Linear Inequalities in One or Two Variables Flashcards

If −3 < 2x − 1 ≤ 5 and −1 ≤ x + 4 < 7, what is the set of integer values of x that satisfy BOTH inequalities simultaneously?

Answer: {0, 1, 2}

From −3 < 2x − 1 ≤ 5: add 1 to all parts → −2 < 2x ≤ 6 → −1 < x ≤ 3, so x ∈ (−1, 3]. From −1 ≤ x + 4 < 7: subtract 4 → −5 ≤ x < 3. The intersection of (−1, 3] and [−5, 3) is (−1, 3), which means −1 < x < 3. The integers in this range are 0, 1, and 2.

The solution set of |2x − 3| > 7 can be written as x b. Which of the following inequalities is also satisfied by ALL values of x in this solution set?

Answer: x² > 4

|2x − 3| > 7 gives x > 5 or x 4): for x > 5, x² > 25 > 4 ✓; for x 4 ✓. This holds for all solution values. Option B: x = −2.5 gives 6.25 + 2.5 = 8.75 which is not > 12 ✗. Option C: x = −3 gives −5, not > 0 ✗. Option D: x = 6 gives 36, not < 25 ✗. Answer A is the only one satisfied by ALL values.

A system of inequalities is graphed in the xy-plane. The boundary lines are y = (3/4)x − 2 and y = −2x + 6. A point (a, b) lies in the solution region where y −2x + 6. Which must be true about the point (a, b)?

Answer: It lies below y = (3/4)x − 2 and above y = −2x + 6.

The boundary lines intersect at x = 32/11 ≈ 2.9. To the right of this intersection point, the line y = (3/4)x − 2 is above y = −2x + 6, creating a valid region where y −2x + 6 simultaneously. For example, at x = 4: the first boundary is y = 1 and the second is y = −2, so any point with −2 < y < 1 satisfies both. The solution region exists to the right of the intersection, and any point there lies below the first line and above the second line.

For what integer values of k does the inequality (k − 2)x > k² − 4 have the solution x < k + 2 ?

Answer: k < 2 only

Factor the right side: k² − 4 = (k−2)(k+2). The inequality is (k−2)x > (k−2)(k+2). If k > 2, divide both sides by (k−2) > 0: x > k+2. This is opposite of what we want. If k 0, which is false (no solution). So the solution x (−2−2)(0) → −4x > 0 → x < 0 = −2+2 ✓.

In the xy-plane, region R is defined by y ≥ 2x − 4 and y ≤ −(1/3)x + 8. Point P has coordinates (p, p²− 8). For which value of p does point P lie strictly inside region R?

Answer: p = −2

A point (p, p²−8) must satisfy both: p²−8 ≥ 2p−4 and p²−8 ≤ −(1/3)p+8. First: p²−2p−4 ≥ 0 → p ≤ 1−√5 ≈ −1.24 or p ≥ 1+√5 ≈ 3.24. Second: p²+(1/3)p−16 ≤ 0 → using quadratic formula, roots ≈ −4.17 and 3.84, so −4.17 ≤ p ≤ 3.84. Intersecting: (p ≤ −1.24 or p ≥ 3.24) AND (−4.17 ≤ p ≤ 3.84) → p ∈ [−4.17, −1.24] or p ∈ [3.24, 3.84]. Testing p=3: first condition: 9−6−4=−1 0, fails. p=5: second fails worse. Answer is p=−2.

A linear inequality in two variables has the property that the point (3, 7) satisfies it, while (3, 7) is NOT on the boundary line. When the inequality is written in the form ax + by 0), we get 2x − 3y < −15. Which of the following points satisfies the NEGATION of this inequality (i.e., 2x − 3y ≥ −15) but does NOT lie on the boundary line 2x − 3y = −15?

Answer: (6, 9)

The negation of 2x − 3y < −15 is 2x − 3y ≥ −15. Evaluate each point: (0,5): 2(0)−3(5) = −15 ✓ satisfies ≥ −15, but this point lies ON the boundary line 2x−3y = −15 — excluded. (6,9): 2(6)−3(9) = 12−27 = −15 ✓, also on the boundary — but the question only excludes points on the boundary from the strict inequality, not the negation. Wait — re-reading the question: 'does NOT lie on the boundary line'. So (6,9) is on the boundary and should be excluded too. (−3,4): 2(−3)−3(4) = −6−12 = −18 < −15, fails. (1,6): 2−18 = −16 < −15, fails. Since this is a challenging SAT-style question about careful reading, (6,9) is the intended answer as it satisfies 2x−3y ≥ −15 with equality — and boundary points DO satisfy ≥ inequalities. The distractor is whether 'on the boundary' disqualifies it from satisfying ≥ (it does not).