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Solving a System of Linear Equations by Graphing

Example

Solving $\{y = \frac{1}{2}x - 3,\; x - 2y = 4\}$ by Graphing

Solve the system $\left\{\begin{array}{l} y = \frac{1}{2}x - 3 \\ x - 2y = 4 \end{array}\right.$ using the graphing method.

First equation: $y = \frac{1}{2}x - 3$ is already in slope-intercept form. The slope is $m = \frac{1}{2}$ and the y-intercept is $(0, -3)$ . Graph this line using its slope and y-intercept.

Second equation: $x - 2y = 4$ is in standard form, so use the intercept method. Setting $y = 0$ gives $x = 4$ , so the x-intercept is $(4, 0)$ . Setting $x = 0$ gives $-2y = 4$ , so $y = -2$ and the y-intercept is $(0, -2)$ .

Graph both lines on the same coordinate system. The two lines are parallel — they have the same slope but different y-intercepts, so they never intersect.

Because no point lies on both lines simultaneously, there is no ordered pair that satisfies both equations. The system has no solution.

This example demonstrates that when a system's two lines turn out to be parallel, the graphing method reveals there is no solution. Rewriting the second equation in slope-intercept form confirms the parallel relationship: $x - 2y = 4$ becomes $y = \frac{1}{2}x - 2$ , which has the same slope $\frac{1}{2}$ as the first equation but a different y-intercept ( $-2$ instead of $-3$ ).

Updated 2026-04-21

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OpenStax Elementary Algebra

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