Questions: Systems of Inequalities

Question 1

A student graphs y > x and y < 4 and shades the region where EITHER inequality is satisfied. What error has the student made?

• A. The student used the wrong type of boundary lines (solid instead of dashed)
• B. The student found the union of the two half-planes instead of their intersection
• C. The student should not shade at all — the solution is just the two boundary lines
• D. The student graphed the inequalities in the wrong order

Answer: (B)

The solution to a system of inequalities is the intersection of all half-planes — every point that satisfies ALL constraints simultaneously. Using the union (shading where ANY one inequality holds) is the core conceptual error. A point like (0, 3) satisfies y < 4 but not y > x, so it should not be in the solution. Only points where both shaded regions overlap belong to the feasible region.

Question 2

When graphing 2x + y < 5, which boundary line is correct?

• A. A solid line through 2x + y = 5, because the boundary is part of the solution
• B. A dashed line through 2x + y = 5, because points on the line do NOT satisfy the strict inequality
• C. A solid line through 2x + y = 5, because all linear boundaries are solid
• D. A dashed line through x = 5/2, because only x-intercepts matter

Answer: (B)

Strict inequalities (< and >) use dashed boundary lines because points on the boundary line make the expression equal (not strictly less than or greater than), so they are NOT solutions. Non-strict inequalities (≤ and ≥) use solid lines because boundary points do satisfy the inequality. The boundary here is 2x + y = 5, drawn as a dashed line.

Question 3

A system of inequalities can have no solution — an empty feasible region — if the constraints are contradictory.

• T. True
• F. False

Answer: True

The feasible region is the intersection of all half-planes. If the constraints conflict — for example, y > 5 and y < 2 simultaneously — no point can satisfy both, and the intersection is empty. This is perfectly valid mathematically. The system has no solution, just as a system of equations can have no solution (parallel lines).

Question 4

The solution to a system of inequalities is the set of most points that satisfy at least one of the inequalities.

• T. True
• F. False

Answer: False

The solution requires satisfying ALL inequalities simultaneously, not just one. A point in the solution must lie in every half-plane at once — it must be in the overlap of all shaded regions. A point that satisfies only some constraints is in the union (not the intersection) of the half-planes and is not a solution to the system.

Question 5

Why does the 'corner test' (using a test point like (0,0)) work for identifying the correct half-plane, and when would you NOT use (0,0) as your test point?

Model answer: The test point strategy works because every linear boundary divides the plane into exactly two half-planes: one where the inequality is satisfied, one where it is not. Plugging any single point into the inequality reveals which side is which. You cannot use (0,0) when the boundary line passes through the origin, because then (0,0) is ON the line — not on either side — and will give an ambiguous result (0 = 0 makes the inequality inconclusive). In that case, choose any other point not on the line, like (1, 0) or (0, 1).

The test point method eliminates the need to memorize direction rules when rearranging inequalities, which often introduces sign errors. Its only limitation is that the test point must not lie on the boundary line itself. When graphing y > x, for instance, the line y = x passes through the origin, so use (1, 0) instead: 0 > 1 is false, confirming (1,0) is in the shaded-wrong half-plane, so shade the other side (above).