Math · College algebra · Worked example

Solve a system of equations by substitution

Solve the system y = 2x − 1 and 3x + 2y = 12.

y = 2 x - 1, 3 x + 2 y = 12

Substitute for y

The first equation already gives y. Replace y in the second equation with 2x − 1.

3 x + 2 (2 x - 1) = 12

Solve for x

Distribute and combine like terms: 7x − 2 = 12, so 7x = 14 and x = 2.

\begin{array}{l} 3 x + 2 (2 x - 1) = 12 \\ 7 x - 2 = 12 \\ x = 2 \end{array}

Find y

Put x = 2 into y = 2x − 1: y = 3.

Check in both original equations

3 = 2(2) − 1 and 3(2) + 2(3) = 12, so (2, 3) satisfies both.

\begin{array}{l} 2 (2) - 1 = 3 \\ 3 (2) + 2 (3) = 12 \end{array}

Result

x = 2 and y = 3.

Your turn

Solve x = 3y + 2 and 2x − 5y = 7.

Show the answer and explanation

x = 11 and y = 3.

Substitute x = 3y + 2 into the second equation: 2(3y + 2) − 5y = 7, so y + 4 = 7 and y = 3. Then x = 3(3) + 2 = 11. Check: 2(11) − 5(3) = 7.

\begin{array}{l} 2 (3 y + 2) - 5 y = 7 \\ y + 4 = 7 \\ y = 3 \end{array}

Keep exploring

In Matrices & linear systems the same system is the rows −2, 1, −1 and 3, 2, 12. Row reduction reaches the same x = 2 and y = 3 without choosing a variable to isolate.

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OpenStax College Algebra 2e — Systems of linear equations: two variables

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