1Cademy - Solving $$\left\{x + 3y > 8,\; y < -\frac{1}{3}x

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A System of Linear Inequalities with No Solution

Example

Solving $\left\{x + 3y > 8,\; y < -\frac{1}{3}x - 2\right\}$ by Graphing

Solve the system $\left\{\begin{array}{l} x + 3y > 8 \\ y < -\frac{1}{3}x - 2 \end{array}\right.$ by graphing.

Step 1 — Graph $x + 3y > 8$ . The boundary line is $x + 3y = 8$ , which has intercepts $x = 8$ and $y = \frac{8}{3}$ . Because the inequality uses $>$ (strict), draw a dashed line. Test $(0, 0)$ : $0 + 3(0) > 8$ gives $0 > 8$ , which is false, so shade the side that does not contain the origin.

Step 2 — Graph $y < -\frac{1}{3}x - 2$ on the same grid. The boundary line is $y = -\frac{1}{3}x - 2$ . Because the inequality uses $<$ (strict), draw a dashed line. Test $(0, 0)$ : $0 < -\frac{1}{3}(0) - 2$ gives $0 < -2$ , which is false, so shade the side that does not contain the origin.

Step 3 — Identify the solution. The two boundary lines are parallel; writing $x + 3y = 8$ in slope-intercept form gives $y = -\frac{1}{3}x + \frac{8}{3}$ , which has the same slope of $-\frac{1}{3}$ as the second line. The line $y = -\frac{1}{3}x + \frac{8}{3}$ lies above $y = -\frac{1}{3}x - 2$ . The first inequality requires shading above $y = -\frac{1}{3}x + \frac{8}{3}$ , and the second requires shading below $y = -\frac{1}{3}x - 2$ . The shaded regions face away from each other and do not overlap. Thus, the system has no solution.

Updated 2026-05-26

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OpenStax Intermediate Algebra

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