Lesson 29: Solving Literal Equations

New Concept

A literal equation is an equation with more than one variable.

What’s next

Now, we will put this idea into practice with worked examples, showing you how to isolate any variable in a formula.

Property

A literal equation is an equation with more than one variable. A formula is a type of literal equation.

Examples

• The distance formula $d = rt$ is a literal equation with three variables.
• The equation for a line, $y = mx + b$, is a literal equation with four variables.
• The formula for the area of a rectangle, $A = lw$, is a literal equation solved for the area.

Explanation

Think of a literal equation as a regular math puzzle, but with more characters! Instead of just solving for 'x', you might be solving for 'y', 'r', or even 'C'. Your goal is still the same: get that one specific letter all by itself on one side of the equals sign. The answer won't be a single number, but a new recipe using the other variables.

Property

As in an equation with one variable, use inverse operations and properties of equalities to solve for a specific variable in a literal equation.

Examples

• To solve $5x + 2y = 12$ for y: $2y = 12 - 5x$, so $y = \frac{12 - 5x}{2}$.
• To solve $P = 2l + 2w$ for w: $P - 2l = 2w$, so $w = \frac{P - 2l}{2}$.
• To solve $A = \frac{1}{2}bh$ for b: $2A = bh$, so $b = \frac{2A}{h}$.

Explanation

Ready to be a variable detective? To solve for a specific letter, you need to isolate it! Use your trusty inverse operations—addition's arch-nemesis is subtraction, and multiplication's is division. Your mission is to systematically move every other term and number to the other side of the equation, leaving your target variable standing alone and proud. It's all about undoing the math!

Property

If the variable being solved for is on both sides of the equation, the first step is to eliminate the variable on one side or the other.

Examples

• To solve $5y + 3b = 2y + 9b$ for y: $3y + 3b = 9b$, then $3y = 6b$, so $y = 2b$.
• To solve $10k - a = 4k + 5a$ for k: $6k - a = 5a$, then $6k = 6a$, so $k = a$.
• To solve $8p - 2z = 3p + 8z$ for p: $5p - 2z = 8z$, then $5p = 10z$, so $p = 2z$.

Explanation

What happens when your target variable shows up to the party on both sides of the equation? You have to play matchmaker! Your first move is to use addition or subtraction to get all the terms with that variable to hang out together on one side. Once they're grouped, you can combine them and then proceed with isolating your chosen variable. It's cleanup time!