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Elimination Method — Definition, Formula & Examples

The elimination method is a technique for solving a system of linear equations by adding or subtracting the equations to cancel out one variable, leaving a single equation in one unknown that you can solve directly.

Given a system of two linear equations in two unknowns, the elimination method (also called the addition method) multiplies one or both equations by suitable constants so that the coefficients of one variable become opposites. Adding the resulting equations eliminates that variable, reducing the system to a single linear equation in one unknown. Back-substitution then yields the complete solution.

How It Works

Start by writing both equations in standard form (ax+by=cax + by = c). Choose which variable to eliminate, then multiply one or both equations by constants so that variable's coefficients are exact opposites. Add the equations together — the chosen variable cancels, and you solve for the remaining one. Finally, substitute that value back into either original equation to find the other variable. This method works for any system of two linear equations and extends naturally to larger systems via Gaussian elimination.

Worked Example

Problem: Solve the system: 2x+3y=122x + 3y = 12 4x3y=64x - 3y = 6
Step 1: Align and inspect coefficients: Both equations are already in standard form. Notice that the yy-coefficients are +3+3 and 3-3, which are already opposites.
2x+3y=124x3y=62x + 3y = 12 \qquad 4x - 3y = 6
Step 2: Add the equations to eliminate y: Add the left sides together and the right sides together. The yy-terms cancel.
(2x+4x)+(3y3y)=12+6    6x=18(2x + 4x) + (3y - 3y) = 12 + 6 \implies 6x = 18
Step 3: Solve for x: Divide both sides by 6.
x=186=3x = \frac{18}{6} = 3
Step 4: Back-substitute to find y: Substitute x=3x = 3 into the first equation and solve for yy.
2(3)+3y=12    6+3y=12    3y=6    y=22(3) + 3y = 12 \implies 6 + 3y = 12 \implies 3y = 6 \implies y = 2
Answer: The solution is (x,y)=(3,2)(x, y) = (3, 2).

Another Example

Problem: Solve the system: 3x+2y=163x + 2y = 16 5x+4y=285x + 4y = 28
Step 1: Choose a variable to eliminate: The yy-coefficients are 2 and 4. Multiply the first equation by 2-2 so the yy-coefficients become 4-4 and +4+4.
2(3x+2y)=2(16)    6x4y=32-2(3x + 2y) = -2(16) \implies -6x - 4y = -32
Step 2: Add the modified first equation to the second: The yy-terms cancel when you add.
(6x+5x)+(4y+4y)=32+28    x=4(-6x + 5x) + (-4y + 4y) = -32 + 28 \implies -x = -4
Step 3: Solve for x and back-substitute: From x=4-x = -4, we get x=4x = 4. Substitute into the original first equation.
3(4)+2y=16    2y=4    y=23(4) + 2y = 16 \implies 2y = 4 \implies y = 2
Answer: The solution is (x,y)=(4,2)(x, y) = (4, 2).

Why It Matters

The elimination method is a core technique in Algebra 1 and Algebra 2, appearing on standardized tests including the SAT and ACT. Engineers and scientists use it routinely when balancing chemical equations, analyzing electrical circuits, or solving mixture problems. It also lays the groundwork for Gauss-Jordan elimination, the matrix-based method used in linear algebra and computer science.

Common Mistakes

Mistake: Multiplying only one side of an equation by the chosen constant.
Correction: You must multiply every term on both sides of the equation. For example, multiplying 3x+2y=163x + 2y = 16 by 2-2 gives 6x4y=32-6x - 4y = -32, not 6x4y=16-6x - 4y = 16.
Mistake: Forgetting to back-substitute to find the second variable.
Correction: After solving for one variable, plug that value into either original equation and solve for the other. Always verify your answer by checking both values in both equations.

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