EST I HS Math · Unit 1 · Foundational Algebra

Lesson 1.6 · Systems of Inequalities & Shaded Regions

On the paper: 1–2 questions per real paper; the 2026 sample has two (Q34, Q46)

Big idea. One linear inequality is half a plane. Two of them overlap in a region. The EST almost never asks you to draw it — it hands you points and asks which one lives inside. So the skill is checking a point in two lines of arithmetic.

1Half-planes and the overlap ◩

🍼 Baby steps

1
y ≤ x − 2
The line y = x − 2, and everything BELOW it (smaller y). Solid line: points on it count.
2
y > −x + 4
Everything ABOVE that line. Dashed line: points on it do not count (strict).
3
both → the wedge where the two shadings overlap
A system means AND. A point must pass every inequality.
4
2x − y ≤ 4 is y ≥ 2x − 4
Rearranging to y-form flips the sign when you divide by −1. Better: do not rearrange — just substitute the point into the original.

🎛 Play with it · shade and test

Flip ≤/≥, slide the lines, then move P. The two lines of arithmetic under the graph are exactly what you do on the exam.

m₁1b₁-2

m₂-1b₂4

y ≤ 1x − 2

y ≥ -1x + 4

The darker overlap (both tints) is the solution set. ≥ shades above the line, ≤ below.

test point Px1y3

3 ≤ 1·1 − 2 = -1 → ✗ false

3 ≥ -1·1 + 4 = 3 → ✓ true

P is NOT a solution — it fails at least one inequality.

2Test the point. Stop at the first failure. 🎯

🍼 Baby steps

1
{ 2x − y ≤ 4 ; y + 4x > 1 } P(−1, 1)?
2026 sample Q46: which point does NOT satisfy the system.
2
2(−1) − 1 = −3 ≤ 4 ✓
First inequality passes.
3
1 + 4(−1) = −3 > 1 ✗
Second fails. (−1, 1) is the answer. You never needed a graph.
4
−3 < 2x − y ≤ 14 (3, 4): 2 → passes; (0, 3): −3 → fails (needs > −3)
May 2021. A double inequality is a strip between two parallel lines — same method, watch the strict end.

Trap · the boundary

A point ON a line passes ≤ / ≥ and fails < / >. The EST plants exactly one such point among the options (May 2021: (0, 3) gives −3, and the sign is strict).

3Integers in a region 🔢

🍼 Baby steps

1
{ x + y < 13 ; y − x > 7 }, a > 1, a and b integers
2026 sample Q34 asks for bᵃ.
2
y > x + 7 and y < 13 − x
Put y between two lines.
3
x + 7 < 13 − x → x < 3
For the strip to be non-empty. With x > 1 and integer: x = 2 = a.
4
x = 2: 9 < y < 11 → y = 10 = b → bᵃ = 10² = 100
Only one integer fits. Squeeze the variable between the two lines, then count.

4Writing the system from words 📝

🍼 Baby steps

1
"must not exceed 250 kg" → 10a + 25b ≤ 250
October 2020: jars of 10 kg and 25 kg. Weight goes with weight.
2
"total sum must be at least $900" → 45a + 120b ≥ 900
Money goes with money. Two quantities, two inequalities, and each option pairs a sign with the wrong quantity.
3
"does not exceed 40 moves, at least 10 horizontal, 800 points" → h + v ≤ 40, h ≥ 10, 5h + 3v ≤ 800
December 2020. Count, minimum, budget — three sentences, three lines.

5Real EST questions, solved twice 📄

Real question 1EST I HS · 2026 official sample · Q46Application

{ 2x − y ≤ 4 ; y + 4x > 1 }

Which of the points below does not satisfy the system of inequalities above?

A.(−1, 6)
−8 ≤ 4 ✓ and 2 > 1 ✓ — a solution.
B.(1, 2)
0 ≤ 4 ✓ and 6 > 1 ✓.
C.(0, 5)
−5 ≤ 4 ✓ and 5 > 1 ✓.
D.(−1, 1)

Safe way · ~60 s

Rearrange: y ≥ 2x − 4 and y > 1 − 4x. Sketch both, shade, plot four points.

⏱ 90-second way · ~20 s

Substitute each point into the ORIGINAL inequalities. (−1, 1): 2(−1) − 1 = −3 ≤ 4 ✓; 1 + 4(−1) = −3 > 1 ✗. Done.

Answer: D

Real question 2EST I HS · 2026 official sample · Q34Reasoning

{ x + y < 13 ; y − x > 7 }

If the point (a, b), where a > 1 and both a and b are integers, lies in the solution set of the system above, what is the value of bᵃ?

A.100
B.225
15² — would need b = 15, but y < 13 − x < 12.
C.1,000
10³ — a = 3 is excluded (x < 3).
D.1,024
2¹⁰ — a and b swapped.

Safe way · ~70 s

Graph y > x + 7 and y < 13 − x; find the lattice points with x > 1 in the wedge.
Only (2, 10). 10² = 100.

⏱ 90-second way · ~25 s

x + 7 < y < 13 − x ⇒ x < 3. With x > 1: x = 2. Then 9 < y < 11 ⇒ y = 10. 10² = 100.

Answer: A

Real question 3EST I · May 2021 · Section II · Q25Application

−3 < 2x − y ≤ 14

Which point could be the solution of the inequality above?

A.(0, 3)
2(0) − 3 = −3, and the left sign is strict: −3 is not > −3.
B.(4, −8)
8 + 8 = 16 > 14.
C.(3, 4)
D.(4, 12)
8 − 12 = −4 < −3.

Safe way · ~40 s

Compute 2x − y for each: −3, 16, 2, −4.
Only 2 sits in (−3, 14].

⏱ 90-second way · ~15 s

Same computation — but note the trap first: (0, 3) lands exactly on the strict boundary.

Answer: C

Real question 4EST I · January 2023 · Q5Reasoning

{ y ≥ 0 ; x − 2 ≤ −3 ; y − x ≤ 3 }

The solution set of the system above is a triangular region. What are the coordinates of its vertices?

A.(−3, 0), (1, 0), (−2, 1)
x = 1 violates x ≤ −1.
B.(−3, 0), (−1, 0), (−1, 2)
C.(0, −3), (−1, 0), (2, 1)
y = −3 violates y ≥ 0.
D.(−3, 0), (−1, 0), (−1, −2)
y = −2 violates y ≥ 0.

Safe way · ~70 s

Lines: y = 0, x = −1, y = x + 3.
Pairwise intersections: y = 0 & x = −1 → (−1, 0); y = 0 & y = x + 3 → (−3, 0); x = −1 & y = x + 3 → (−1, 2).

⏱ 90-second way · ~25 s

Every vertex must satisfy all three. Kill options with a point that breaks y ≥ 0 or x ≤ −1: A has x = 1, C has y = −3, D has y = −2. B survives.

Answer: B

Real question 5EST I · October 2020 · Section II · Q15Application

A craftsman is looking for two kinds of paint from a wholesaler. The first kind a is packaged in 10 kg jars, the second b in 25 kg jars. The 10 kg jar costs $45 and the 25 kg one costs $120. The load must not exceed 250 kg and the total sum must be at least $900 in order to get a discount. Which system of inequalities verifies the given information?

A.10a + 25b ≥ 250 ; 45a + 120b ≤ 900
Both signs reversed.
B.10a + 25b ≤ 250 ; 45a + 120b ≤ 900
"At least $900" is ≥.
C.10a + 25b ≥ 250 ; 45a + 120b ≥ 900
"Must not exceed 250 kg" is ≤.
D.10a + 25b ≤ 250 ; 45a + 120b ≥ 900

Safe way · ~40 s

Weights: 10a + 25b; "must not exceed" → ≤ 250.
Money: 45a + 120b; "at least" → ≥ 900.

⏱ 90-second way · ~12 s

Two phrases, two signs: not exceed = ≤, at least = ≥. Only one option has that pair.

Answer: D

⏱ The 90-second lane

One question = 90 seconds. These are the moves that finish this topic in one line.

1Point questions: substitute into the ORIGINAL inequalities. Stop at the first ✗ (for "not a solution") or first ✓✓ (for "is a solution").
2Never rearrange to y = … unless you must graph; dividing by a negative flips the sign and costs 20 seconds.
3Boundary points: ≤/≥ include the line, </> exclude it. One option always sits on a strict boundary.
4Integers in a region: squeeze one variable between the two lines (x + 7 < y < 13 − x), get the range of the other, count.
5Vertices of a region: intersect the boundary lines pairwise — or kill options with one coordinate that breaks a constraint.
6From words: "not exceed / at most" → ≤, "at least" → ≥, one inequality per quantity (weight, money, count).
7Which quadrant has no solutions: pick the sign of x and y in that quadrant and see if the inequalities can both hold.

Take with you

▸A system of inequalities is an AND: the overlap of half-planes.
▸The exam tests points, not drawings — two lines of arithmetic per point.
▸Words map to signs: not exceed ≤, at least ≥; each quantity gets its own line.

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