In short
When you start expanding an equation, watch the two sides like a hawk. The instant the left-hand side and the right-hand side look like the same expression — same coefficients, same constants, same structure — you can stop. There is no "solving" left to do. Every x works. Write "identity, infinitely many solutions" and move on. The sibling 0 = 0 article explains why — this article trains the habit of seeing it early, before you waste five lines of algebra on a question that needs none.
You're solving a row of textbook problems on the bus to coaching. Equation 7 reads 5(x - 1) - 2 = 5x - 7. Your hand is already moving — distribute, transpose, subtract 5x, simplify, end at 0 = 0. Five lines of careful algebra.
Now imagine a friend next to you who took two seconds longer at the start. They expanded the left side: 5x - 5 - 2 = 5x - 7. They glanced at the right side: 5x - 7. They paused. Same thing. They wrote "identity — infinitely many solutions" and moved on to question 8.
Both of you are correct. But the friend is going to finish the worksheet first, and — more importantly — they have just trained an eye that pays back forever. This article is about that eye. The pattern is simple to state and easy to miss: when expanding both sides reveals the same expression, the equation is an identity, and the answer is "every x".
What "the same expression" means
Two expressions in x are the same if, after simplification, every coefficient and every constant matches. Not approximately, not nearly — exactly.
- 3x + 6 and 3x + 6 — same.
- 3x + 6 and 6 + 3x — same (addition is commutative).
- 3x + 6 and 3(x + 2) — same (distributive law in disguise).
- 3x + 6 and 3x + 5 — not the same. One constant is off by one. That single digit is the difference between "every x works" and "no x works".
Why: an equation has solutions only at the special x values that make LHS and RHS agree numerically. An identity is the case where LHS and RHS are the same algebraic object — they agree at every x because they were never different to begin with.
So your job, when you start a question, is partly arithmetic and partly visual. Expand. Simplify. Then look. If the two sides match exactly, the question was secretly an identity, dressed up in brackets and constants.
Spot the twin — a visual
The visual habit is exactly this: expand left, expand right, then compare shapes. Same coefficient of x? Same constant? Same sign? If yes on all three, twin. If even one of them is different, it is not a twin and you must continue solving normally.
Three worked examples
Example 1: $3(x + 2) = 3x + 6$
Expand the LHS. Distribute the 3:
Read the RHS. It is already simplified:
Compare. LHS = 3x + 6. RHS = 3x + 6. Twin!
Why: both sides are the same algebraic expression. The bracket on the left was just the distributive law in costume. There is nothing to "solve" — the equation is an identity and every real x satisfies it.
Answer: identity, infinitely many solutions (x \in \mathbb{R}).
Example 2: $\dfrac{2x + 4}{2} = x + 2$
Simplify the LHS. Divide each term in the numerator by 2:
Read the RHS.
Compare. LHS = x + 2. RHS = x + 2. Twin!
Why: dividing the numerator term-by-term is just the distributive law for division. After the simplification, both sides are literally the same expression — so the equation is an identity. Plug in any x: x = 7 gives \frac{18}{2} = 9 on the left and 7 + 2 = 9 on the right. Try x = -5: \frac{-6}{2} = -3 on the left and -5 + 2 = -3 on the right. Always equal.
Answer: identity, infinitely many solutions.
Example 3: $5(x - 1) - 2 = 5x - 7$
Expand the LHS. Distribute the 5, then combine the constants:
Read the RHS.
Compare. LHS = 5x - 7. RHS = 5x - 7. Twin!
Why: this one is sneakier because the LHS has two simplification steps — the distribution gives 5x - 5, and only after combining the trailing -2 do you arrive at 5x - 7. The disguise is thicker, but the punchline is the same: both sides are the same expression, so the equation is an identity.
Answer: identity, infinitely many solutions.
Compare this with the close cousin 5(x - 1) - 2 = 5x - 8. Expanding gives 5x - 7 on the left and 5x - 8 on the right — not twins (constants differ). That equation has no solution because the two sides can never agree.
Why early recognition saves time
Suppose you do not spot the twin and grind through Example 3 mechanically. The work goes:
Six lines, four arithmetic steps, and you arrive at 0 = 0, which (as the sibling article explains) tells you it was an identity all along. The conclusion is correct, but you spent four extra steps confirming what was already visible at line 3.
The twin-spotting habit collapses those six lines into three: expand, compare, declare. Why: those four extra steps are not just wasted ink — every extra step is another chance for a sign error, a transposition mistake, or a careless cancellation. Confident readers stop at the earliest line where the answer is settled. That economy is what mathematicians call fluency, and CBSE markers reward it because it shows you understood the structure rather than just executed a routine.
There is also a deeper payoff. Recognising twins trains your eye to see when two expressions are secretly the same — a skill that returns in factorisation (when you spot a^2 - b^2 = (a - b)(a + b)), in trigonometric identities (when you spot \sin^2 \theta + \cos^2 \theta hiding inside a longer expression), and in calculus (when you spot that two antiderivatives differ only by a constant). The habit is small here and enormous later.
A two-second checklist before you start solving
Before you transpose anything, run this check:
- Expand any brackets on both sides. Do not transpose yet.
- Combine like terms on each side independently. Do not move anything across the equals sign.
- Compare the two simplified expressions.
- If they are identical (same coefficient of x, same constant) → identity, infinite solutions, stop.
- If they have the same coefficient of x but different constants → no solution, stop. (See the no-solution sibling.)
- If they have different coefficients of x → proceed normally, you will get a unique answer.
That checklist takes about two seconds and tells you, before the algebra begins, which of the three branches you are on. Most textbook questions land on branch 3 — that's the normal "solve for x" case. But identities and no-solution traps appear regularly in NCERT exercises and exam papers precisely to test whether you noticed.
Where this appears in your syllabus
The CBSE Class 8 mathematics curriculum (NCERT Chapter 2, Linear Equations in One Variable) explicitly tests identity recognition. Exercise problems include equations that look ordinary but reduce to identities, and the marking scheme awards full marks only if you write "identity" or "infinitely many solutions" rather than leaving the answer blank or scribbling "0 = 0" without an interpretation.
In Class 9 and Class 10, the same trichotomy returns when you study linear equations in two variables and systems of them — coincident lines, parallel lines, intersecting lines. The vocabulary expands but the underlying check is the same: are the two sides really the same object in disguise?
A quick mental drill
Glance at each equation below and decide without doing any algebra whether it is likely an identity. Trust your structural eye, then verify.
- 7(x + 3) = 7x + 21 → expand left: 7x + 21. Twin. Identity.
- 7(x + 3) = 7x + 22 → expand left: 7x + 21. Constant differs. No solution.
- 7(x + 3) = 6x + 21 → expand left: 7x + 21. Coefficient of x differs. Unique solution exists; solve normally.
- \dfrac{6x - 9}{3} = 2x - 3 → simplify left: 2x - 3. Twin. Identity.
After a week of this drill, you will spot identities at sight, and the algebra you save adds up to real time on every test.
References
- NCERT, Mathematics Textbook for Class VIII, Chapter 2 — Linear Equations in One Variable.
- Wikipedia, Identity (mathematics).
- Wikipedia, Distributive property — the engine that lets a bracketed expression and its expansion be the same object.
- Khan Academy, Number of solutions to linear equations.
- CBSE, Class 8 Mathematics syllabus.