Equivalent Expressions Pattern - Simplifying Fractions

Simplifying Algebraic Fractions

This pattern tests your ability to add, subtract, or simplify rational expressions — fractions where the numerator and denominator contain polynomials. The key skills are finding a common denominator, distributing negative signs carefully, and canceling common factors after factoring.

The Core Techniques

Adding or subtracting rational expressions: Just like numeric fractions, you need a common denominator. If the denominators have no common factors, the LCD is their product.

$\dfrac{a}{P} - \dfrac{b}{Q} = \dfrac{aQ - bP}{PQ}$

Factoring and canceling: If the numerator and denominator share a common factor, you can simplify by canceling it.

$\dfrac{(x+3)(2x+1)}{(x+3)(x-5)} = \dfrac{2x+1}{x-5}$ (provided $x \neq -3$)

Worked Examples

Example 1. Which expression is equivalent to $\dfrac{3}{2x - 1} - \dfrac{6}{4x + 1}$?

A) $\dfrac{9}{(2x-1)(4x+1)}$
B) $\dfrac{-3}{(2x-1)(4x+1)}$
C) $\dfrac{3}{2x+2}$
D) $\dfrac{24x - 3}{(2x-1)(4x+1)}$

The LCD is $(2x - 1)(4x + 1)$. Rewrite each fraction:
$\dfrac{3(4x+1) - 6(2x-1)}{(2x-1)(4x+1)}$
Expand the numerator: $12x + 3 - 12x + 6 = 9$
Result: $\dfrac{9}{(2x-1)(4x+1)}$
Gotcha: Option B gets $-3$ in the numerator from a sign error: $-6(2x - 1)$ gives $-12x + 6$ (the $-1$ becomes $+6$), not $-12x - 6$. Distributing a negative across a subtraction flips it to addition.
The answer is A.

Example 2. Which expression is equivalent to $\dfrac{1}{x - 4} - \dfrac{3}{3x + 6}$?

A) $\dfrac{18}{(x-4)(3x+6)}$
B) $\dfrac{-6}{(x-4)(3x+6)}$
C) $\dfrac{1}{x+5}$
D) $\dfrac{6x - 6}{(x-4)(3x+6)}$

LCD: $(x - 4)(3x + 6)$
Numerator: $1 \cdot (3x + 6) - 3 \cdot (x - 4) = 3x + 6 - 3x + 12 = 18$
Result: $\dfrac{18}{(x-4)(3x+6)}$
Gotcha: Option C comes from the error of subtracting numerators and denominators directly — you cannot do $\dfrac{1-3}{(x-4)-(3x+6)}$. Fractions don't work that way.
The answer is A.

Example 3. The expression $\dfrac{20z^2 + 31z + 12}{(4z+3)(z+5)}$ can be simplified. If $z > 0$, which expression is equivalent?

Factor the numerator. We need two numbers that multiply to $20 \times 12 = 240$ and add to $31$. Those are $16$ and $15$.
$20z^2 + 16z + 15z + 12 = 4z(5z + 4) + 3(5z + 4) = (4z + 3)(5z + 4)$
Now cancel the common factor:
$\dfrac{(4z+3)(5z+4)}{(4z+3)(z+5)} = \dfrac{5z+4}{z+5}$
Gotcha: You must factor the numerator completely before canceling. If you just tried to cancel individual terms, you'd get the wrong answer.

Example 4. Which expression is equivalent to $\dfrac{(2z+1)(5z+9) + (2z+1)(3z-1)}{(2z+1)(z-4)}$?

The numerator has a common factor of $(2z + 1)$:
$(2z+1)\bigl[(5z+9) + (3z-1)\bigr] = (2z+1)(8z + 8) = (2z+1) \cdot 8(z+1)$
Now simplify: $\dfrac{(2z+1) \cdot 8(z+1)}{(2z+1)(z-4)} = \dfrac{8(z+1)}{z-4}$
Gotcha: Don't expand everything — look for a common binomial factor first. Expanding creates a mess; factoring keeps it clean.

Example 5. For $a \neq 2$, which expression is equivalent to $\dfrac{3(a-2)(a+7) + 4(a+7)}{(a-2)}$?

Factor $(a + 7)$ from the numerator:
$(a+7)\bigl[3(a-2) + 4\bigr] = (a+7)(3a - 6 + 4) = (a+7)(3a - 2)$
Simplify: $\dfrac{(a+7)(3a-2)}{(a-2)}$
Since $(a - 2)$ is not a factor of the numerator, this fraction doesn't simplify further. The answer is $\dfrac{(a+7)(3a-2)}{(a-2)}$.
Gotcha: Don't cancel $(a - 2)$ unless it appears as a factor in both the numerator and denominator. Here it only appears in the denominator.

Example 6. Which expression is equivalent to $8s^4t + \dfrac{8s^4}{t}$?

Rewrite with a common denominator of $t$:
$\dfrac{8s^4t \cdot t}{t} + \dfrac{8s^4}{t} = \dfrac{8s^4t^2 + 8s^4}{t}$
Factor the numerator: $\dfrac{8s^4(t^2 + 1)}{t}$

What to Do on Test Day

• Common denominator: When adding or subtracting fractions with polynomial denominators, the LCD is usually the product of the two denominators (unless they share a factor).
• Distribute the negative: In $\dfrac{a}{P} - \dfrac{b}{Q}$, the subtraction applies to the entire second numerator: $aQ - bP$. The most common error is mishandling the sign: $-b(Q) = -bQ$, and if $Q$ has a minus sign inside, it flips to plus.
• Factor before canceling: Never cancel individual terms. Only cancel common factors of the entire numerator and denominator. Factor first, then cancel.
• Look for common binomial factors: If the same binomial (like $(2z + 1)$) appears in multiple terms of the numerator, factor it out.
• Standalone term + fraction: Convert the standalone term to a fraction with denominator $1$, then find the LCD. Example: $8s^4t = \dfrac{8s^4t \cdot t}{t}$ when the other fraction has denominator $t$.
• Domain restrictions: Watch for conditions like $a \neq 2$ — these tell you the original denominator is $(a - 2)$ and help confirm you have the right factoring.