Bluebook SAT Test Linear Inequalities in One or Two Variables 4

Question 1

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

Accepted Answer

{0, 1, 2}

Answer

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.

Question 2

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

Accepted Answer

x² > 4

Answer

|2x − 3| > 7 gives x > 5 or x < −2. Test each option against every value in this set. Option A (x² > 4): for x > 5, x² > 25 > 4 ✓; for x < −2, x² > 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.

Question 3

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 < (3/4)x − 2 AND y > −2x + 6. Which must be true about the point (a, b)?

Accepted Answer

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

Answer

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 < (3/4)x − 2 AND 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.

Question 4

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

Accepted Answer

k < 2 only

Answer

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 < 2, divide by (k−2) < 0, flipping the sign: x < k+2 ✓. If k = 2, both sides equal 0, giving 0 > 0, which is false (no solution). So the solution x < k+2 occurs precisely when k < 2. Note: k = −2 is valid since (−2−2)x > (−2−2)(0) → −4x > 0 → x < 0 = −2+2 ✓.

Question 5

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?

Accepted Answer

p = −2

Answer

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. Testing p=−2: −4.17 ≤ −2 ≤ −1.24 ✓. Check: (−2, 4−8)=(−2,−4). First: −4 ≥ −4−4=−8 ✓. Second: −4 ≤ (2/3)+8=8.67 ✓. p=−2 works. But is p=3 in: 9−6−4=−1, not ≥ 0, so fails first inequality. p=4: second condition 16+4/3−16=4/3>0, fails. p=5: second fails worse. Answer is p=−2.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Linear Inequalities in One or Two Variables 4