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1.1.1

Absolute Value, Simplifying & Units

EST I HS Math · Unit 1 · Foundational Algebra

Lesson 1.1 · Absolute Value, Simplifying & Units

On the paper: Foundational Algebra is 27–32% of the 50 questions (≈ 14–16 marks)

Big idea. |x − a| is not a formula. It is the distance between x and a on the number line. Read it that way and every absolute-value question on the EST becomes 'how far, and which side'.

1What |x| really says 📏

Forget "make it positive". Say it like this: |x| is how far x is from 0, and |x − a| is how far x is from a. Distance is never negative — that is the only reason the answer is "positive".

🍼 Baby steps

  1. 1

    |7| = 7 |−7| = 7

    Both are 7 steps from 0. Direction is thrown away; only the length stays.

  2. 2

    |x − 3| = 2

    Read: "x is 2 steps from 3". Two places do that: 3 − 2 = 1 and 3 + 2 = 5.

  3. 3

    x = 1 or x = 5

    So an equation |A| = d (with d > 0) always splits into two: A = d and A = −d.

  4. 4

    |x − 3| ≤ 2 → 1 ≤ x ≤ 5

    "At most 2 steps from 3" is the segment between 1 and 5. One piece.

  5. 5

    |x − 3| ≥ 2 → x ≤ 1 or x ≥ 5

    "At least 2 steps from 3" is everything outside — two rays, never one piece.

🎛 Play with it · the distance picture

Slide the centre a and the distance d. Switch =, ≤, ≥ and watch where the solutions live. The integer count is a real EST question type.

-10-9-8-7-6-5-4-3-2-1012345678910d = 2.5d = 2.5a = 2-0.54.5

|x − 2| = 2.5

x = -0.5 or x = 4.5

Two answers, both at distance 2.5 from 2: 2 − 2.5 and 2 + 2.5.

Rule · the three shapes

|A| = d → A = d or A = −d (two points) · |A| ≤ d → −d ≤ A ≤ d (one segment) · |A| ≥ d → A ≤ −d or A ≥ d (two rays).

Trap · the negative on the right

|A| = −3 has no solution — a distance cannot be −3. But |A| ≥ −3 is true for every x, and |A| ≤ −3 for none. The EST asks exactly this (March 2021: "which equation has a real solution?").

2Pull the constant out — the 10-second move ⚡

The paper never writes |x − 2|. It writes |2x − 4| and hopes you square it. Do not. A constant factor comes straight out of an absolute value: |2x − 4| = 2|x − 2|.

🍼 Baby steps

  1. 1

    |2x − 4| = 5

    Spot the common factor inside: 2x − 4 = 2(x − 2).

  2. 2

    2|x − 2| = 5

    |2 · (x − 2)| = |2| · |x − 2| = 2|x − 2|. Only the sign is lost, so a positive factor just walks out.

  3. 3

    |x − 2| = 2.5

    Divide by 2. Now it is pure distance: "x is 2.5 from 2".

  4. 4

    x = −0.5 or x = 4.5

    2 − 2.5 and 2 + 2.5. Both are 2.5 away from 2 — which is what the December 2020 paper actually asked.

  5. 5

    |−2b − 3| = |2b + 3| = 2|b + 1.5|

    A negative factor comes out as its size: |−2| = 2. So |−2b − 3| ≤ 7 is |b + 1.5| ≤ 3.5, centre −1.5.

90-second lane · why this wins

Every "how many integer values" and "distance between the solutions" question is answered by the centre and the distance alone. Centre a, distance d: solutions a ± d, their difference is 2d, their sum is 2a, and the integers in |x − a| ≤ d are counted on the segment.

3Nested bars and 'A² = B²' 🧅

Two disguises the EST likes. Both peel in one move.

🍼 Peel from the outside

  1. 1

    | |x| − 2 | = 1

    Treat the inside |x| as one thing, call it u. Then |u − 2| = 1.

  2. 2

    u = 1 or u = 3

    "u is 1 from 2".

  3. 3

    |x| = 1 → x = ±1 |x| = 3 → x = ±3

    Now peel the inner bars. Four answers in total — so the graph of f meets y = 1 in 4 points.

🍼 A² = B² means A = ±B

  1. 1

    (3x − 2)² − (x + 3)² = 0

    Do not expand. Two squares equal each other.

  2. 2

    3x − 2 = x + 3 or 3x − 2 = −(x + 3)

    Squares are equal when the insides are equal or opposite. This is |3x − 2| = |x + 3| in disguise.

  3. 3

    x = 2.5 or x = −0.25

    Two linear equations, no quadratic formula, no calculator. The paper then asks |2.5 − (−0.25)| = 2.75 = 11/4.

4Units: square the factor when the unit is squared 📐

🍼 Baby steps

  1. 1

    1 m = 3.28 ft

    A length factor.

  2. 2

    1 m² = 3.28² ft² = 10.76 ft²

    Area is length × length, so the factor is squared too. This single step is the whole June 2021 question.

  3. 3

    1400 ft² ÷ 10.76 = 130 m²

    Going from the small unit (ft²) to the big unit (m²) means fewer of them — divide.

  4. 4

    19 cm/s × 3 s = 57 cm; 57 ÷ 29.9 = 1.9

    Chain conversions: multiply by rate, then divide by "how much one new unit is worth". (Dec 2020, light-nanoseconds.)

Trap · dividing by 3.28 once

1400 ÷ 3.28 = 427. That number sits in option A of the real paper, waiting. A squared unit needs a squared factor.

5Real EST questions, solved twice 📄

Real question 1EST I HS · 2026 official sample · Q44Knowledge

What is the value of x if 4|2x + 1| = 10 and x < 0?

  1. A.1.75

    Positive — the question says x < 0.

  2. B.0.75

    The positive branch, x = 0.75, rejected by x < 0.

  3. C.−1.75
  4. D.−2.0

    Comes from 2x + 1 = −3, i.e. forgetting to divide 10 by 4.

Safe way · ~40 s

  1. Divide by 4: |2x + 1| = 2.5.
  2. Split: 2x + 1 = 2.5 → x = 0.75; 2x + 1 = −2.5 → x = −1.75.
  3. Keep the one with x < 0: x = −1.75.

⏱ 90-second way · ~15 s

  1. |2x + 1| = 2.5 → 2|x + 0.5| = 2.5 → |x + 0.5| = 1.25.
  2. Centre −0.5, distance 1.25, the LEFT one: −0.5 − 1.25 = −1.75.

Answer: C

Real question 2EST I · December 2020 · Section I · Q9Application

If a is a solution of the equation |2x − 4| = 5, what is the distance between a and the point of coordinate 2 on the number line?

  1. A.0.5

    That is the smaller solution itself (−0.5, sign dropped), not its distance from 2.

  2. B.2.5
  3. C.4.5

    The larger solution, not a distance.

  4. D.5

    The 5 on the right-hand side — still divided by nothing.

Safe way · ~50 s

  1. 2x − 4 = 5 → x = 4.5; 2x − 4 = −5 → x = −0.5.
  2. Distance from 2: |4.5 − 2| = 2.5 and |−0.5 − 2| = 2.5.
  3. Either solution gives 2.5.

⏱ 90-second way · ~10 s

  1. |2x − 4| = 2|x − 2| = 5 → |x − 2| = 2.5.
  2. That IS "the distance from x to 2". Done: 2.5.

Answer: B

Real question 3EST I · October 2020 · Section II · Q33 (grid-in on the paper; options are ours)Application

If |2b − 1| ≤ 3, how many possible integer values of b are there?

  1. A.3

    Forgot one endpoint — both ends are included because the sign is ≤.

  2. B.4
  3. C.5

    Counted 0 twice, or included −2 which gives |−5| = 5 > 3.

  4. D.6

    Counted integers of 2b (−3..3 has 7) and then dropped one.

Safe way · ~45 s

  1. −3 ≤ 2b − 1 ≤ 3.
  2. Add 1: −2 ≤ 2b ≤ 4. Divide by 2: −1 ≤ b ≤ 2.
  3. Integers: −1, 0, 1, 2 → four.

⏱ 90-second way · ~15 s

  1. 2|b − 0.5| ≤ 3 → |b − 0.5| ≤ 1.5 → segment from −1 to 2.
  2. Count on the segment: −1, 0, 1, 2 → 4.

Answer: B

Real question 4EST I · December 2021 · Section I · Q11Reasoning

The graphs of f(x) = | |x| − 2 | and g(x) = 1 in the same system will have:

  1. A.0 points of intersection

    f does reach the value 1 — it reaches every value ≥ 0.

  2. B.1 point of intersection

    Only if you solved |x| − 2 = 1 and forgot the ± on |x|.

  3. C.2 points of intersection

    The outer bars alone give 2 values of |x|; the inner bars double them.

  4. D.4 points of intersection

Safe way · ~60 s

  1. Set | |x| − 2 | = 1 → |x| − 2 = 1 or |x| − 2 = −1.
  2. |x| = 3 → x = ±3; |x| = 1 → x = ±1.
  3. Four x-values, four intersection points.

⏱ 90-second way · ~20 s

  1. Peel outside-in: outer bars give 2 values of |x|, each positive, and each positive |x| gives 2 values of x.
  2. 2 × 2 = 4 — no need to name them.

Answer: D

Real question 5EST I · June 2021 · Section II · Q30Application

Harvey wants to buy a new apartment. He estimates that he will need its area to be around 1400 square feet. But the provider measures its units in square meters. If 1 meter is nearly 3.28 feet, how many square meters approximately does Harvey need?

  1. A.427

    1400 ÷ 3.28 — divided by the length factor once. The unit is squared.

  2. B.130
  3. C.4592

    1400 × 3.28 — multiplied, and only once.

  4. D.15062

    1400 × 3.28² — right factor, wrong direction (that would be m² → ft²).

Safe way · ~45 s

  1. 1 m = 3.28 ft, so 1 m² = 3.28 × 3.28 = 10.7584 ft².
  2. Square metres = 1400 ÷ 10.7584 = 130.1.

⏱ 90-second way · ~12 s

  1. Squared unit → squared factor. Big unit → fewer of them → divide.
  2. On the calculator: 1400 ÷ 3.28 x² = → 130.

Answer: B

⏱ The 90-second lane

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

  • 1Read |A| = d as "A is d away from 0". Never square both sides.
  • 2Pull the constant out first: |2x − 4| = 2|x − 2|, |−3x + 6| = 3|x − 2|. Then the centre and distance are visible.
  • 3Centre a, distance d ⇒ solutions a ± d, difference 2d, sum 2a. "Sum of the solutions" needs no solving.
  • 4≤ is one segment (count integers on it, both ends in). ≥ is two rays.
  • 5|A| = negative → no solution. |A| ≥ negative → every x. |A| ≤ negative → no x.
  • 6A² = B² ⇒ A = ±B. Two linear equations instead of a quadratic.
  • 7Nested bars: peel from the outside, count 2 × 2.
  • 8Squared unit ⇒ squared factor. Big unit ⇒ divide. Type 3.28 x² not 3.28.

Take with you

  • ▸|x − a| is the distance from x to a. Every rule (=, ≤, ≥) is a picture on the number line.
  • ▸Constants walk out of the bars as their size: |k·A| = |k|·|A|.
  • ▸The EST asks the same family every year: solve, count integers, distance between solutions, nested bars, which has a real solution.
  • ▸Units: area and volume conversions square or cube the factor; the wrong single-factor answer is always one of the options.

Keep reading — this lesson marks itself done automatically.