1Cademy - Solving $$\{2x + y = 7,\; x - 2y = 6\}$$ by Elimination

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Solving a System of Linear Equations by Elimination

Example

Solving $\{2x + y = 7,\; x - 2y = 6\}$ by Elimination

Solve the system $\left\{\begin{array}{l} 2x + y = 7 \\ x - 2y = 6 \end{array}\right.$ using the elimination method.

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

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

$2(2x + y) = 2(7) \implies 4x + 2y = 14$

The system becomes $\left\{\begin{array}{l} 4x + 2y = 14 \\ x - 2y = 6 \end{array}\right.$

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

$4x + 2y + x - 2y = 14 + 6$

$5x = 20$

The $y$ -terms cancel because $2y + (-2y) = 0$ .

Step 4 — Solve for the remaining variable. Divide both sides by $5$ :

$x = 4$

Step 5 — Substitute back into an original equation. Substitute $x = 4$ into the second equation $x - 2y = 6$ :

$4 - 2y = 6$

$-2y = 2$

$y = -1$

Step 6 — Write the solution as an ordered pair: $(4, -1)$ .

Step 7 — Check in both original equations:

First equation: $2(4) + (-1) = 8 - 1 = 7$ . Since $7 = 7$ is true ✓
Second equation: $4 - 2(-1) = 4 + 2 = 6$ . Since $6 = 6$ is true ✓

Both equations are satisfied, confirming that $(4, -1)$ is the solution. This is the same system previously solved by graphing and by substitution, and all three methods produce the same answer — demonstrating that the elimination method is another valid algebraic approach to solving systems of linear equations.

Updated 2026-04-24

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