1Cademy - Solving $$\left\{ 3x - 2y = -2,\; 5x - 6y = 10 \right\}$$ by Elimination

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Solving a System of Linear Equations by Elimination

Example

Solving $\left\{ 3x - 2y = -2,; 5x - 6y = 10 \right\}$ by Elimination

Solve the system $\left\{\begin{array}{l} 3x - 2y = -2 5x - 6y = 10 \end{array}\right.$ using the elimination method.

Step 1 — Write both equations in standard form. Both equations are already in the form $Ax + By = C$ , so no rewriting is needed.

Step 2 — Make the coefficients of one variable opposites. Neither pair of coefficients are already opposites. To eliminate $y$ , note that the coefficients of $y$ are $-2$ and $-6$ . Multiplying the first equation by $-3$ transforms its coefficient of $y$ from $-2$ to $+6$ , which is the opposite of $-6$ : $-3(3x - 2y) = -3(-2) \implies -9x + 6y = 6$

The system becomes: $\left\{\begin{array}{l} -9x + 6y = 6 5x - 6y = 10 \end{array}\right.$

Step 3 — Add the equations to eliminate $y$ . Adding the left sides and right sides: $-9x + 6y + 5x - 6y = 6 + 10$ $-4x = 16$ The terms containing $y$ cancel because $6y + (-6y) = 0$ .

Step 4 — Solve for the remaining variable. Divide both sides by $-4$ : $x = -4$

Step 5 — Substitute back into an original equation. Substitute $x = -4$ into the first equation $3x - 2y = -2$ : $3(-4) - 2y = -2 \implies -12 - 2y = -2$ Add $12$ to both sides: $-2y = 10$ Divide both sides by $-2$ : $y = -5$