Linear Inequalities in One or Two Variables Flashcards

If −2 ≤ (1 − 3x)/4 < 5, which of the following represents all possible values of x?

Answer: −19/3 < x ≤ 3

Multiply all three parts by 4: −8 ≤ 1 − 3x −19/3, or equivalently −19/3 < x ≤ 3. Choice B swaps which endpoint is strict. Choice C fails to flip the inequalities when dividing by a negative. Choice D uses 'or' instead of 'and,' which is a compound intersection, not a union.

For which integer value of k does the inequality kx > k² have no solution for x?

Answer: 0

Rewrite as kx > k². If k > 0, divide both sides by k to get x > k, which has infinitely many solutions. If k 0, which is always false — no value of x satisfies it. Therefore k = 0 is the only value that yields no solution.

The system of inequalities y ≥ −2x + 4 and y < 3x − 1 has at least one solution for all values of x satisfying which condition?

Answer: x > 1

For both inequalities to be satisfied simultaneously, there must exist a y with −2x + 4 ≤ y 1. At x = 1 exactly, we'd need y ≥ 2 and y 1.

If 3 < 2x + 1 < 11 and −1 < y < 4, which of the following correctly describes all possible values of x + y?

Answer: 0 < x + y < 9

First isolate x: subtract 1 from all parts of 3 < 2x + 1 < 11 to get 2 < 2x < 10, then divide by 2: 1 < x < 5. Now add the inequalities for x and y: (1) + (−1) < x + y < (5) + (4), giving 0 < x + y < 9. Choice A incorrectly uses the bounds of y alone. Choice C mistakenly adds 1 (from 2x + 1) instead of using the isolated x range. Choice D uses non-strict inequalities, but all bounds are strict.

Which of the following points lies in the solution region of 2x − 3y > 6 in the xy-plane?

Answer: (4, −1)

Substitute each point into 2x − 3y > 6. (A): 2(0) − 3(−2) = 6; this equals 6, not greater than 6 — it lies on the boundary line, not in the open region. (B): 2(3) − 3(0) = 6; also on the boundary. (C): 2(4) − 3(−1) = 8 + 3 = 11 > 6 ✓ — this point is inside the region. (D): 2(1) − 3(−1) = 2 + 3 = 5 < 6 — this is in the wrong half-plane. The boundary points in A and B are designed to trap students who ignore the strict inequality.

How many ordered pairs (x, y) satisfy both x − 2y ≤ 6 and 2x − 4y ≥ 14?

Answer: None

Notice the left sides are proportional: the second inequality is exactly 2 times the first. Multiply the first inequality by 2: 2x − 4y ≤ 12. The second inequality requires 2x − 4y ≥ 14. A single expression (2x − 4y) cannot simultaneously be ≤ 12 and ≥ 14, since 12 < 14. These constraints are contradictory, so the system has no solution. Students who see 'parallel lines' and guess 'infinitely many' or 'one' are misapplying rules from systems of equations rather than inequalities.