Clear the denominators to escape the fraction — then check for the fake roots it leaves behind.
Point 4 of 5 in this lesson: 9.5.4 The full procedure — three steps, and the check is a gate
Up to now, fractions with letters were things you tidied: you reduced them, you added them, you never asked them to be equal to anything. A rational equation changes the question. It puts an = sign down and asks: for which value of x is this actually true? The unknown is hiding downstairs, in a denominator — and the first thing we'll do is coax it out into the open.
By the end you'll be able to: recognize a rational equation; clear its denominators by multiplying both sides by the LCD; solve the plain polynomial equation left behind; and — the part everyone forgets — check each answer in the original to throw out the fake (extraneous) roots the clearing step can invent. Throughout, we keep the Stage 9 colors: a numerator is amber, a denominator is blue, a cancelled factor is green, and an excluded or extraneous value is red.
You met rational expressions in 9.1 through 9.4: things like x+1x−2. An expression has no = sign — there's nothing to "solve," only to simplify. A rational equation is what you get when two expressions are set equal and the unknown lives in a denominator:
An equation asks a question (which x?) and expects an answer. An expression is just a quantity you rewrite. The moment a variable appears in a denominator and there's an =, you have a rational equation — and the new danger from 9.1 follows it everywhere: the bottom must never be 0.
Before you solve anything, glance at the denominators and write down the values that are off‑limits — the ones that would make a bottom zero. In 2x = 3x+1, the bottoms x and x+1 forbid x = 0 and x = −1. Keep that little no‑go list in the corner of your eye; it's exactly what the final check will compare against.
Tap each item. Sort it into "expression" (just simplify) or "rational equation" (solve for x). The denominators light blue.
Here's the whole trick. Fractions are awkward to solve with, so we get rid of them. Multiply both sides of the equation by the LCD — the least common denominator of every fraction in sight. Each denominator cancels, and what's left is an ordinary polynomial equation (linear or quadratic) that you already know how to crush.
Solve 2x = 3x+1. The denominators are x and x+1, so the LCD is x(x+1). Multiply both sides by it:
Solve xx+2 = 3x−2. The off‑limits values are x = −2 and x = 2. The LCD is (x+2)(x−2); multiply both sides and the bottoms vanish:
x(x−2) = 3(x+2) → x²−2x = 3x+6 → x² − 5x − 6 = 0 → (x−6)(x+1) = 0
So the candidates are x = 6 and x = −1. Neither is on the no‑go list, so both are real solutions. (Quick check: 68 = 34 ✓, and −11 = −1 = 3−3 ✓.)
You must multiply every single term by the LCD — including any plain number or polynomial that isn't a fraction. Forget the "+ 1" or the "2" on one side and the whole equation tips out of balance. (More on this in 9.5.4.)
Pick an equation, then step through it one move at a time: find the LCD, clear the bottoms, solve.
Multiplying both sides by the LCD is a legal move only when the LCD isn't zero. But the LCD is built out of denominators — the very things that are forbidden from being zero. If a candidate root happens to be one of those off‑limits values, the multiplication we did was illegal for that value, and the "root" it produced is a phantom: an extraneous root. The clearing step invented it. It is not a solution.
Multiplying an equation by an expression that equals 0 can turn a false statement into a true one (anything times 0 is 0, so both sides "agree"). That's how a value that breaks the original sneaks into the cleared equation as a candidate. The cure is simple and non‑negotiable: substitute every candidate back into the original and reject any that zero a denominator.
Solve xx−3 = 2 + 3x−3. The off‑limits value is x = 3. The LCD is x−3; clear it (multiplying the lone 2 as well):
x = 2(x−3) + 3 → x = 2x − 6 + 3 → x = 2x − 3 → x = 3
The only candidate is x = 3 — but that's exactly the forbidden value! Substituting it back puts a 0 under both fractions in the original. So x = 3 is extraneous; we reject it. This equation has no solution.
Solve x²x−1 = 1x−1. Off‑limits: x = 1. LCD is x−1; clearing gives x² = 1, so x² − 1 = 0 → (x−1)(x+1) = 0, with candidates x = 1 and x = −1.
| candidate | denominator x−1 | verdict |
|---|---|---|
| x = 1 | 1 − 1 = 0 | extraneous — reject |
| x = −1 | −1 − 1 = −2 ≠ 0 | valid ✓ |
Only x = −1 survives the gate. The candidate x = 1 looked like a root but zeros the denominator, so out it goes.
Pick an equation. The widget shows the candidate roots, then substitutes each into the original denominators. A zero on the bottom means the root is a fake.
Now assemble everything into a routine you can run on any rational equation. There are three steps, but the third is not optional bookkeeping — it is a gate the answers must pass through.
Run it once, slowly, on 1x−2 + 3 = xx−2. Step 0: off‑limits is x = 2. Step 1 (clear, × (x−2)): 1 + 3(x−2) = x. Step 2 (solve): 1 + 3x − 6 = x → 3x − 5 = x → 2x = 5 → x = 52. Step 3 (check): 52 is not 2, so the denominator x−2 = ½ ≠ 0. The root passes. Solution: x = 2.5.
Write the excluded values before you solve, off to the side. Then the check is a five‑second glance: "is my candidate one of these red numbers?" If yes, reject it; if no, keep it. The check costs almost nothing and saves you from the most common wrong answer in the chapter.
The two sides of 2/x = 3/(x+1) are drawn as curves. Slide x; the solution is where they cross. The dashed red lines are the forbidden vertical asymptotes — a denominator is zero there.
Rational equations earn their keep on rate problems, where a quantity is divided by something you don't yet know — a time, a speed, a number of workers. The classic is the shared‑work problem: if pipe A fills a tank in a hours, in one hour it does 1a of the job. Rates of parallel workers add.
Pipe A fills a tank in 3 h; pipe B fills it in 6 h. Open both. How long, t hours, to fill it together? In one hour the pipes together do 13 + 16 of the tank, and that must equal 1t of the tank:
A rate answer must make physical sense. Two pipes working together must beat the faster pipe (2 h < 3 h ✓). A negative time or a speed that zeros a denominator gets thrown out for the same reason an extraneous root does — it doesn't fit the original situation. Always end by reading the number back into words.
Pipe A fills in a hours, pipe B in b hours. Slide them; the widget writes the equation 1/a + 1/b = 1/t and solves for the together‑time t.
A rational equation has the unknown in a denominator and an = sign begging for an answer. To solve it, clear the fractions by multiplying every term on both sides by the LCD; the bottoms cancel and a plain polynomial equation remains. Solve that. Then — because multiplying by the LCD can invent fakes — check each candidate in the original and throw out any that zero a denominator. Clear, solve, check: the check is the gate, never skip it.
You've now closed the loop on Stage 9: from meeting rational expressions in 9.1, to reducing and the LCD in 9.2, to 9.3 and 9.4's arithmetic, to solving here. Next you'll meet these same fractions as graphs — and that "forbidden vertical line" from the crossing widget becomes a curve's asymptote.
For each one: note the excluded values, clear the denominators, solve, and check. Some have a rejected root — or no solution at all.
Six questions to lock it in. Tap the answer you think is right.
This lesson develops CCSS A‑REI.A.2 — "solve simple rational equations in one variable, and give examples showing how extraneous solutions may arise" — together with A‑CED.A.1 (creating equations, here from work/rate situations) and the broader A‑REI.B reasoning about equivalent equations. The #1 misconception is skipping the check and reporting an extraneous root as a real solution; a close second is multiplying only some terms by the LCD (forgetting the lone constants or polynomials). The antidote is to make the check a non‑negotiable gate, not an afterthought: have the student write the excluded values before solving, and treat "clear → solve → check" as one inseparable move. Reinforce why the fake roots appear — multiplying both sides by an expression that can equal zero is the step that breaks equivalence — so the check feels like understanding, not ritual.