Absolute value and distance on a number line

1. Absolute value is distance from zero, so it is never negative.
2. \lvert -7 \rvert = 7 (the number -7 is 7 units from 0).
3. \lvert 3 \rvert = 3.
4. Add: 7 + 3 = 10.

Answer: 10.

Check: on a number line, walk 7 units (from -7 to 0) and then 3 more units — 10 units in all. ✓

Tip: Absolute value bars work like parentheses: evaluate inside each set of bars first, then do the operation between them. \lvert -7 \rvert + \lvert 3 \rvert is not the same as \lvert -7 + 3 \rvert = 4.

1. Evaluate the absolute values: \lvert -7 \rvert = 7 and \lvert 9 \rvert = 9.
2. Replace them: 2 \cdot 7 - 9.
3. Multiply before subtracting: 2 \cdot 7 = 14.
4. Subtract: 14 - 9 = 5.

Answer: 5.

Check: the bars only affect what is inside them. 2\lvert -7 \rvert means 2 times the distance of -7 from zero, which is 2 \times 7 = 14; then 14 - 9 = 5. ✓

Tip: Absolute value bars act like parentheses: simplify inside each set first, then follow the normal order of operations (multiply before you subtract). Never merge two separate sets of bars into one.

1. Place the scores on a number line: Ana at -4, Beth at 3, par at 0.
2. Distance = \lvert 3 - (-4) \rvert = \lvert 3 + 4 \rvert = 7.

Their scores were 7 strokes apart.

Check: from -4 up to par is 4 strokes; from par up to +3 is 3 strokes; 4 + 3 = 7. ✓

Tip: When two values are on opposite sides of zero (under and over par, below and above sea level, owing and having money), add their sizes. Subtracting 4 - 3 = 1 treats both scores as if they were on the same side.

1. A rise is the change from the start to the end: \text{end} - \text{start}.
2. 15 - (-12) = 15 + 12 = 27.
3. Distance view: \lvert 15 - (-12) \rvert = 27.

The temperature rose 27°F.

Check (on a thermometer): from -12 up to 0 is 12 degrees, and from 0 up to 15 is 15 degrees; 12 + 15 = 27. ✓

Tip: A temperature that crosses zero changes by the sum of the two distances from zero. Subtracting 15 - 12 = 3 treats -12 as if it were +12.

1. Sea level is 0. Mount Whitney is at +14{,}505 ft; Badwater Basin is at -282 ft.
2. Distance = \lvert 14{,}505 - (-282) \rvert.
3. 14{,}505 - (-282) = 14{,}505 + 282 = 14{,}787.

The vertical distance is 14,787 feet.

Check: from Badwater up to sea level is 282 ft, and from sea level up to the summit is 14{,}505 ft; 282 + 14{,}505 = 14{,}787. ✓

Tip: "Above" and "below" a reference point (sea level, ground level, zero balance) are on opposite sides of zero — add their sizes to get the distance.

1. First bars: 3 - 8 = -5, and \lvert -5 \rvert = 5.
2. Second bars: \lvert -4 \rvert = 4.
3. Subtract: 5 - 4 = 1.

Answer: 1.

Check: both absolute values are positive (5 and 4), and 5 is one more than 4, so the result must be 1. ✓

Tip: The minus sign between the bars is a real subtraction; it stays. Only the signs inside the bars disappear after you take the absolute value.

1. "6 units from -2" means \lvert x - (-2) \rvert = 6, or \lvert x + 2 \rvert = 6.
2. Move right: -2 + 6 = 4.
3. Move left: -2 - 6 = -8.

Answer: -8 and 4.

Check: \lvert 4 - (-2) \rvert = \lvert 6 \rvert = 6 ✓ and \lvert -8 - (-2) \rvert = \lvert -6 \rvert = 6 ✓. The center -2 is exactly halfway between them: \dfrac{-8 + 4}{2} = -2 ✓.

Tip: There are always two points at a given distance from a center c: c - d and c + d. They are balanced around c, not around 0.

1. Read the points. The small ticks are halves.
   • M is halfway between -4 and -3: M = -3.5.
   • N is halfway between 2 and 3: N = 2.5.
2. Distance = \lvert 2.5 - (-3.5) \rvert = \lvert 2.5 + 3.5 \rvert = 6.

Answer: 6 units.

Check: from M to 0 is 3.5 units and from 0 to N is 2.5 units; 3.5 + 2.5 = 6. ✓

Tip: Always check the scale. Here there are two ticks per unit, so 12 tick spaces between M and N means 6 units, not 12.

1. The rule \lvert T - 98.6 \rvert \le 1.2 says T is at most 1.2 degrees from 98.6, in either direction.
2. Lowest value: 98.6 - 1.2 = 97.4.
3. Highest value: 98.6 + 1.2 = 99.8.
4. So 97.4 \le T \le 99.8.

Answer: from 97.4°F to 99.8°F.

Check: both endpoints are exactly 1.2 from the target: \lvert 97.4 - 98.6 \rvert = 1.2 ✓ and \lvert 99.8 - 98.6 \rvert = 1.2 ✓. The range is 2.4 degrees wide, twice the tolerance. ✓

Tip: A tolerance applies both below and above the target, and "within" includes the endpoints. Line up the decimal points carefully: 1.2 is twelve tenths, not twelve hundredths.

1. Street level = 0, up is positive.
   • P1: -12 ft
   • P4: -42 ft
   • Roof: +85 ft (not needed)
2. Distance between P1 and P4: \lvert -12 - (-42) \rvert = \lvert -12 + 42 \rvert = \lvert 30 \rvert = 30.

Answer: 30 feet.

Check: both levels are below the street, so subtract their depths: 42 - 12 = 30. ✓

Tip: Same side of zero → subtract the sizes. The roof height is extra information; GED problems often include a number you do not need, so decide which two positions matter before you calculate.

1. Before - bill = after: \text{before} - 212.40 = -37.15.
2. Undo the subtraction by adding 212.40 to the ending balance: \text{before} = -37.15 + 212.40.
3. 212.40 - 37.15 = 175.25, and the result is positive because 212.40 > 37.15.

Answer: $175.25.

Check: 175.25 - 212.40 = -37.15 ✓. The bill used up the whole $175.25 and then went $37.15 below zero: 175.25 + 37.15 = 212.40 ✓.

Tip: Working backwards means doing the opposite operation. If you get a negative "before" balance here, the bill would have made the account even lower — always plug your answer back into the story.

1. Trip 1: from 3 down to -2: \lvert -2 - 3 \rvert = \lvert -5 \rvert = 5 floors.
2. Trip 2: from -2 up to 6: \lvert 6 - (-2) \rvert = \lvert 8 \rvert = 8 floors.
3. Total distance: 5 + 8 = 13 floors.

Answer: 13 floors.

Check: on a vertical number line, count the floors: 3 \to 0 is 3, 0 \to -2 is 2 (trip 1 = 5); -2 \to 0 is 2, 0 \to 6 is 6 (trip 2 = 8). 5 + 8 = 13 ✓.

Tip: "Total distance traveled" adds the size of every move. "Net change" only compares start and finish (6 - 3 = 3). Read the question to see which one it wants.

1. Distance = \left\lvert 1\tfrac12 - \left(-2\tfrac34\right) \right\rvert = 1\tfrac12 + 2\tfrac34.
2. Common denominator 4: 1\tfrac12 = 1\tfrac24.
3. Add: 1\tfrac24 + 2\tfrac34 = 3\tfrac54 = 3 + 1\tfrac14 = 4\tfrac14.

Answer: 4\tfrac14 units.

Check (decimals): -2.75 and 1.5: 1.5 - (-2.75) = 4.25 = 4\tfrac14 ✓.

Tip: Opposite sides of zero → add the sizes. Then add fractions only after rewriting them with the same denominator; adding tops and bottoms (\tfrac34 + \tfrac12 \to \tfrac46) is never correct.

1. Set the ocean surface at 0; up is positive.
   • Diver: -45 ft
   • Pelican before the dive: +28 ft (not needed for the final answer)
   • Pelican after the dive: -3 ft
2. The question asks about the diver and the pelican after the dive: -45 and -3.
3. Distance = \lvert -3 - (-45) \rvert = \lvert -3 + 45 \rvert = \lvert 42 \rvert = 42.

Answer: 42 feet.

Check: both points are below the surface, so the distance is the difference of their depths: 45 - 3 = 42. ✓

Tip: When both points are on the same side of zero, subtract their sizes; on opposite sides, add them. Extra information (the 28 ft) is a common GED distractor.

1. Write the rule with absolute value: a length L is accepted when \lvert L - 2.5 \rvert \le 0.05.
2. Find the acceptable range: 2.5 - 0.05 = 2.45 and 2.5 + 0.05 = 2.55. So 2.45 \le L \le 2.55.
3. Test each length (distance from 2.5):
   • 2.45: \lvert 2.45 - 2.5 \rvert = 0.05 \le 0.05 → accepted (on the boundary)
   • 2.53: \lvert 0.03 \rvert = 0.03 → accepted
   • 2.549: \lvert 0.049 \rvert = 0.049 → accepted
   • 2.56: \lvert 0.06 \rvert = 0.06 > 0.05 → rejected

Answer: 2.56 inches.

Check: 2.56 > 2.55, the largest allowed length, so it is outside the range. ✓

Tip: A tolerance of \pm 0.05 allows 0.05 in each direction (a range 0.10 wide). "Within" includes the boundary values.

1. Read the scale. Labels are every 100 m; the unlabeled ticks are halfway, at multiples of 50 m.
   • Helicopter: halfway between 300 and 400, so +350 m.
   • Submarine: at the -200 label, so -200 m.
2. Distance = \lvert 350 - (-200) \rvert = \lvert 350 + 200 \rvert = 550 m.

Answer: 550 meters.

Check: from the submarine up to sea level is 200 m, and from sea level up to the helicopter is 350 m; 200 + 350 = 550. ✓

Tip: On a scale with unlabeled ticks, work out what one tick is worth before reading a point. Points on opposite sides of sea level are separated by the sum of their distances from 0.

1. First bars: multiply first, 2 \cdot 6 = 12; then 5 - 12 = -7; so \lvert 5 - 2 \cdot 6 \rvert = 7.
2. Apply the outside negative: -\lvert 5 - 2 \cdot 6 \rvert = -7.
3. Second bars: -4 + 1 = -3, so \lvert -4 + 1 \rvert = 3, and 3\lvert -4 + 1 \rvert = 9.
4. Combine: -7 + 9 = 2.

Answer: 2.

Check: the first term must be \le 0 (negative outside the bars) and the second must be \ge 0. -7 and 9 fit, and -7 + 9 = 2. ✓

Tip: Three separate rules work together here: order of operations inside the bars, absolute value of the result, then the sign or coefficient outside. Go one layer at a time.

1. Substitute: -\lvert -3 \rvert + \lvert (-3) - 4 \rvert.
2. First term: \lvert -3 \rvert = 3, and the outside negative gives -3.
3. Second term: (-3) - 4 = -7, and \lvert -7 \rvert = 7.
4. Combine: -3 + 7 = 4.

Answer: 4.

Check: \lvert x - 4 \rvert is the distance from x to 4; from -3 to 4 is 7 units ✓. And -\lvert x \rvert is never positive, so the first term must be -3 ✓.

Tip: Put the substituted value in parentheses. The negative sign outside \lvert x \rvert stays negative even when x is negative: -\lvert -3 \rvert = -3, not +3.

1. Positions (ground = 0, up positive): Tunnel A = -340 ft, Tunnel B = -1{,}215 ft. The tower height (+60 ft) is not needed.
2. Distance: \lvert -1{,}215 - (-340) \rvert = \lvert -875 \rvert = 875 ft.
3. Time = \dfrac{\text{distance}}{\text{speed}} = \dfrac{875}{125} = 7 minutes.

Answer: 7 minutes.

Check: 125 \times 7 = 875 ft, and 340 + 875 = 1{,}215 ft below ground ✓.

Tip: Both tunnels are on the same side of zero, so subtract their depths. Decide which numbers the question needs before you start — the tower height is a planted distractor.