You learned the elimination method and the rule felt simple: multiply one equation by a clever number, line up coefficients, subtract, and one variable disappears. So when you next sit down with a system, your hand reaches for the easy move — pick one equation, scale it up, subtract.

And then nothing cancels. The variable you wanted dead is still standing on the page, mocking you. You stare at the result, double-check the multiplication, and there is no error. The trap is older and quieter than that. You needed to multiply BOTH equations, not just one. This article is about that trap, why it happens, and how to spot it before you waste five minutes solving in the wrong direction.

In short

The elimination method needs the coefficients of one variable to match (or be exact opposites) before you add or subtract. Sometimes scaling one equation is enough — but only when one coefficient already divides the other cleanly. When neither coefficient is a multiple of the other (think coefficients 3 and 5, or 2 and 7), you must multiply both equations, each by a different factor, so that both sides land on the LCM of the two coefficients. Always check first: does one coefficient divide the other? If yes, scale only one. If no, scale both. Skipping this check is the single most-marked elimination mistake on CBSE Class 10 boards — top three, every year.

The trap, in slow motion

Look at this innocent-looking system:

3x + 2y = 7 \quad \cdots (1)
5x - 3y = 8 \quad \cdots (2)

You want to eliminate x. The coefficients are 3 and 5. You think — multiply equation (1) by 5 to make the x-coefficient also 5… wait, that gives 15x, not 5x. Okay, multiply equation (1) by 5:

15x + 10y = 35 \quad \cdots (1')
5x - 3y = 8 \quad \cdots (2)

Now subtract? (1') - (2) gives 10x + 13y = 27. The x did not cancel — equation (1') has 15x but equation (2) still has 5x. You scaled one equation and forgot the other equation also has to grow.

Why: subtraction kills x only when both coefficients of x are the same number. 15x - 5x = 10x \neq 0. The whole point of multiplication in elimination is to engineer that "same number" on both sides. One multiplication only does half the engineering when neither coefficient already divides the other.

The fix is to also scale equation (2), by a factor that brings its x-coefficient up to 15 as well. Since 5 \times 3 = 15, multiply equation (2) by 3:

15x + 10y = 35 \quad \cdots (1')
15x - 9y = 24 \quad \cdots (2')

Now (1') - (2') really does kill x:

0 + 19y = 11 \implies y = \frac{11}{19}

Both equations had to be scaled — equation (1) by 5, equation (2) by 3 — so that both ended up with 15x on the left. 15 is the LCM of 3 and 5. That number is the secret.

When one multiplication IS enough

Compare the trap above to this system:

2x + 3y = 5 \quad \cdots (1)
4x + 5y = 11 \quad \cdots (2)

To eliminate x, look at the coefficients 2 and 4. Notice that 4 = 2 \times 2 — one is already a clean multiple of the other. So scale equation (1) by 2 and leave equation (2) alone:

4x + 6y = 10 \quad \cdots (1')
4x + 5y = 11 \quad \cdots (2)

Subtract (1') - (2):

0 + y = -1 \implies y = -1

One multiplication was enough because the LCM of 2 and 4 is just 4 — equation (2) already sat at the LCM, so it needed no scaling.

Why: when one coefficient cleanly divides the other (like 2 \mid 4), the larger coefficient is already at the LCM. Only the smaller equation needs a boost. When neither divides the other (like 3 and 5), neither equation is at the LCM, so both have to be lifted there.

This is the diagnostic question to ask before any elimination: "Does one coefficient divide the other?" If yes — scale one equation. If no — scale both.

How to find the right multipliers

Here is the recipe, no guessing involved.

  1. Pick the variable you want to kill. Look at its two coefficients, say a_1 and a_2.
  2. Compute the LCM of a_1 and a_2. Call it L.
  3. The multiplier for equation (1) is L / a_1. The multiplier for equation (2) is L / a_2.
  4. Multiply each ENTIRE equation (every term — left, right, and the constant) by its multiplier.
  5. Now both equations have the same coefficient L on the targeted variable. Subtract (or add, if signs are opposite) to cancel it.

Test this with the trap system above. Eliminate x: coefficients are 3 and 5. LCM is 15. Multiplier for (1) is 15/3 = 5. Multiplier for (2) is 15/5 = 3. Apply: equation (1) \times 5 gives 15x + 10y = 35. Equation (2) \times 3 gives 15x - 9y = 24. Both have 15x. Subtract — x dies. The recipe handles every case, including the easy one: if one coefficient divides the other (say a_1 = 2, a_2 = 4), then L = 4, the multiplier for (2) is 4/4 = 1 (no scaling needed), and only equation (1) gets multiplied.

Multiplication factor finder for elimination A diagram showing two equations stacked on the left, the LCM of their x-coefficients in the middle, and the multipliers for each equation on the right. The first equation 3x + 2y = 7 with coefficient 3, second equation 5x − 3y = 8 with coefficient 5, LCM 15, multiplier 5 for first equation, multiplier 3 for second equation. 3x + 2y = 7 5x − 3y = 8 x-coefficients: 3 and 5 LCM 15 multiplier = 15 ÷ 3 × 5 multiplier = 15 ÷ 5 × 3 After scaling — both equations have 15x on the left 15x + 10y = 35 15x − 9y = 24 subtract → 19y = 11 y = 11/19
The multiplier finder. Read off each equation's coefficient of the variable you want to kill, compute their LCM, and divide LCM by each coefficient to get the multiplier for that equation. Apply each factor to its ENTIRE equation. Now the targeted coefficient matches, and elimination works in one stroke.

A subtle but vital reminder: when you multiply an equation by a factor, every term gets multiplied — the x-term, the y-term, and the constant on the right. Forgetting to scale the right-hand side is the second-most-common slip after this article's main misconception.

Three worked examples

Example 1: The trap — $3x + 2y = 7$, $5x - 3y = 8$

Coefficients of x are 3 and 5. Neither divides the other. Both equations need scaling.

LCM of 3 and 5 is 15. Multiplier for (1): 15/3 = 5. Multiplier for (2): 15/5 = 3.

Multiply equation (1) by 5:

15x + 10y = 35 \quad \cdots (1')

Multiply equation (2) by 3:

15x - 9y = 24 \quad \cdots (2')

Why: now both rows carry 15x. Subtraction will cancel that column cleanly. If you had multiplied only equation (1) and left equation (2) alone, the x-terms would have been 15x and 5x — not matched, no cancellation.

Subtract (1') - (2'):

(15x - 15x) + (10y - (-9y)) = 35 - 24
19y = 11
y = \frac{11}{19}

Back-substitute into equation (1):

3x + 2 \cdot \frac{11}{19} = 7
3x = 7 - \frac{22}{19} = \frac{133 - 22}{19} = \frac{111}{19}
x = \frac{37}{19}

Solution: (x, y) = \left(\dfrac{37}{19}, \dfrac{11}{19}\right).

Example 2: The "easy" case — $2x + 3y = 5$, $4x + 5y = 11$

Coefficients of x are 2 and 4. Notice 4 = 2 \times 2 — one cleanly divides the other. Only equation (1) needs scaling.

LCM of 2 and 4 is 4. Multiplier for (1): 4/2 = 2. Multiplier for (2): 4/4 = 1 (do nothing).

Multiply equation (1) by 2:

4x + 6y = 10 \quad \cdots (1')
4x + 5y = 11 \quad \cdots (2)

Why: equation (2) was already sitting at the LCM. Multiplying it by 1 would change nothing. The trap of "always multiply both" wastes work in this case — the diagnostic question saved you a step.

Subtract (1') - (2):

0 + y = -1 \implies y = -1

Back-substitute into (1): 2x + 3(-1) = 5 \implies 2x = 8 \implies x = 4.

Check in (2): 4(4) + 5(-1) = 16 - 5 = 11. Confirmed.

Solution: (x, y) = (4, -1).

Example 3: The general LCM rule in action — $4x + 6y = 10$, $6x + 9y = 12$

Coefficients of x are 4 and 6. Does 4 divide 6? No. Does 6 divide 4? No. So both equations need scaling.

LCM of 4 and 6 is 12. Multiplier for (1): 12/4 = 3. Multiplier for (2): 12/6 = 2.

Multiply equation (1) by 3:

12x + 18y = 30 \quad \cdots (1')

Multiply equation (2) by 2:

12x + 18y = 24 \quad \cdots (2')

Why: every term in each equation is multiplied — including the constants on the right (10 \times 3 = 30 and 12 \times 2 = 24). Forgetting to scale the right-hand side gives a wrong system.

Subtract (1') - (2'):

0 + 0 = 6
0 = 6

That is a contradiction — meaning the original system has no solution. The two lines are parallel. (Check: the ratio 4/6 = 6/9 = 2/3 matches for the x- and y-coefficients but the right-hand-side ratio 10/12 \neq 2/3.) The LCM machinery still ran cleanly; it just told you the system is inconsistent. See Systems of Linear Equations for the geometric meaning.

A common variation — opposite signs

Sometimes the cleanest move is to make the matched coefficients equal-and-opposite so you can ADD instead of subtract. The factor-finding logic is identical — only the sign on one of the multipliers flips. For 3x + 2y = 7 and 5x - 3y = 8, you could instead eliminate y (coefficients 2 and -3). LCM of 2 and 3 is 6. Multiply (1) by 3 to get 9x + 6y = 21. Multiply (2) by 2 to get 10x - 6y = 16. ADD: 19x = 37, so x = 37/19. Same solution, different variable killed first. Either choice works — pick whichever has the smaller LCM and the friendlier signs.

Common confusions

References

  1. NCERT Class 10 Mathematics, Chapter 3: Pair of Linear Equations in Two Variables — the standard CBSE treatment, where the LCM-based elimination rule is introduced explicitly.
  2. Khan Academy — Elimination strategies — short videos on choosing the right multipliers when neither coefficient already matches.
  3. Wikipedia — System of linear equations — broader context, with the elimination method as a special case of Gaussian elimination.
  4. Paul's Online Notes — Linear Systems with Two Variables — clean examples of the both-equations-multiplied case.