1Cademy - Solving $$\left\{3x - 2y \geq 12,\; y \geq \frac{3}{2}x + 1\right\}$$ by Graphing

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A System of Linear Inequalities with No Solution

Example

Solving $\left\{3x - 2y \geq 12,\; y \geq \frac{3}{2}x + 1\right\}$ by Graphing

Solve the system $\left\{\begin{array}{l} 3x - 2y \geq 12 \\ y \geq \frac{3}{2}x + 1 \end{array}\right.$ by graphing.

Step 1 — Graph $3x - 2y \geq 12$ . The boundary line is $3x - 2y = 12$ , which has intercepts $x = 4$ and $y = -6$ . Because the inequality uses $\geq$ (non-strict), draw a solid line. Test $(0, 0)$ : $3(0) - 2(0) \geq 12$ gives $0 \geq 12$ , which is false, so shade the side that does not contain the origin.

Step 2 — Graph $y \geq \frac{3}{2}x + 1$ on the same grid. The boundary line is $y = \frac{3}{2}x + 1$ . Because the inequality uses $\geq$ (non-strict), draw a solid line. Test $(0, 0)$ : $0 \geq \frac{3}{2}(0) + 1$ gives $0 \geq 1$ , 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 $3x - 2y = 12$ in slope-intercept form gives $y = \frac{3}{2}x - 6$ , which has the same slope of $\frac{3}{2}$ as the second line. The line $y = \frac{3}{2}x + 1$ lies above $y = \frac{3}{2}x - 6$ . Since the first inequality requires shading below $y = \frac{3}{2}x - 6$ and the second requires shading above $y = \frac{3}{2}x + 1$ , the shaded regions face away from each other. Because there is no region that satisfies both inequalities simultaneously, the system has no solution.

Updated 2026-05-26

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OpenStax Intermediate Algebra

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