1Cademy - Solving $$\left\{y \leq -\frac{1}{4}x + 2,\; x + 4y \leq 4\right\}$$ by Graphing

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Solving a System of Linear Inequalities by Graphing

Example

Solving $\left\{y \leq -\frac{1}{4}x + 2,; x + 4y \leq 4\right\}$ by Graphing

Solve the system $\left\{\begin{array}{l} y \leq -\frac{1}{4}x + 2 x + 4y \leq 4 \end{array}\right.$ by graphing.

Step 1 — Graph $y \leq -\frac{1}{4}x + 2$ . The boundary line is $y = -\frac{1}{4}x + 2$ . Because the inequality uses the non-strict $\leq$ symbol, draw a solid line. Test the point (0, 0): $0 \leq -\frac{1}{4}(0) + 2$ simplifies to $0 \leq 2$ , which is true. Shade the side that contains the origin.

Step 2 — Graph $x + 4y \leq 4$ on the same grid. The boundary line is $x + 4y = 4$ . Because the inequality uses the non-strict $\leq$ symbol, draw a solid line. Test the point (0, 0): $0 + 4(0) \leq 4$ simplifies to $0 \leq 4$ , which is true. Shade the side that contains the origin.

Step 3 — Identify the solution. The two boundary lines are parallel; writing $x + 4y = 4$ in slope-intercept form gives $y = -\frac{1}{4}x + 1$ , which has the same slope of $-\frac{1}{4}$ as the first line. The line $y = -\frac{1}{4}x + 1$ lies below $y = -\frac{1}{4}x + 2$ . Since both shaded regions are below their respective boundary lines, the overlapping region is the area below and including the lower line, $y = -\frac{1}{4}x + 1$ . The solution is the doubly-shaded region, which corresponds exactly to the solution for $x + 4y \leq 4$ alone.

Step 4 — Verify with a test point. Choose a test point from the overlapping region and substitute it into both inequalities to confirm both are true.

Updated 2026-06-17

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OpenStax Intermediate Algebra

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