Elimination method

A pair of linear equations in two unknowns is really one puzzle hiding inside two clues. The elimination method solves it by a clever trick: scale the two equations so that one variable has matching coefficients, then add or subtract the equations so that variable vanishes entirely, leaving a single equation in one unknown. Solve that, back-substitute, and both values fall out. This is the second of the two standard algebraic techniques for the chapter Pair of Linear Equations in Two Variables, sitting alongside substitution. Elimination shines exactly where substitution gets clumsy — when neither equation has a lone variable with coefficient $1$ to isolate cheaply. Board exams test it directly on number pairs and, more demandingly, on word problems (ages, speeds, fractions) that must first be translated into a linear pair before elimination even begins.

The core insight

Scale two linear equations so one variable's coefficients match, then add or subtract to cancel that variable and collapse the pair into a single solvable equation.

The core idea

Elimination is a short, repeatable drill rather than a single formula to memorise. Given $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$ , the steps are:

Pick a target. Choose which variable, $x$ or $y$ , you want gone first — usually whichever has smaller or more convenient coefficients.

Scale to match. Multiply one or both equations by constants so that the target variable's coefficient becomes numerically equal in both. The LCM of the two coefficients is the natural multiplier to use, since it keeps numbers as small as possible.

Add or subtract. If the matched coefficients carry the same sign, subtract one equation from the other; if they carry opposite signs, add them. Either operation cancels the target variable, leaving one equation in the remaining unknown.

Solve, then back-substitute. Solve the single-variable equation, then plug that value into either original equation to recover the other variable. Always verify both values in both original equations — this catches almost every arithmetic slip. Multiplying the general pair by $a_2$ and $a_1$ respectively matches the $x$ -coefficients at $a_1a_2$ , and subtracting eliminates $x$ :

a_1a_2x + a_2b_1y = a_2c_1, \qquad a_1a_2x + a_1b_2y = a_1c_2 \ \Rightarrow\ (a_2b_1 - a_1b_2)\,y = a_2c_1 - a_1c_2

where: $a_1,b_1,c_1$ are the coefficients and constant of the first equation, $a_2,b_2,c_2$ those of the second, and $x,y$ are the two unknowns being solved for. Elimination is most efficient when a variable's coefficients are already equal or simple multiples of one another — no fractions creep in until the final division. That makes it the natural first choice for many textbook pairs, while substitution wins when a coefficient of $1$ is already sitting in front of a variable. Compare the two methods on the same pair, $2x + y = 7$ and $3x - y = 8$ . Substitution would isolate $y = 7 - 2x$ from the first equation and plug it into the second. Elimination notices instead that the $y$ -coefficients are already $+1$ and $-1$ — a perfect match with no scaling needed — so simply adding the equations gives $5x = 15 \Rightarrow x = 3$ in one line. Neither method is universally faster; the skill being tested is recognising, at a glance, which coefficients make elimination the shorter path.

Figure

A four step flow from scaling equations to matching coefficients to adding or subtracting to solving for both variables

Derivation of the elimination formula

Derivation ASolving a general linear pair by elimination

The goal is to eliminate one variable at a time from the general pair $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$ using only multiplication and addition or subtraction, and to arrive at closed-form expressions for both $x$ and $y$ purely in terms of the six given coefficients, with no guessing of multipliers required.

Step 1. Eliminate $y$ first. Multiply the first equation by $b_2$ and the second by $b_1$ , so the $y$ -coefficient becomes $b_1b_2$ in both: $a_1b_2x + b_1b_2y = b_2c_1$ and $a_2b_1x + b_1b_2y = b_1c_2$ .

Step 2. Subtract the second scaled equation from the first, since the $y$ -terms now carry the same sign: $(a_1b_2 - a_2b_1)x = b_2c_1 - b_1c_2$ . This gives $x$ alone, provided $a_1b_2 - a_2b_1 \neq 0$ :

x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1}

Step 3. Repeat the same scaling idea to eliminate $x$ instead. Multiply the first equation by $a_2$ and the second by $a_1$ , matching the $x$ -coefficient at $a_1a_2$ in both, then subtract to get $(a_2b_1 - a_1b_2)y = a_2c_1 - a_1c_2$ , so

y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1}

Step 4. Read the result. Both $x$ and $y$ come out as ratios with the same denominator, $a_1b_2 - a_2b_1$ . When this denominator is zero the two equations are parallel (no unique solution), which is why a well-posed pair always has $a_1b_2 \neq a_2b_1$ . In practice a student never memorises these formulas — the point of the derivation is to show that the scale-then-subtract drill always terminates in a clean answer, which justifies using it freely on any specific pair.

Worked example

Solve using elimination: $3x + 4y = 10$ and $2x - 2y = 2$ .

Step 1. Choose to eliminate $y$ . Its coefficients are $4$ and $-2$ ; their LCM is $4$ .

Step 2. Scale to match. Multiply the second equation by $2$ : $4x - 4y = 4$ . Keep the first equation unchanged: $3x + 4y = 10$ .

Step 3. Add the two equations, since the $y$ -coefficients $+4$ and $-4$ have opposite signs:

(3x+4y) + (4x-4y) = 10 + 4 \implies 7x = 14 \implies x = 2

Step 4. Back-substitute $x=2$ into the first original equation: $3(2) + 4y = 10 \implies 6 + 4y = 10 \implies 4y = 4 \implies y = 1$ .

Step 5. Verify in both originals: $3(2)+4(1) = 10$ and $2(2)-2(1) = 2$ , both correct. The solution is $x = 2,\ y = 1$ .

Common traps

Scaling only one equation when both need adjusting to reach the matched coefficient — check that every term, including the constant, gets multiplied, not just the variable you are targeting.
Choosing add instead of subtract, or vice versa. Same-sign matched coefficients need subtraction; opposite-sign matched coefficients need addition. A quick sign check before combining saves a full redo.
Forgetting to multiply the constant term while scaling an equation, which silently breaks the whole solution even though the variable coefficients look correctly matched.
Stopping after finding one variable. Elimination hands you one unknown directly; always back-substitute for the second and verify both values in the original equations, not just the scaled ones.
Jumping straight into elimination on fractional forms like $\frac{1}{x-y}$ terms in word problems, without first substituting a new variable — this is the single most common reason students get stuck midway on speed and distance questions.

Ask the tutor: Stuck on which multipliers to use or whether to add or subtract for a specific pair of equations? Share it with the AI tutor for a step-by-step elimination walkthrough.

How do I decide which variable to eliminate first?

Look at the coefficients of both variables in both equations and pick whichever pair is easier to match — usually the one with smaller numbers or an existing common factor. If one variable already has equal or simply related coefficients (like $4$ and $-2$ ), eliminate that one first, since the multiplier needed is small and the arithmetic stays clean. There is no fixed rule; either choice eventually gives the same answer.

What if, after eliminating, I get an equation like 0 equals 5 or 0 equals 0?

These are special cases the same algebra always reveals. If the variable terms cancel and you are left with a false statement like $0=5$ , the two lines are parallel and never meet — the pair has no solution. If instead you get a true statement like $0=0$ , the two equations represent the same line, so there are infinitely many solutions. Both outcomes are legitimate answers, not mistakes; report them as such rather than searching for a numeric $x,y$ .

Why does elimination sometimes need a substitution step first, as with speed and distance word problems?

Some word problems naturally produce equations with variables in denominators, such as $\frac{1}{x-y}$ and $\frac{1}{x+y}$ terms from upstream and downstream speeds. Eliminating directly on these fractions is messy, so the standard move is to substitute $u = \frac{1}{x-y}$ and $v = \frac{1}{x+y}$ first, turning the pair into an ordinary linear pair in $u$ and $v$ . Solve that with elimination as usual, then invert $u$ and $v$ at the end to recover $x$ and $y$ .

Elimination method

The core idea

Derivation of the elimination formula

Worked example

Common traps

Before this

Where this leads