Lesson 43: Simplifying Rational Expressions

New Concept

Algebra is the language of modern mathematics. It uses symbols and rules to explore relationships and solve for unknowns in a systematic, powerful way.

What’s next

This course builds your skills step-by-step. To begin, we'll explore rational expressions, learning how to simplify them and identify their undefined values.

Property

Rational expressions are fractions with a variable in the denominator. Since division by 0 in mathematics is undefined, the denominator of a rational expression must not equal 0.

Examples

• In the expression $\frac{x+8}{4x}$, the denominator is $4x$. The expression is undefined when $4x=0$, which happens at $x=0$.
• For $\frac{2x+1}{x-4}$, the denominator $x-4$ equals zero when $x=4$. So, the expression is undefined at $x=4$.
• For $\frac{(x+5)(x-3)}{3x+21}$, the denominator $3x+21$ is zero when $3x=-21$, so the expression is undefined at $x=-7$.

Explanation

Think of the denominator as a bridge; if it becomes zero, the bridge collapses and the expression is undefined! To find these “weak spots,” set the denominator equal to zero and solve for the variable. This tells you which values you must avoid to keep the expression mathematically sound and prevent a total meltdown.

Property

To simplify a rational expression, find the greatest common factor (GCF) for the numerator and denominator. Factor both completely, then divide out any common factors.

Examples

• $\frac{2x^2-10x}{4x^2+6x} = \frac{2x(x-5)}{2x(2x+3)} = \frac{x-5}{2x+3}$
• $\frac{2x^2-14x}{4x-28} = \frac{2x(x-7)}{4(x-7)} = \frac{2x}{4} = \frac{x}{2}$
• $\frac{4x+28}{3x^2+21x} = \frac{4(x+7)}{3x(x+7)} = \frac{4}{3x}$

Explanation

Simplifying these expressions is like a cleanup mission! Your goal is to find identical pieces (factors) on the top and bottom and cancel them out. By factoring everything first, you can easily spot the matching parts. Once they’re gone, you're left with a much neater and simpler expression that is easier to work with.

Property

Determine values that cause the denominator to equal zero before the fraction is simplified.

Examples

• In $\frac{2x(x-7)}{4(x-7)}$, the original denominator is zero at $x=7$. After simplifying to $\frac{x}{2}$, this is hidden. The final answer must state $x \neq 7$.
• In $\frac{x(x+5)}{3x(x-1)}$, the original denominator is zero at $x=0$ and $x=1$. After simplifying to $\frac{x+5}{3(x-1)}$, the $x=0$ restriction disappears. You must state both $x \neq 0$ and $x \neq 1$.

Explanation

This is a golden rule! Always find the forbidden values from the original expression, not the simplified one. If you simplify first, you might accidentally cancel out a factor that was causing a division-by-zero error, effectively hiding the evidence. The original denominator tells the whole truth about where the expression is undefined, so check it first!