Bluebook SAT Test Linear Inequalities in One or Two Variables 6

Question 1

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

Accepted Answer

−19/3 < x ≤ 3

Answer

Multiply all three parts by 4: −8 ≤ 1 − 3x < 20. Subtract 1: −9 ≤ −3x < 19. Divide by −3, which flips both inequality signs: 3 ≥ x > −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.

Question 2

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

Accepted Answer

0

Answer

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

Question 3

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

Accepted Answer

x > 1

Answer

For both inequalities to be satisfied simultaneously, there must exist a y with −2x + 4 ≤ y < 3x − 1. This is only possible when the lower bound is strictly less than the upper bound: −2x + 4 < 3x − 1. Solving: 5 < 5x, so x > 1. At x = 1 exactly, we'd need y ≥ 2 and y < 2, which is impossible — so x ≥ 1 is too broad. The condition is strictly x > 1.

Question 4

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

Accepted Answer

0 < x + y < 9

Answer

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.

Question 5

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

Accepted Answer

(4, −1)

Answer

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.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Linear Inequalities in One or Two Variables 6