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1.3.1

Linear Inequalities

EST I HS Math · Unit 1 · Foundational Algebra

Lesson 1.3 · Linear Inequalities

On the paper: 2–4 questions on every real EST I paper; the 2026 spec lists inequality laws, linear inequalities and their graphs

Big idea. An inequality is solved exactly like an equation, with one exception: multiply or divide by a negative and the sign turns around. Then the question is almost never 'what is x' — it is 'which integer', 'the greatest value of', or 'at most how many'.

1Same moves as an equation — except one ↔️

🍼 Baby steps

  1. 1

    2z − 7(z − 1) ≤ 1

    December 2020. Expand first, exactly as for an equation.

  2. 2

    2z − 7z + 7 ≤ 1 → −5z ≤ −6

    Collect. Now the coefficient of z is negative.

  3. 3

    z ≥ 6/5 = 1.2

    Divide by −5 and FLIP: ≤ becomes ≥. The paper then asks the least integer z: 2.

  4. 4

    alternative: 7 − 1 ≤ 7z − 2z → 6 ≤ 5z → z ≥ 1.2

    Move the z-terms to the side where they stay positive and there is nothing to flip. Same answer, no risk.

🎛 Play with it · the flip

Keep a < b and slide k through zero. When k is negative the two points swap order — that is the whole rule.

-12-8-404812before: 1 < 313after × -2: -2 > -6 — the sign FLIPPED

1 < 3 ⟹ -2·(1) > -2·(3) ⟹ -2 > -6

Rule: adding or subtracting anything, or multiplying/dividing by a positive number, keeps the sign. Multiplying or dividing by a negative number reverses it. Nothing else ever flips it.

Rule · what flips and what does not

Add/subtract anything: no flip. Multiply/divide by a positive: no flip. Multiply/divide by a negative: flip. Swapping the two sides: flip (a < b is b > a).

2Compound inequalities: work on all three parts at once 🎯

🍼 Baby steps

  1. 1

    27 < 4x − 3 ≤ 31

    December 2024. Two inequalities sharing a middle. Do the same thing to all three parts.

  2. 2

    30 < 4x ≤ 34

    Add 3 everywhere.

  3. 3

    7.5 < x ≤ 8.5

    Divide by 4 everywhere (positive: no flip).

  4. 4

    x = 8

    The only integer in (7.5, 8.5]. Notice: the answer was never "x" — it was "which integer".

  5. 5

    3 < 2x + 7 ≤ 15 → −2 < x ≤ 4 → smallest x = −1

    October 2021. The left end −2 is EXCLUDED (strict), so the smallest integer is −1, and x + 3 = 2.

Trap · strict vs inclusive at the ends

< excludes the endpoint, ≤ includes it. In "which integer" questions the answer is usually sitting right at an endpoint — check which sign guards it.

3'Highest value of another expression' 🔝

🍼 Baby steps

  1. 1

    5 − (3/2)x ≥ 3

    October 2020 asks: what is the highest value of (9/8)x + 1?

  2. 2

    2 ≥ (3/2)x → x ≤ 4/3

    Solve for the range of x first.

  3. 3

    (9/8)x + 1 is increasing in x

    Positive coefficient → it is biggest when x is biggest.

  4. 4

    at x = 4/3: (9/8)(4/3) + 1 = 3/2 + 1 = 2.5

    Plug the endpoint in. If the coefficient were negative, you would plug the SMALLEST x instead.

4Words: at least, at most, does not exceed 📝

🍼 Baby steps

  1. 1

    at least 90 → ≥ 90 at most 23 → ≤ 23

    "at least" includes the number. "at most" / "does not exceed" / "no more than" include it too.

  2. 2

    more than → > fewer than / less than → <

    These exclude the number.

  3. 3

    42 + 2w ≥ 90 → w ≥ 24

    October 2021: 90 hours over three weeks, 42 done, the rest split equally over two weeks. 24 hours each.

  4. 4

    6y ≤ 23 → y ≤ 3.83 → y = 3

    2026 sample: copybooks at $6 with $23. "Maximum number" → round DOWN to a whole item. 3.83 copybooks do not exist.

5Real EST questions, solved twice 📄

Real question 1EST I · December 2024 · Q20Knowledge

27 < 4x − 3 ≤ 31

The only integer that satisfies the inequality above is:

  1. A.8
  2. B.11

    Solved 4x − 3 = 31 + ... — 11 would give 4(11) − 3 = 41, far outside.

  3. C.28

    Read the middle as x, not 4x − 3.

  4. D.32

    Nowhere near: 4(32) − 3 = 125.

Safe way · ~35 s

  1. Add 3 to all parts: 30 < 4x ≤ 34.
  2. Divide by 4: 7.5 < x ≤ 8.5.
  3. Only integer: 8.

⏱ 90-second way · ~10 s

  1. Test the options in 4x − 3: 8 → 29 ✓ (between 27 and 31). First hit wins.

Answer: A

Real question 2EST I · October 2020 · Section II · Q14Application

5 − (3/2)x ≥ 3

What is the highest value of (9/8)x + 1?

  1. A.2.5
  2. B.3.5

    Used x = 2 (forgot the 3/2 factor).

  3. C.4.5

    Flipped the sign the wrong way and picked an x that is not allowed.

  4. D.5.5

    Evaluated (9/8)x + 1 at x = 4.

Safe way · ~45 s

  1. 2 ≥ (3/2)x → x ≤ 4/3.
  2. (9/8)x + 1 increases with x, so use the largest x = 4/3.
  3. (9/8)(4/3) + 1 = 1.5 + 1 = 2.5.

⏱ 90-second way · ~15 s

  1. Largest x = 2 ÷ 1.5 = 4/3. The 8 and the 4 cancel: (9 × 4)/(8 × 3) = 3/2. Plus 1 → 2.5.

Answer: A

Real question 3EST I · December 2020 · Section II · Q6Application

If 2z − 7(z − 1) ≤ 1 and z is an integer, what is the least possible value of z?

  1. A.−2

    Forgot to flip when dividing by −5 (z ≤ 1.2 → "least" makes no sense, so a guess).

  2. B.0

    Sign slip in expanding −7(z − 1).

  3. C.2
  4. D.4

    An integer that works, but not the least.

Safe way · ~40 s

  1. 2z − 7z + 7 ≤ 1 → −5z ≤ −6.
  2. Divide by −5, flip: z ≥ 1.2.
  3. Least integer: 2.

⏱ 90-second way · ~15 s

  1. Move z to the positive side: 7 − 1 ≤ 5z → z ≥ 1.2 → 2. No flip needed.

Answer: C

Real question 4EST I · October 2021 · Section I · Q11Application

To prepare for a boxing tournament, Jamison needs to work out for at least 90 hours during the next 3 weeks. In the first week he practised 42 hours and decided to divide the rest of the hours equally over the remaining weeks. How many hours at least should he practise each remaining week?

  1. A.12 hours

    (90 − 42) ÷ 4, or 48 ÷ 4.

  2. B.24 hours
  3. C.46 hours

    90 − 42 − 2, a slip.

  4. D.48 hours

    The total that remains — not divided between the two weeks.

Safe way · ~35 s

  1. 42 + 2w ≥ 90.
  2. 2w ≥ 48 → w ≥ 24.

⏱ 90-second way · ~10 s

  1. Remaining hours 90 − 42 = 48, over 2 weeks → 24 each. "At least" here just means "the minimum that works".

Answer: B

Real question 5EST I HS · 2026 official sample · Q26Knowledge

A notebook costs $5 and a copybook costs $6.

What is the maximum number of copybooks that can be purchased with $23?

  1. A.10

    Ignores the price entirely.

  2. B.4

    4 copybooks cost $24 — over budget by $1. Rounding 3.83 UP.

  3. C.3
  4. D.1

    Would leave $17 unspent — not the maximum.

Safe way · ~25 s

  1. 6y ≤ 23 → y ≤ 3.83.
  2. Whole copybooks only: y = 3.

⏱ 90-second way · ~8 s

  1. 23 ÷ 6 = 3.8 → floor → 3. "Maximum" with a budget always rounds DOWN.

Answer: C

⏱ The 90-second lane

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

  • 1Move the variable to the side where its coefficient stays positive — then you never divide by a negative and never need to flip.
  • 2Compound inequality: do the same thing to all three parts at once.
  • 3"Which integer satisfies" → after solving, look at the endpoints and check < vs ≤.
  • 4Testing the options in the original inequality is often faster than solving — first hit wins.
  • 5"Greatest value of (expression)": positive coefficient → plug the largest x; negative coefficient → plug the smallest x.
  • 6at least / at most / does not exceed / no more than → ≥ or ≤ (endpoint included). more than / fewer than → strict.
  • 7"Maximum number you can buy" → divide and round DOWN. "Minimum number needed" → divide and round UP.
  • 8Averages: to get an average of A over n tests you need a TOTAL of nA. Work in totals, not averages.

Take with you

  • ▸One rule differs from equations: a negative multiplier flips the sign. Avoid it by keeping x positive.
  • ▸The EST asks for an integer, an endpoint, or the extreme of another expression — solve, then read the last line.
  • ▸Word problems: translate "at least / at most", solve, and round in the direction the situation forces.

Keep reading — this lesson marks itself done automatically.