How to solve a system of equations by substitution

The substitution method isolates one variable in one equation and replaces it in the other equation to solve for the remaining variable. This method applies best when at least one equation has a variable with a coefficient of 1 or -1.

The setup

Identify the two equations in the system. Inspect both equations to find a variable with a coefficient of 1 or -1 to minimize fractional arithmetic.

The steps

  1. Isolate one variable in one equation. 2. Substitute the resulting expression into the other equation in place of that variable. 3. Solve the new single-variable equation. 4. Substitute the solved value back into the isolated expression from step 1 to find the second variable.

Checking the result

Substitute both numerical values back into the original unmanipulated equations. Both equations must evaluate to true algebraic statements.

Common errors

Failing to distribute a negative sign or coefficient across the entire substituted expression. Stopping after finding the first variable and forgetting to solve for the second.

Worked example

Solve the system: 2x+y=72x + y = 7 and 3x2y=03x - 2y = 0.

Isolate y in the first equation: y=72xy = 7 - 2x. Substitute this expression into the second equation: 3x2(72x)=03x - 2(7 - 2x) = 0. Distribute the -2: 3x14+4x=03x - 14 + 4x = 0. Combine like terms: 7x14=07x - 14 = 0. Solve for x: 7x=147x = 14, so x=2x = 2. Substitute x back into the isolated equation to find y: y=72(2)=74=3y = 7 - 2(2) = 7 - 4 = 3. The solution is (2,3)(2, 3).

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References: OpenStax College Algebra, Chapter 7: Systems of Equations and Inequalities · Khan Academy, Algebra 1: Systems of equations

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