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Absolute Value and Distance on a Number Line

Evaluate and compare absolute values and find distances between points, including negative numbers and tolerances.

Practice this skill 20 min Absolute value and distance on a number line Leer en español
Key idea

What you'll learn

Absolute value is a number's distance from zero. In this lesson you will learn to evaluate absolute value expressions, compare them, and use absolute value to find the distance between two points on a number line — including when one or both points are negative.

How it shows up on the GED:

  • Evaluate expressions such as \lvert 3 - 8 \rvert - \lvert -4 \rvert (often without a calculator).
  • Find the distance between two numbers, or click a point that is a given distance from another point.
  • Real-world word problems: temperature changes across zero, elevations above and below sea level, bank balances that went negative, and manufacturing tolerances ("within 0.05 inch").

Absolute value also prepares you for the coordinate plane: the distance between two points on a horizontal or vertical line is an absolute value. If you can picture a number line, you can solve every question in this lesson.

Words to know

  • Absolute value \lvert x \rvert — the distance from x to 0 on a number line; never negative (\lvert -6 \rvert = 6, \lvert 6 \rvert = 6).
  • Opposites — two numbers the same distance from 0 on opposite sides (-8 and 8).
  • Integer — a whole number or its opposite (\ldots, -2, -1, 0, 1, 2, \ldots).
  • Distance between two points — how far apart they are, always positive or zero (from -2 to 5 is 7).
  • Sea level — elevation 0; above is positive, below is negative (-282 ft).
  • Overdrawn — a bank balance below $0, written as a negative number.
  • Tolerance — the largest allowed distance from a target value (a 2.5-inch pin with tolerance 0.05 inch).
  • Boundary — the endpoint of an allowed range (2.45 and 2.55 for the pin above).

All Math words in the glossary

What absolute value means

The absolute value of a number is how far it is from 0, ignoring direction.

\lvert 5 \rvert = 5 \qquad \lvert -5 \rvert = 5 \qquad \lvert 0 \rvert = 0 \qquad \lvert -2.75 \rvert = 2.75 \qquad \left\lvert -\tfrac{3}{4} \right\rvert = \tfrac34
Number line from -6 to 6. Points at -5 (red) and 5 (blue). Brackets show each is 5 units from 0, so |-5| = 5 and |5| = 5. −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 |−5| = 5 units |5| = 5 units Opposites are the same distance from 0

Key facts:

  • Absolute value is never negative, because distance is never negative.
  • Opposite numbers have the same absolute value: \lvert -9 \rvert = \lvert 9 \rvert = 9.
  • If \lvert x \rvert = 5, then x = 5 or x = -5 (two answers).
  • \lvert x \rvert = -3 has no solution — no distance is negative.

Bars act like parentheses. Do the arithmetic inside first, then take the absolute value:

  • \lvert 2 - 9 \rvert = \lvert -7 \rvert = 7
  • \lvert -4 \rvert + \lvert 3 \rvert = 4 + 3 = 7, but \lvert -4 + 3 \rvert = \lvert -1 \rvert = 1
  • \lvert -3 \rvert \cdot \lvert 4 \rvert = 3 \cdot 4 = 12

Signs outside the bars stay. -\lvert -6 \rvert = -(6) = -6. The bars change only what is inside them.

Comparing and ordering with absolute value

To compare expressions with absolute value, simplify each side to one number, then compare on the number line.

Example: \lvert -9 \rvert vs \lvert 6 \rvert: 9 > 6, so \lvert -9 \rvert > \lvert 6 \rvert.

Example: -\lvert 5 \rvert vs -\lvert -2 \rvert: these are -5 and -2. Since -5 is farther left, -5 < -2.

Size vs value. Absolute value measures size (how far from zero). A negative number can have a bigger size but a smaller value:

  • -10 < 3 (value), but \lvert -10 \rvert > \lvert 3 \rvert (size).
  • A debt of $500 (-500) is a bigger debt than $200 (-200): \lvert -500 \rvert > \lvert -200 \rvert, even though -500 < -200.
Number line from -11 to 4. Points at -10, -2 and 3. Below the line, bars from 0 show their sizes: |-10| = 10, |-2| = 2, |3| = 3. By value, -10 < -2 < 3; by size (absolute value), 2 < 3 < 10. −11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 −10 −2 3 |−10| = 10 |−2| = 2 |3| = 3 value: −10 < −2 < 3; size: |−2| < |3| < |−10|

Real-world language often talks about size: "the larger loss", "the greater depth", "the colder temperature". Decide whether the question wants the value (with sign) or the size (absolute value).

NumberValue on number lineAbsolute value
-10far left10
-2a little left of 02
3right of 03

Distance between two points on a number line

The distance between two numbers a and b is the absolute value of their difference:

d = \lvert b - a \rvert = \lvert a - b \rvert

The order of subtraction does not matter, because the absolute value removes the sign.

Same side of zero → subtract the sizes. Between -9 and -2: \lvert -2 - (-9) \rvert = \lvert 7 \rvert = 7 (also 9 - 2 = 7).

Number line from -10 to 2. Points at -9 and -2, both left of zero. A bracket between them is labeled 7 units: |-2 - (-9)| = 7, the same as 9 - 2 = 7. −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2 7 units Same side of 0: subtract the sizes, 9 − 2 = 7

Opposite sides of zero → add the sizes. Between -4 and 3: \lvert 3 - (-4) \rvert = 7 (also 4 + 3 = 7). See the figure below.

Working backwards — find a point at a given distance. The points d units from c are c + d (right) and c - d (left). Points 6 units from -1: -1 + 6 = 5 and -1 - 6 = -7.

Number line from -8 to 6. The center point -1 is marked in blue. Arrows of 6 units go left to -7 and right to 5 (red points). The two points 6 units from -1 are -7 and 5. −8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 − 6 + 6 −1 − 6 = −7 and −1 + 6 = 5

Tolerance problems use the same idea. "Within t of target T" means

\lvert x - T \rvert \le t \quad\Longleftrightarrow\quad T - t \le x \le T + t

For a target of 10 oz with tolerance 0.2 oz, acceptable weights run from 9.8 to 10.2 oz.

Number line from 9.6 to 10.4 ounces. The target 10 is marked. The acceptable range from 9.8 to 10.2 is shaded green with closed dots at both ends, 0.2 on each side of 10. 9.6 9.7 9.8 9.9 10 10.1 10.2 10.3 10.4 0.2 0.2 |x − 10| ≤ 0.2 means 9.8 ≤ x ≤ 10.2 oz

Real-world reference points: sea level, 0°F, a $0 balance and ground level are all "zero". Write values on one side as positive and the other side as negative, then use \lvert b - a \rvert.

Number line from -6 to 6 labeled at every integer. Points are marked at -4 and 3. A bracket above the line spans from -4 to 3 and is labeled 7 units: 4 units from -4 to 0 plus 3 units from 0 to 3. −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 −4 3 7 units

The distance from -4 to 3 is 7 units: 4 units from -4 to 0 plus 3 units from 0 to 3. In symbols, \lvert 3 - (-4) \rvert = 7.

Formula

Formulas to memorize (not on the formula sheet)

These are not on the GED formula sheet.

Absolute value:

\lvert x \rvert = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases}

(If x is negative, -x is positive: \lvert -7 \rvert = -(-7) = 7.)

Distance between two points on a number line:

d = \lvert b - a \rvert
  • a, b — the coordinates (numbers) of the two points
  • d — the distance, always \ge 0

Points at distance d from c: c + d and c - d.

Tolerance: \lvert x - T \rvert \le t means T - t \le x \le T + t.

Worked example

Example 1 (easy): evaluate an expression

Evaluate \lvert -12 \rvert - \lvert 5 \rvert + \lvert -3 \cdot 2 \rvert.

Three distance bars measured from 0: |-12| is a bar 12 units long to the left of 0, |5| is 5 units to the right, and |-6| (from -3 × 2 = -6) is 6 units to the left. All three absolute values are positive: 12 - 5 + 6 = 13. −12 −10 −8 −6 −4 −2 0 2 4 5 6 |−12| = 12 |5| = 5 |−3 · 2| = |−6| = 6
  1. \lvert -12 \rvert = 12.
  2. \lvert 5 \rvert = 5.
  3. Inside the last bars: -3 \cdot 2 = -6, so \lvert -6 \rvert = 6.
  4. Combine left to right: 12 - 5 + 6 = 13.

Answer: 13.

Check: all three absolute values are positive (12, 5, 6), and 12 - 5 = 7, 7 + 6 = 13. ✓

Worked example

Example 2: distance between −7 and 4

Find the distance between -7 and 4 on a number line.

  1. d = \lvert 4 - (-7) \rvert = \lvert 4 + 7 \rvert = \lvert 11 \rvert = 11.
  2. The other order gives the same result: \lvert -7 - 4 \rvert = \lvert -11 \rvert = 11.
Number line from -8 to 5. Points at -7 and 4. Above, a bracket from -7 to 4 is labeled 11 units. Below, it is split into 7 units from -7 to 0 and 4 units from 0 to 4. −8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 11 units 7 4

Answer: 11 units.

Check: opposite sides of zero, so add the sizes: 7 + 4 = 11. ✓

Worked example

Example 3 (word problem): submarine and helicopter

A submarine is 250 ft below the surface of the sea. A rescue helicopter hovers 1{,}200 ft above the surface. How far apart are they vertically?

  1. Use sea level as 0: submarine = -250, helicopter = 1{,}200.
  2. d = \lvert 1{,}200 - (-250) \rvert = \lvert 1{,}200 + 250 \rvert = 1{,}450.
Vertical number line for elevation with sea level at 0 ft. The helicopter is at +1,200 ft and the submarine at -250 ft. Brackets: 1,200 ft from sea level up to the helicopter and 250 ft from the submarine up to sea level, for a total of 1,450 ft. −400 −200 0 200 400 600 800 1,000 1,200 sea level helicopter +1,200 submarine −250 1,450 1,200 250

Answer: 1,450 ft.

Check: 250 ft up to the surface plus 1{,}200 ft up to the helicopter = 1{,}450 ft. ✓

Worked example

Example 4 (work backwards): find the points

Which numbers are exactly 9 units from -2 on the number line?

  1. Move 9 units right: -2 + 9 = 7.
  2. Move 9 units left: -2 - 9 = -11.
Number line from -12 to 8. The center -2 is marked in blue. Arrows of 9 units go left to -11 and right to 7 (red points). The numbers 9 units from -2 are -11 and 7. −12 −10 −8 −6 −4 −2 0 2 4 6 8 −11 7 9 left 9 right −2 − 9 = −11 and −2 + 9 = 7

Answer: 7 and -11.

Check: \lvert 7 - (-2) \rvert = \lvert 9 \rvert = 9 ✓ and \lvert -11 - (-2) \rvert = \lvert -9 \rvert = 9 ✓.

Worked example

Example 5 (GED-hard): tolerance at a machine shop

A bolt must be 3.75 inches long, with a tolerance of 0.02 inch. Which of these bolts are acceptable: 3.728, 3.74, 3.771, 3.77?

  1. Rule: \lvert L - 3.75 \rvert \le 0.02, so 3.75 - 0.02 \le L \le 3.75 + 0.02, which is 3.73 \le L \le 3.77.
  2. Test each bolt:
    • 3.728: \lvert 3.728 - 3.75 \rvert = 0.022 > 0.02 → rejected
    • 3.74: \lvert -0.01 \rvert = 0.01 → accepted
    • 3.771: \lvert 0.021 \rvert = 0.021 > 0.02 → rejected
    • 3.77: \lvert 0.02 \rvert = 0.02 → accepted (boundary)
Number line from 3.72 to 3.78 inches with ticks every 0.005. The acceptable range 3.73 to 3.77 (3.75 plus or minus 0.02) is shaded green with closed endpoints. Bolts 3.74 and 3.77 (green) are inside the range; bolts 3.728 and 3.771 (red) are just outside. 3.72 3.73 3.74 3.75 3.76 3.77 3.78 0.02 0.02 target 3.75 3.728 ✗ 3.74 ✓ 3.77 ✓ 3.771 ✗ Accepted: 3.73 ≤ L ≤ 3.77 (endpoints included)

Answer: 3.74 in and 3.77 in are acceptable.

Check: pad to three places — 3.730 \le 3.740 \le 3.770 and 3.770 \le 3.770, while 3.728 < 3.730 and 3.771 > 3.770. ✓

Calculator tip

TI-30XS MultiView tips

Most absolute value questions are designed for the calculator-free section. You rarely need an absolute value key: work out the inside, then drop any negative sign.

When the calculator is allowed:

  • Use the (−) key for negative numbers and the subtraction key for "minus". Distance from -45.5 to 120.25: 120.25 − (−) 45.5 enter → 165.75.
  • If the result is negative, the distance is the same number without the sign.
  • For tolerance problems, compute T - t and T + t once, write the range on your erasable note board, and compare each choice to it.

When not to use it: problems like \lvert -7 \rvert + \lvert 3 \rvert or "distance from -5 to 3" are faster with a quick number-line sketch — and a sketch catches sign mistakes that a calculator will not.

Common mistake

Common mistakes

  • Leaving the sign inside. Wrong: \lvert -7 \rvert + \lvert 3 \rvert = -4. Right: 7 + 3 = 10.
  • Combining before taking absolute values. Wrong: \lvert -7 \rvert + \lvert 3 \rvert = \lvert -7 + 3 \rvert = 4. Right: each set of bars is evaluated on its own.
  • Dropping a sign outside the bars. Wrong: -\lvert -4 \rvert = 4. Right: -\lvert -4 \rvert = -4.
  • Subtracting sizes across zero. Wrong: distance from -12 to 15 is 15 - 12 = 3. Right: 15 - (-12) = 27.
  • Adding sizes on the same side. Wrong: distance from -45 to -3 is 48. Right: \lvert -3 - (-45) \rvert = 42.
  • Measuring from zero instead of from the given point. Wrong: "7 units from -3" is 7. Right: -3 + 7 = 4 or -3 - 7 = -10.
Test strategy

GED test strategy

  • Sketch a number line on your erasable note board for every distance question. Mark zero and both points — the sketch tells you whether to add or subtract.
  • Eliminate negative answers for any distance, rise, or "how far" question. Distances are never negative.
  • Translate words into signs: below, under, overdrawn, loss, colder → negative; above, deposit, gain, warmer → positive.
  • Watch for extra information. Multi-step word problems often include numbers you do not need. Underline the two positions the question actually asks about.
  • Plug in to check "find the point" answers: the distance from the given point must equal the stated distance.
  • Tolerance: write the range T - t to T + t, pad the decimals, and remember that "within" includes the endpoints.

Check yourself

Absolute value and distance on a number lineNo calculator

What is the value of \lvert -7 \rvert + \lvert 3 \rvert?

Answer choices
Hint

Find each absolute value first (its distance from zero), then add.

Check yourself

Absolute value and distance on a number lineNo calculatorHot spot

Click the point on the number line that represents -\lvert -4 \rvert.

Click on the figure to mark your answer (or focus it and use the arrow keys).

Number line from -6 to 6 with a tick mark and label at every integer. −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6
Hint

Work from the inside out: first find \lvert -4 \rvert, then apply the negative sign that is outside the bars.

Summary

  • \lvert x \rvert is the distance from x to 0, so it is never negative; opposites have equal absolute values.
  • Treat bars like parentheses: simplify inside first; signs outside the bars stay.
  • Distance between a and b is \lvert b - a \rvert: add sizes across zero, subtract sizes on the same side.
  • The points d units from c are c + d and c - d.
  • Tolerance "within t of T" means \lvert x - T \rvert \le t, the range T - t to T + t, endpoints included.
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