1Cademy - Solving $$\left\{3x + y = 5,\; 2x - 3y = 7\right\}$$ by Elimination

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Solving a System of Linear Equations by Elimination

Example

Solving $\left\{3x + y = 5,\; 2x - 3y = 7\right\}$ by Elimination

Solve the system $\left\{\begin{array}{l} 3x + y = 5 \\ 2x - 3y = 7 \end{array}\right.$ using the elimination method.

Step 1 — Write both equations in standard form. Both equations are already in the standard form $Ax + By = C$ .

Step 2 — Make the coefficients of one variable opposites. To eliminate $y$ , multiply the first equation by $3$ so that the $y$ -coefficients become $3$ and $-3$ :

$3(3x + y) = 3(5) \implies 9x + 3y = 15$

The system becomes $\left\{\begin{array}{l} 9x + 3y = 15 \\ 2x - 3y = 7 \end{array}\right.$

Step 3 — Add the equations to eliminate one variable. Adding the left and right sides together:

$9x + 3y + 2x - 3y = 15 + 7$

$11x = 22$

The $y$ -terms cancel because $3y + (-3y) = 0$ .

Step 4 — Solve for the remaining variable. Divide both sides by $11$ :

$x = 2$

Step 5 — Substitute back into an original equation. Substitute $x = 2$ into the first original equation $3x + y = 5$ :

$3(2) + y = 5$

$6 + y = 5$

$y = -1$

Step 6 — Write the solution as an ordered pair: $(2, -1)$ .

Step 7 — Check in both original equations:

First equation: $3(2) + (-1) = 6 - 1 = 5$ . Since $5 = 5$ is true ✓
Second equation: $2(2) - 3(-1) = 4 + 3 = 7$ . Since $7 = 7$ is true ✓

Because both equations are satisfied, the solution of the system is $(2, -1)$ .

Updated 2026-05-07

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OpenStax Intermediate Algebra

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