How to solve a system of equations by elimination

The elimination method solves a system of linear equations by adding a multiple of one equation to another to cancel out a variable. It applies best when both equations are in standard form Ax+By=CAx + By = C.

The setup

Align both equations in standard form Ax+By=CAx + By = C. Ensure like variables and constants are vertically aligned in columns.

The steps

  1. Multiply one or both equations by constants so that the coefficients of one variable are exact opposites. 2. Add the two equations together to eliminate that variable. 3. Solve the resulting single-variable equation. 4. Substitute this value back into either original equation to find the remaining variable.

Checking the result

Substitute the derived (x,y)(x, y) coordinate pair into the original equation not used in step 4. Both sides must evaluate to the identical numerical value.

Common errors

Failing to multiply the constant term on the right side of the equals sign is a frequent error. Another is subtracting equations instead of adding opposites, which frequently leads to sign distribution errors.

Worked example

Solve the system 2x+3y=122x + 3y = 12 and 5xy=135x - y = 13.

Align equations: 2x+3y=122x + 3y = 12 and 5xy=135x - y = 13. Multiply the second equation by 3 to get 15x3y=3915x - 3y = 39. Add this to the first equation: (2x+15x)+(3y3y)=(12+39)(2x + 15x) + (3y - 3y) = (12 + 39). This yields 17x=5117x = 51. Solve for x: x=3x = 3. Substitute x=3x = 3 into the first equation: 2(3)+3y=122(3) + 3y = 12 yields 6+3y=126 + 3y = 12. Subtract 6 from both sides: 3y=63y = 6, which gives y=2y = 2. The solution is (3,2)(3, 2).

FAQ

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References: OpenStax Intermediate Algebra, Chapter 3 · Khan Academy, Systems of Equations Unit

See also