Inequalities & Systems of Inequalities: PSAT/NMSQT Practice Problems

Rewrite the first inequality in slope-intercept form: y − x ≥ 1 → y ≥ x + 1. This is the region above the line y = x + 1.

The second inequality, y ≤ −2x, is the region below (or left of) the steep line y = −2x.

The solution to the system is wherever both shaded regions overlap. Testing a point in Section C, such as (−2, 3): 3 ≥ −2 + 1 = −1 ✓ and 3 ≤ −2(−2) = 4 ✓ — both true.

Testing a point in each other section shows it fails at least one inequality: in Section A, e.g. (2, 0), 0 ≥ 3 is false. In Section B, e.g. (0, 3), 3 ≤ 0 is false. In Section D, e.g. (0, −3), −3 ≥ 1 is false.

The correct answer is C, the region where both shaded areas overlap.

Check each ordered pair against both inequalities — a solution must satisfy both at once.

(4, −2): −2 > 4 − 4 = 0? No — fails the first inequality.

(0, 2): 2 > 0 − 4 = −4? Yes. 0 + 2 = 2 < 5? Yes. Both true!

(5, 3): 3 > 5 − 4 = 1? Yes. But 5 + 3 = 8 < 5? No — fails the second inequality.

(0, −5): −5 > 0 − 4 = −4? No — fails the first inequality.

Only (0, 2) satisfies both inequalities. The correct answer is B.

Rewrite the first inequality in slope-intercept form. Dividing by −2 flips the inequality sign: x − 2y ≤ −2 → −2y ≤ −2 − x → y ≥ (1/2)x + 1. This is the region above the line y = (1/2)x + 1.

The second inequality, y < −x + 2, is the region below the dashed line y = −x + 2 (dashed because the inequality is strict).

Testing a point in Section R, such as (−3, 1): 1 ≥ (1/2)(−3) + 1 = −0.5 ✓ and 1 < −(−3) + 2 = 5 ✓ — both true.

Testing points in the other sections shows they each fail at least one inequality: Section Q (e.g. (1, 3)) fails the second inequality; Section P (e.g. (4, 0)) fails the first; Section S (e.g. (1, −3)) fails the first.

The correct answer is C, Section R.

Rewrite both inequalities in slope-intercept form. For the first: 2 − y < 2x → −y < 2x − 2 → y > 2 − 2x (dividing by −1 flips the sign).

For the second: −x ≤ 4 − y → y − x ≤ 4 → y ≤ x + 4.

Focus on the first inequality, y > 2 − 2x, since it's the more restrictive one for negative x. Quadrant III requires both x < 0 and y < 0.

But whenever x is negative, −2x is positive, so 2 − 2x is always greater than 2. That means y must be greater than a number bigger than 2 — so y itself must always be positive whenever x is negative.

This makes it impossible for both x < 0 and y < 0 to hold at the same time, so no point in Quadrant III can ever satisfy the first inequality. Quadrants I, II, and IV can each be checked with a sample point (like (10,10), (−0.5, 3.4), and (5, −1) respectively) to confirm they do contain solutions.

The correct answer is B, Quadrant III.

The boundary line is horizontal, which means it represents a constant y-value — so the inequality must be in terms of y, not x. That immediately rules out choices A and B.

The dashed line sits at y = −2. Since the line is dashed (not solid), the inequality is strict — it does not include equality, ruling out any "≥" or "≤" version.

The shaded region lies above the line, which corresponds to y-values greater than −2.

The correct answer is C, y > −2.

First find the slope using the two labeled points, (−3, 0) and (0, 2): slope = (2 − 0) / (0 − (−3)) = 2/3.

With a y-intercept of 2, the boundary line is y = (2/3)x + 2. Multiplying every term by 3 to clear the fraction: 3y = 2x + 6 → 3y − 2x = 6. This matches the "3y − 2x" form in choices C and D — ruling out A and B, which come from a completely different slope.

The line is solid, so the inequality includes equality (≥ or ≤). The shaded region lies above the line — for example, the point (0, 3) is shaded, and 3 ≥ (2/3)(0) + 2 = 2 is true.

Substituting (0, 3) into the equation form confirms it: 3(3) − 2(0) = 9, and 9 ≥ 6 ✓.

The correct answer is C, 3y − 2x ≥ 6.

Find the slope using the two labeled points, (−1, 0) and (0, −1): slope = (−1 − 0) / (0 − (−1)) = −1. With a y-intercept of −1, the boundary line is y = −x − 1, which rearranges to x + y = −1.

The line is solid, so the inequality includes equality — ruling out choices A and B.

The shaded region is on the side of the line away from the origin — the lower-left region. Testing the point (−3, −3), which is clearly shaded: −3 + (−3) = −6, and −6 ≤ −1 is true.

Testing the origin (0, 0), which is clearly unshaded: 0 + 0 = 0, and 0 ≤ −1 is false — correctly excluded.

The correct answer is C, x + y ≤ −1.

The line passes through the origin and (1, 2), so its slope is 2/1 = 2, giving the equation y = 2x, or equivalently 2x − y = 0. This matches the "2x − y" form in choices A and B — the "x − 2y" choices come from a line with the opposite (reciprocal) slope, so they can be ruled out immediately.

The line is solid, so the inequality includes equality. The shaded region is to the left of this steep line — testing a shaded point like (−2, 3): 2(−2) − 3 = −7, and −7 ≤ 0 is true.

Testing an unshaded point like (2, 0) (to the right of the line): 2(2) − 0 = 4, and 4 ≤ 0 is false — correctly excluded.

The correct answer is B, 2x − y ≤ 0.