = 2w$

Divide by 2:

$w = \frac{P - 2l}{2}$

Example 2.5.16: Area of a Trapezoid

$A = \frac{1}{2}(b_1 + b_2)h$ , solve for $h$

Trapezoid labeled bases b1 and b2 and height h. — **Figure 2.5G**

Solution

Multiply both sides by 2:

$2A = (b_1 + b_2)h$

Divide by $(b_1 + b_2)$ :

$h = \frac{2A}{b_1 + b_2}$

Try It

Common Mistakes

🚩 Dividing only one term instead of the entire expression

🚩 Dropping parentheses when dividing by a binomial

🚩 Forgetting that dividing by a fraction means multiplying by its reciprocal

🚩 Losing negative signs

🚩 Stopping before the variable is fully isolated

Try It: Summary

Key Takeaways

Literal equations follow the same rules as numeric equations.
Identify the variable you are solving for first.
Use inverse operations to isolate the variable.
Divide or multiply by the entire expression attached to the variable.
Simplify carefully and keep the equation balanced.

License

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Launch – What Is a Literal Equation?

Solving for a Variable in Geometry Formulas

Working with Coefficients and Fractions

Solving for a Variable in a Linear Equation

Solving for Other Variables in Science Formulas

More Complex Formulas

Common Mistakes

License

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