In short

Multiplying both sides of an equation by a constant like 7 is safe — the new equation has exactly the same solutions as the old one. But multiplying both sides by an expression with x inside it, like (x - 3) or x, is not safe. Any value of x that makes that multiplier zero becomes a "free" solution of the new equation, even if it had nothing to do with the original. These ghost solutions are called extraneous solutions — they pass the transformed equation but fail the original. The fix is one habit: always plug your final candidates back into the original equation, not the rearranged one. Skipping this check is one of the most reliable ways to lose marks on rational equations in JEE Mains.

You already know the do-the-same-to-both-sides rule from linear equations in one variable. Add a number to both sides — fine. Subtract — fine. Multiply by a non-zero number — fine. Divide by a non-zero number — fine. The reason these moves are safe is that each one is reversible: whatever you did, you can undo, and undoing brings you back to exactly the same equation you started with. Same equation, same solutions.

Multiplying both sides by an expression that contains x — like (x - 3) — looks like it belongs to the same family of moves. It does not. It is the move that smuggles in fake answers, and the trap is so reliably set that JEE Mains question setters love it.

This article is about exactly when this happens, why it happens, and the one habit that protects you from it.

The mechanism

Start with the safe case. If you have an equation A = B and you multiply both sides by a non-zero constant k, you get kA = kB. Going backward, divide both sides by k — which is allowed because k \neq 0 — and you recover A = B. The two equations are equivalent: they have the same solutions.

Why: multiplication by a non-zero number is reversible. Division by that same number undoes it exactly. No solutions are created and none are destroyed.

Now consider what happens if k is actually an expression involving x, say k = x - 3. Then "k \neq 0" is no longer something you can check in advance — it depends on which x you are plugging in. For most values of x, the multiplier is non-zero and the move is safe. But for x = 3, the multiplier is zero, and multiplying both sides by zero is the algebraic equivalent of erasing the equation.

Look at what happens to any equation when you multiply both sides by zero:

A = B \;\;\xrightarrow{\;\times 0\;}\;\; 0 = 0.

The new equation 0 = 0 is true for every value of x — even values that had nothing to do with the original. The information about which x satisfied A = B has been wiped out. This is what "multiplying by zero loses all info" means.

So when you multiply both sides by (x - 3), you are multiplying by something that is non-zero for most x (safe) and zero exactly at x = 3 (unsafe). The transformed equation inherits the safe behaviour for most x, and at x = 3 it inherits the universal-truth behaviour of 0 = 0. So x = 3 becomes a "free solution" of the new equation, regardless of whether it was a solution of the old one.

That free solution is the extraneous one.

Original solutions are a subset of multiplied-equation solutionsA Venn diagram of two overlapping circles. The smaller left circle is labelled solutions of the original equation A equals B. The larger right circle is labelled solutions of the multiplied equation k times A equals k times B. The smaller circle sits entirely inside the larger one. The crescent-shaped region of the larger circle that lies outside the smaller circle is shaded and labelled extraneous solutions, the values of x where k of x equals zero. A caption below explains that multiplying by an expression with x inside can only add solutions, never remove them. original solutions A = B extraneous solutions where k(x) = 0 solutions of multiplied equation kA = kB Multiplying both sides by k(x) can only add solutions, never remove them.
Every solution of the original equation $A = B$ is still a solution of $kA = kB$ — the small circle sits inside the big one. But the big circle can also contain extras: any $x$ where $k(x) = 0$. Those extras are the extraneous solutions, and they are exactly what the verification step is designed to catch.

The key picture: the solution set of the multiplied equation is always a superset of the original. So solving the multiplied equation gives you all the real solutions, possibly with some extras. You then have to filter the extras out by hand.

The classic example

Here is the equation that traps thousands of students every year.

The classic trap

Solve \dfrac{x^2 - 9}{x - 3} = 6.

Multiply both sides by (x - 3) to clear the denominator:

x^2 - 9 = 6(x - 3).

Expand the right side: 6(x - 3) = 6x - 18. So

x^2 - 9 = 6x - 18 \;\;\implies\;\; x^2 - 6x + 9 = 0 \;\;\implies\;\; (x - 3)^2 = 0 \;\;\implies\;\; x = 3.

It looks like the answer is x = 3. But pause — go back to the original equation and substitute:

\frac{(3)^2 - 9}{3 - 3} \;=\; \frac{0}{0}.

The denominator is zero. The original equation is undefined at x = 3. So x = 3 cannot possibly be a solution.

The candidate x = 3 is extraneous. It snuck in because the move "multiply both sides by (x - 3)" is exactly the move that multiplies by zero when x = 3. The original equation has no solution at all.

Notice how clean the trap is. The algebra is correct at every step. There is no arithmetic error. The factoring is right. And yet the final answer is a number that does not solve the original equation. The only thing that catches it is the verification.

For sanity, here is a case where multiplying by an x-expression does not introduce trouble.

Multiplying by $x$ — no spurious solution this time

Solve \dfrac{1}{x} + 1 = \dfrac{2}{x}.

Multiply both sides by x:

1 + x = 2 \;\;\implies\;\; x = 1.

Check in the original: \dfrac{1}{1} + 1 = 1 + 1 = 2, and \dfrac{2}{1} = 2. Both sides equal 2. The candidate x = 1 is genuine.

Why no extraneous solution this time? Because the multiplier x is zero exactly at x = 0, and the transformed equation 1 + x = 2 does not have x = 0 as a solution either. The danger is real but it did not get triggered — the "free zero" x = 0 failed the new equation too, so it never appeared as a candidate to be filtered.

So the rule is not "multiplying by a variable expression always creates extras." It is "multiplying by a variable expression can create extras, and the extras are always values where the multiplier is zero."

And here is a case where the danger does fire — the would-be extra is exactly the multiplier's zero, and it shows up as a candidate.

A zero gets manufactured

Solve \dfrac{x}{x - 2} = \dfrac{2}{x - 2}.

Multiply both sides by (x - 2):

x = 2.

That is the only candidate from the new equation. Now check it in the original:

\frac{2}{2 - 2} \;=\; \frac{2}{2 - 2} \;=\; \frac{2}{0}.

Both sides are undefined. The candidate x = 2 is extraneous, because the original equation is not even defined there. The original equation has no solution.

Compare: the new equation x = 2 is perfectly happy with x = 2. The new equation has nothing wrong with it. The original does. That mismatch is exactly the extraneous-solution phenomenon.

The fix

There is exactly one habit that protects you, and it is short.

Always verify candidate solutions in the original equation, not the rearranged one.

This sentence is doing a lot of work. The word "original" is load-bearing — checking in the rearranged equation will tell you the candidate satisfies the rearranged equation, which it will, by construction. That check is meaningless. The candidate has to satisfy the equation you were actually asked to solve.

A useful sub-habit, which makes the verification almost automatic, is to note the forbidden values before you multiply. If the original equation has \dfrac{1}{x - 3} in it, then x = 3 is forbidden — the original equation is undefined there. Any candidate that turns out to equal 3 is automatically extraneous and must be discarded. You do not even need to plug it in; the forbidden-value list already disqualifies it.

So the full procedure for any rational equation looks like this:

  1. Note forbidden values. Read the original equation and write down every value of x that makes any denominator zero. Those values are off-limits.
  2. Multiply both sides by the LCM of the denominators to clear fractions. This is where extraneous solutions can sneak in.
  3. Solve the resulting polynomial equation by your usual methods.
  4. Filter. Discard any candidate that appears in the forbidden-values list.
  5. Verify. Plug each surviving candidate into the original equation as a final check.

The discipline of step 1 alone catches most extraneous solutions before they cost you anything.

Why this matters for JEE

Rational equations are everywhere in the JEE Mains paper — in algebra, in trigonometry, in coordinate geometry. A typical mistake pattern looks like this: a student clears denominators, factors a quadratic, gets two roots, circles both, and moves on. One of the roots makes an original denominator zero, so the correct answer was the other root alone. Half the marks are lost on a problem the student actually knew how to solve.

The reverse mistake is to refuse to multiply by variable expressions at all and try to solve fractional equations by hand-juggling — much slower, much more error-prone. Multiplying through is the right move. You just have to follow it with the verification step.

A clean way to think about the entire phenomenon: multiplying both sides by k(x) takes you from your equation to a bigger equation — one with at least as many solutions, possibly more. Solving the bigger equation is fine; you just have to come back down to the smaller original equation at the end and throw out anything that does not belong. The check is not a paranoid double-check — it is a structural part of the method. Skipping it is like solving a system of equations and never substituting back to make sure your answer works in both equations.

For the more general phenomenon — including the opposite trap, where dividing both sides by x destroys real solutions — see extraneous and lost roots. That article walks through both directions of the danger and the broader pattern of "non-reversible algebraic moves."

The takeaway

Multiplying both sides by a constant is safe. Multiplying both sides by an expression with x in it is conditionally safe — it is safe wherever the expression is non-zero, and it manufactures a free fake solution at every value of x where the expression is zero. The safe-looking algebraic step has actually enlarged the solution set, and the extras need to be hunted down and discarded.

The hunting tool is a single sentence: plug every candidate into the original equation. If it works, keep it. If it fails — including failing because some denominator is zero — throw it out. That one check, performed reflexively at the end of every rational equation, prevents nearly every extraneous-solution mistake you will ever make.

References

  1. Extraneous and missing solutions — Wikipedia overview of how non-reversible operations create or destroy roots.
  2. Rational equation — background on equations involving rational expressions.
  3. NCERT Class 8 Mathematics, Chapter 2: Linear Equations in One Variable — official Indian curriculum source.
  4. Khan Academy: Equations with rational expressions — worked examples with verification step emphasised.
  5. Paul's Online Math Notes: Rational Expressions — clear treatment of the extraneous-solution check for JEE-style problems.