A system of inequalities asks for every point that satisfies all of them at once. You already know how to graph one linear inequality as a shaded half-plane, and from systems of equations that a solution must work in every equation.
Why it works
Each inequality's solutions fill a half-plane. A point solves the system only if it is in the first half-plane and in the second, so the solution set is where the shadings overlap, the same “and” idea as in compound inequalities. Points in only one shading are not solutions. Boundaries follow the same rules as before: dashed for $<$ and $>$, solid for $\le$ and $\ge$, and a point on a dashed boundary is not included.
Key idea
Graph each inequality (boundary dashed or solid, shade the side that works). The solution set is the overlap. Check with a test point that lies in the overlap: it must make every inequality true.
Worked examples
Worked example: Graph and check a system
Describe the solutions of $y \le -x + 4$ and $y > 2x - 2$.
- First: solid line $y = -x + 4$, shade below.
- Second: dashed line $y = 2x - 2$, shade above.
- The lines cross where $-x + 4 = 2x - 2$, at $x = 2$, $y = 2$. The overlap is the wedge to the left of that point, between the lines.
- Test $(0, 0)$: $0 \le 4$ is true and $0 > -2$ is true, so $(0, 0)$ is a solution.
Answer: The wedge left of $(2, 2)$, including the solid boundary but not the dashed one.
Worked example: A budget
Pens cost \$2 and notebooks \$5. You can spend at most \$30 and want at least $3$ notebooks. Can you buy $6$ pens and $4$ notebooks?
- Let $p$ = pens and $n$ = notebooks: $2p + 5n \le 30$ and $n \ge 3$.
- Test $(6, 4)$: $2(6) + 5(4) = 32$, and $32 \le 30$ is false.
Answer: No: it meets the notebook goal but goes over budget.
Common mistake
Reporting every shaded point, from either inequality, as a solution. That is the union, an “or.” A system means “and,” so only the overlap counts. Check any point you choose in every inequality.