I’m revising algebraic fractions to strengthen my fundamentals, and I’m getting tripped up on when it’s okay to cancel factors and how to keep track of the domain. I love when a factor neatly cancels out, but then I panic about whether I just accidentally erased an important restriction.
Example: (x^2 – 9)/(x – 3). I factor to ((x – 3)(x + 3))/(x – 3) and I want to cancel to get x + 3. But I know x ≠ 3 in the original. Do I have to keep writing x ≠ 3 even though the simplified expression looks like it’s defined at x = 3? Is there a clear rule of thumb for this?
I tried a slightly bigger one too: (x^2 – 1)/(x^2 – x) ÷ ((x – 1)/x). My attempt:
– Factor: ((x – 1)(x + 1))/(x(x – 1)) ÷ ((x – 1)/x)
– Flip and multiply: ((x – 1)(x + 1))/(x(x – 1)) * (x/(x – 1))
– Then I canceled things and ended up with (x + 1)/(x – 1). But now I’m confused: the original expression had x ≠ 0 and x ≠ 1 from the denominators. The simplified form only shows x ≠ 1. Do I still need to keep x ≠ 0 in the final answer because it was excluded in the original?
Follow-up: when I’m adding things like 1/(x – 3) + 2/(x^2 – 9), I factor the second denominator to (x – 3)(x + 3) and pick the LCD as (x – 3)(x + 3). I rewrote 1/(x – 3) as (x + 3)/(x^2 – 9), and 2/(x^2 – 9) already matches the LCD, but I keep second-guessing how the numerators combine. Is there a reliable step-by-step way to build the new numerators so I don’t miss a factor or mess up a sign?
TL;DR: When exactly is canceling a factor valid, how do I correctly carry over domain restrictions after simplifying, and what’s a foolproof way to set up the numerators when finding a common denominator?
Ready to make maths more enjoyable, accessible, and fun? Join a friendly community where you can explore puzzles, ask questions, track your progress, and learn at your own pace.
By becoming a member, you unlock:
3 Responses
Short version: you can cancel common factors, but you can’t cancel domain restrictions. Make a “bad x’s list” first by setting every denominator you ever see (before you cancel or flip) equal to 0, and also make sure any divisor isn’t zero. Those excluded x-values stay excluded forever, even if they disappear after simplification. And only cancel factors, not terms: (x^2 − 9)/(x − 3) is fine because x^2 − 9 = (x − 3)(x + 3); something like (x + 3)/(x + 1) doesn’t let you cancel the x’s. For your first example, (x^2 − 9)/(x − 3) = x + 3 with x ≠ 3. No, you didn’t magically make x = 3 legal; you just hid the hole.
Worked example (your division one). Start with (x^2 − 1)/(x^2 − x) ÷ ((x − 1)/x). Bad x’s list from denominators and “divisor not zero”: x^2 − x = x(x − 1) ⇒ x ≠ 0,1; and (x − 1)/x ≠ 0 ⇒ x ≠ 1; plus x ≠ 0 from its denominator. So overall: x ≠ 0,1. Now simplify: factor and flip-multiply:
((x − 1)(x + 1))/(x(x − 1)) * (x/(x − 1)) → cancel x and one (x − 1) → (x + 1)/(x − 1). Final answer: (x + 1)/(x − 1), with x ≠ 0,1 kept from the original. The cancellations don’t restore those values.
For adding fractions, the foolproof move is: new numerator = old numerator × (LCD ÷ old denominator). Keep parentheses if there’s subtraction. Example: 1/(x − 3) + 2/(x^2 − 9). Factor x^2 − 9 = (x − 3)(x + 3), so LCD = (x − 3)(x + 3). First numerator becomes 1 × (x + 3) = x + 3. Second numerator is 2 × 1 = 2. Sum: (x + 3) + 2 = x + 5 over the LCD, so (x + 5)/(x^2 − 9). Domain: exclude roots of the LCD (and anything you had before): x ≠ 3, −3.
The rule of thumb is: you can cancel only common factors, never parts of a sum, and you must record any x-values that made a denominator zero before you canceled. Those excluded values don’t magically come back just because the factor disappeared; they become a “hole” (a removable discontinuity) in the simplified expression. So (x^2 − 9)/(x − 3) = (x − 3)(x + 3)/(x − 3) simplifies to x + 3, but only for x ≠ 3. A neat way to say it is: (x^2 − 9)/(x − 3) equals x + 3 on the domain x ≠ 3. Same idea for your bigger example. Start by collecting restrictions: from x^2 − x in the denominator, x ≠ 0,1; and since you’re dividing by (x − 1)/x, you also need that fraction not to be zero or undefined, which again gives x ≠ 0,1. After flipping and canceling, you get (x + 1)/(x − 1), but you still keep x ≠ 0,1 from the original. In short: simplify freely, but carry the original domain along for the ride.
For building numerators with a common denominator, use this reliable recipe: 1) factor every denominator; 2) choose an LCD using each distinct factor to its highest power; 3) for each fraction, multiply top and bottom by whatever factor(s) it’s missing to reach the LCD-use parentheses so signs don’t wander off; 4) combine the numerators; 5) state the domain as “all x except the zeros of the original denominators.” Your example: 1/(x − 3) + 2/(x^2 − 9). Factor x^2 − 9 = (x − 3)(x + 3), so LCD = (x − 3)(x + 3) and the domain excludes x = 3, −3. The first term is missing (x + 3), so it becomes (x + 3)/(x^2 − 9); the second already matches the LCD, so it stays 2/(x^2 − 9). Add: (x + 3 + 2)/(x^2 − 9) = (x + 5)/(x^2 − 9), with x ≠ 3, −3. If you’re subtracting instead of adding, keep the whole adjusted numerator in parentheses before distributing the minus-future-you will thank you.
Hope this helps!
My rule of thumb is: you may cancel only common factors (not terms), and you must record all forbidden x-values before you cancel-those are the x that make any denominator zero (and, in a division A ÷ B, also any x that make B = 0). After canceling, the simplified formula is equal to the original on that recorded domain, but any canceled factor’s zeros remain excluded; they become “holes” in the graph. Think of it like simplifying a road map: you can redraw the route more cleanly, but if a bridge was closed on the original, the closure still exists even if the new drawing doesn’t show it. Example 1: (x^2 − 9)/(x − 3) = ((x − 3)(x + 3))/(x − 3) simplifies to x + 3, but you must keep x ≠ 3; the original is undefined at 3 even though the simplified formula would give 6 there (that’s a hole). Example 2: (x^2 − 1)/(x^2 − x) ÷ ((x − 1)/x): before simplifying, exclude x = 0, 1 (from x(x − 1) in a denominator, from the divisor’s denominator x ≠ 0, and because the divisor can’t be 0, x − 1 ≠ 0). Now factor, flip, and cancel to get (x + 1)/(x − 1), but keep x ≠ 0, 1; there’s a hole at 0 and a vertical asymptote at 1. For adding 1/(x − 3) + 2/(x^2 − 9): factor the LCD as (x − 3)(x + 3); multiply the first fraction by (x + 3)/(x + 3) so its numerator becomes x + 3; the second already matches the LCD so its numerator is 2; then add to get (x + 3) + 2 = x + 5 over (x − 3)(x + 3), with x ≠ ±3. Reliable workflow: factor everything, list all exclusions up front, build each new numerator by multiplying by the exact “missing factor” (carry parentheses and signs-e.g., 3 − x = −(x − 3)), combine, then simplify and report the final formula together with the original exclusions.