1Cademy - Solving $$\left\{y \geq 3x + 1,\; -3x + y \geq -4\right\}$$ by Graphing

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Solving a System of Linear Inequalities by Graphing

Example

Solving $\left\{y \geq 3x + 1,\; -3x + y \geq -4\right\}$ by Graphing

Solve the system $\left\{\begin{array}{l} y \geq 3x + 1 \\ -3x + y \geq -4 \end{array}\right.$ by graphing.

Step 1 — Graph $y \geq 3x + 1$ . The boundary line is $y = 3x + 1$ . Because the inequality uses the non-strict $\geq$ symbol, draw a solid line. Test the point $(0, 0)$ : $0 \geq 3(0) + 1$ simplifies to $0 \geq 1$ , which is false. Shade the side that does not contain the origin.

Step 2 — Graph $-3x + y \geq -4$ on the same grid. The boundary line is $-3x + y = -4$ . Because the inequality uses the non-strict $\geq$ symbol, draw a solid line. Test the point $(0, 0)$ : $-3(0) + 0 \geq -4$ simplifies to $0 \geq -4$ , which is true. Shade the side that contains the origin.

Step 3 — Identify the solution. The two boundary lines are parallel; writing $-3x + y = -4$ in slope-intercept form gives $y = 3x - 4$ , which has the same slope of $3$ as the first line. The line $y = 3x + 1$ lies above $y = 3x - 4$ . Since both shaded regions are above their respective boundary lines, the overlapping region is the area above and including the higher line, $y = 3x + 1$ . The solution is the doubly-shaded region, which corresponds exactly to the solution for $y \geq 3x + 1$ 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 2026-04-28

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OpenStax Intermediate Algebra

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