Now put your home at 5 instead of zero. To find how far you are from home, compare your position x with 5. The subtraction x − 5 says which side of home you are on and how far. Taking its absolute value keeps the distance. This makes |x − 5| the distance from the center 5. The same idea describes a manufactured part allowed to be a little too small or a little too large. That allowed difference is its tolerance. You name the center, measure the distance, and compare it with the permitted amount. This lets one statement cover departures in both directions.
The allowable resistances form [646, 714], with both endpoint values included.
Reminder
Subtracting a negative. x − (−6) = x + 6. Keep parentheses around a negative center before simplifying.
Endpoint notation. A filled dot and bracket include an end. A hollow dot and parenthesis leave it out: [7, 11] includes 7 and 11; (7, 11) leaves both out.
Set-builder notation. {x | x ≥ 5} and {x : x ≥ 5} both mean all x such that x ≥ 5. A colon helps distinguish set notation from absolute value bars.
Why it works. Subtracting the center makes its own position count as zero difference: at x = 5, x − 5 = 0. At x = 8 the difference is 3; at x = 2 it is −3. Both positions are three units away. Reversing the subtraction reverses the direction, while the bars keep the distance. Thus |x − 5| = |5 − x|. To stay within an allowed distance r of a center c, you cannot go below c − r or above c + r. This follows by walking r units each way from c.
RuleDistance between two real numbers A and B is |A − B| = |B − A|. Here A and B name the two positions. For an allowed distance r ≥ 0 from center c, at most r means |x − c| ≤ r, with ends c − r and c + r included. Strictly less than r means |x − c| < r, with the ends excluded. Exactly r means |x − c| = r. At least r means |x − c| ≥ r. More than r means |x − c| > r. The sign ± is read plus or minus: c ± r names the lower limit c − r and upper limit c + r. If r = 0, the two end positions coincide at c; an exact zero distance has only that one position.
The same idea, five ways
Say it
Say the distance between x and c. Read |x − c| as the absolute value of x minus c.
Write it
Subtract the center, then measure the size of the difference.
In math
|x − c| = |c − x|
|x − c| ≤ r
c − r ≤ x ≤ c + r
[c − r, c + r]
{x : c − r ≤ x ≤ c + r}
Like
Move the home address before measuring how far away a house is.
See itThe center 5 and actual position 8 are three units apart.
The same idea, other ways
As a walk from the center
Start at 5 and walk 3 right to 8. The difference 8 − 5 is 3, so |8 − 5| = 3. A walk 3 left reaches 2 and gives |2 − 5| = 3.
A three-unit walk from 5 reaches 8.
With opposite differences
The output is the size of the difference. In the table, the columns under 2 and 8 both show 3 because these positions are equally far from 5.
Subtracting the center before taking absolute value measures distance from that center.
As permitted error
A part can be slightly smaller or larger than its intended size. A tolerance puts a limit on the distance between the measured size and the intended center.
The permitted bolt lengths run from 3.48 to 3.52 cm.
Words
Absolute value statement, r > 0
Street picture
Exactly r; at a distance of r
|x − c| = r
Only c − r and c + r
At most r; no more than r; within r, ends included
|x − c| ≤ r
Between the ends, both included
Less than r; strictly within r
|x − c| < r
Between the ends, both left out
At least r; no closer than r
|x − c| ≥ r
Outside either end, ends included
More than r; farther than r
|x − c| > r
Outside either end, ends left out
c ± r; margin of error r
|x − c| ≤ r
Between c − r and c + r, ends included
.1Distance from a center
Subtract the address of home before measuring how far away you are. Either order of subtraction gives the same distance.
Distance from c is |x − c|.
|x − c| = |c − x|.
|x| ≤ 4 means within or including four units of zero: −4 ≤ x ≤ 4, the interval [−4, 4].
From 5 to 1 is a four-unit trip.
Worked exampleCompare two subtraction orders
Find the distance between 1 and 5.
The four-unit trip from 5 ends at 1.
What it asks. Find the length of the trip between positions 1 and 5.
Plan. Subtract the positions and measure the size of that difference, then reverse the subtraction to check.
Keep the subtraction inside intact.
Distance has no negative sign.
|1 − 5| = |−4| = 4.Moving from 5 to 1 changes position by −4 but travels four units.
|5 − 1| = |4| = 4.Reversing the trip changes its direction but keeps its length.
Answer
The distance is 4.
Check The four unit gaps are 1 to 2, 2 to 3, 3 to 4, and 4 to 5.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: The distance between 1 and 5 is −4.
−4 is the difference in position for one direction of travel, not the length.
✓ Instead: |1 − 5| = |−4| = 4.
Tips and tricks
Either subtraction order gives the same distance once the bars are applied.
.2Absolute tolerance
A tolerance stated in centimeters tells you directly how far a measured length may differ from the intended center. Think of a marked ruler with a short acceptable stretch on each side of 3.5 cm. Two hundredths can be added or subtracted because lengths have decimal parts too.
At most 0.02 from 3.5: |x − 3.5| ≤ 0.02.
The boundary values are included when the maximum error is allowed.
The filled ends include the acceptable limits 3.48 and 3.52 cm.
Worked exampleA bolt within two hundredths
A bolt must be within 0.02 cm of 3.5 cm, ends included. Write the condition and limits.
All lengths on the shaded stretch satisfy the tolerance.
What it asks. Find the acceptable bolt lengths, including their two limits.
Plan. Write distance from 3.5, then walk 0.02 cm below and above the center.
Within, ends included means at most.
Write 3.5 as 3.50 when subtracting hundredths.
|x − 3.5| ≤ 0.02.The bars measure the distance of the actual length from the intended 3.5 cm center; at most permits the maximum difference.
3.50 − 0.02 = 3.48 and 3.50 + 0.02 = 3.52.The two positions exactly two hundredths from 3.5 mark the acceptable ends.
Write 3.48 ≤ x ≤ 3.52, or [3.48, 3.52].Every length between the ends is close enough, and both ends are allowed.
Answer
|x − 3.5| ≤ 0.02
3.48 ≤ x ≤ 3.52 cm
Interval: [3.48, 3.52] cm
Check |3.48 − 3.5| = 0.02 and |3.52 − 3.5| = 0.02, so both ends sit exactly at the limit and are allowed.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Only x ≤ 3.52 is required.
That would allow lengths such as 3 cm, whose distance from 3.5 is 0.5 cm.
✓ Instead: Both limits are needed: 3.48 ≤ x ≤ 3.52.
Tips and tricks
Keep the same measurement unit in the center, allowed distance, and answer.
.3Percent tolerance
A percent tolerance makes the permitted difference depend on the stated rating. Five percent of a larger rating is a larger amount. A resistor is a part that limits how much electricity flows, and ohms measure its resistance. A capacitor stores electric charge, an electrical amount. Its capacitance is stored charge per unit of voltage, the electrical push. Only the measurements and units matter for this calculation.
Nominal value means the intended rating; actual value means the measured value.
Tolerance distance = size of rating × percent ÷ 100.
Resistance and capacitance are different measurements; use the units given.
A tolerance ±p% permits the same percentage below and above the rating. Common stated tolerances include ±1%, ±5%, and ±10%.
Center: 680 ohms
5% = 5100
Allowed distance: 34 ohms
646 ≤ R ≤ 714
A percentage tolerance must first become a distance in the stated units.
Reminder
Percent. 2% = 2100 = 0.02. Multiply the center's size by this fraction.
Worked exampleA two percent tolerance
A 250-unit rating allows 2% variation. Find the limits.
The rating's 2% tolerance permits [245, 255].
What it asks. Find the smallest and largest allowed values for a 250-unit rating.
Plan. Convert 2% of 250 to units, then subtract and add that amount.
Percent is a fraction out of 100.
Both ends are permitted.
r = 250 × 2100 = 500100 = 5 units.Multiplying the rating by two hundredths gives its permitted distance.
Write |x − 250| ≤ 5.The measured value may differ from the rating by no more than 5 units.
250 − 5 = 245 and 250 + 5 = 255, so 245 ≤ x ≤ 255.Walking the allowed distance each way locates the included ends.
Answer
Allowed difference: 5 units.
Allowed values: [245, 255].
Check 5 ÷ 250 = 0.02 = 2%, so each endpoint uses exactly the permitted variation.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: A 2% tolerance always means two units.
Two percent is a fraction of the rating, so its unit amount depends on that rating.
✓ Instead: For a 250-unit rating, 250 × 2100 = 5 units.
Tips and tricks
Convert the percent to a measured distance before writing the bars.
.4Negative centers and plus or minus
The home address can be negative. Subtract that entire address before measuring. Plus or minus, written ±, gives two directions from a center. For a tolerance it names a lower and an upper allowed limit. For a short list such as ±2, it names the numbers 2 and −2.
Distance from −2 is |x − (−2)| = |x + 2|.
680 ± 34 gives limits 646 and 714.
±2 means 2 and −2.
A margin of error is the largest stated distance allowed from a reported value.
A percentage point is a change of 1 in a percentage number: 52% to 53% changes by one percentage point. A margin stated in percentage points uses subtraction of these numbers, rather than a percent of the center.
Three units right from the negative center −2 reaches 1.
Reminder
Subtracting a negative. 1 − (−2) = 1 + 2 = 3.
Worked exampleWithin three of negative two
Describe all numbers within 3 of −2, including the ends.
The acceptable numbers lie between −5 and 1, including both ends.
What it asks. Find the permitted positions near a center left of zero.
Plan. Subtract the negative center correctly, then locate the two end positions by walking 3 each way.
x minus negative 2 becomes x + 2.
The allowed ends are included.
Write |x − (−2)| ≤ 3, which is |x + 2| ≤ 3.The difference must be measured from center −2, and the distance can be at most 3.
The left end is −2 − 3 = −5. The right end is −2 + 3 = 1.These positions are exactly three units each way from the center.
All positions between the ends work: −5 ≤ x ≤ 1, or [−5, 1].They are no farther from −2 than the included two ends.
Answer
|x + 2| ≤ 3
−5 ≤ x ≤ 1
Interval: [−5, 1]
Check |−5 + 2| = |−3| = 3 and |1 + 2| = 3. At the center −2 the distance is zero, which is also allowed.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Within 3 of −2 is written |x − 2| ≤ 3.
That expression measures from positive 2.
✓ Instead: Write |x + 2| ≤ 3.
Tips and tricks
Put a negative center in parentheses until the subtraction is simplified.
Strategy: step by step
Choose a letter for the actual value, such as x, a resistance R, a score S, or a temperature T. Identify the center c.
Find the permitted distance r. For a percent tolerance, multiply the size of the stated rating by the percent divided by 100.
Read the words carefully. At most, no more than, and ends included use an equality bar under the comparison. Strictly less than and strictly more than leave the end out.
Write |actual value − center|, then the comparison and distance.
Locate the two positions c − r and c + r. Nearby means the stretch between them. Farther away means the left side or the right side beyond them. Exactly that distance means the two end positions.
Keep the units. Use a closed, filled dot or a bracket for an included end; use an open, hollow dot or a parenthesis for an excluded end.
Strategy
Translate a distance statement
1
Is the distance given as a percent?
YesMultiply the rating's size by percent ÷ 100 first.
NoUse the stated distance and units.
↓
2
Does the wording say exactly or at a distance of?
YesUse = and the two positions at that distance.
NoCompare whether the value is nearer or farther.
↓
3
Does the wording say at most or strictly less?
YesUse the stretch between the two end positions.
NoAt least or more than permits positions beyond either end.
↓
4
Are the end positions allowed?
YesUse an equals bar in the comparison and filled dots or brackets.
NoUse a strict sign and hollow dots or parentheses.
Identify the actual value, center, and permitted distance.
Write the absolute value of actual value minus center.
Choose the sign using the exact words.
Use the street endpoints center minus distance and center plus distance to describe allowed positions.
Worked exampleA resistor's permitted resistance
A resistor has nominal resistance 680 ohms, ±5%. Express its allowable actual resistance R using absolute value and an interval.
Filled ends include the permitted resistance limits 646 and 714 ohms.
What it asks. Find how far the measured resistance may depart from 680, and write all the allowed values.
Plan. Convert five percent to an allowed distance. Walk that distance below and above 680, including both ends.
Nominal value is the stated rating. Actual value is the measured number.
±5% permits the same percentage below and above the rating.
Let R be the actual resistance in ohms. The center is 680 ohms.R names the measurement that varies, while 680 is the fixed rating.
5% = 5100 = 0.05.Percent means out of one hundred.
680 × 5 = 3400; then 3400 ÷ 100 = 34 ohms.Five hundredths of 680 is the permitted distance from the rating.
Write |R − 680| ≤ 34.The distance from the actual measurement to its rating can be at most 34 ohms.
The lower end is 680 − 34 = 646. The upper end is 680 + 34 = 714.Walking 34 units either way from the center finds the two furthest allowed values.
Write 646 ≤ R ≤ 714, or the interval [646, 714].Every value between these limits is close enough, and the equals bar includes both limits.
Answer
|R − 680| ≤ 34
646 ≤ R ≤ 714 ohms
Interval: [646, 714] ohms
Check 646 is 34 below 680, and 714 is 34 above it. Both distances are 34. Also 34680 = 120 = 0.05 = 5%, so each endpoint uses the permitted percentage.
Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.
Rung 1Rung 1: exactly six from a negative center
Describe all numbers exactly 6 from −1.
Only the two marked positions are exactly six units from −1.
What it asks. Find the two positions whose distance from −1 is exactly 6.
Plan. Write the distance statement, then walk six units each way from the center.
Subtracting −1 gives x + 1.
Exactly uses an equals sign.
Write |x − (−1)| = 6, or |x + 1| = 6.The center is −1 and the required distance is 6.
Walk left to −1 − 6 = −7 and right to −1 + 6 = 5.On a straight line these are the two positions at that exact positive distance.
Answer
|x + 1| = 6
x = −7 or x = 5
Set: {−7, 5}
Check |−7 + 1| = |−6| = 6 and |5 + 1| = |6| = 6.
Rung 2Rung 2: at most two from nine
Describe all numbers at most 2 from 9.
Filled ends show that distance exactly two is allowed.
What it asks. Find the stretch of nearby positions, including the allowed limit.
Plan. Walk two units each way from 9 and keep the stretch between the ends.
At most permits equality.
Nearby positions lie between the two ends.
Write |x − 9| ≤ 2.At most 2 means distance 2 or less from center 9.
The two end positions are 9 − 2 = 7 and 9 + 2 = 11.They are the furthest permitted positions on the two sides.
Keep 7 ≤ x ≤ 11, or [7, 11].Every position between them is no farther than 2, including the two ends.
Answer
|x − 9| ≤ 2
7 ≤ x ≤ 11
Interval: [7, 11]
Check The center 9 has distance 0 and both ends have distance 2, so all meet the condition.
Rung 3Rung 3: strictly less than half a unit
Describe all numbers less than 0.5 from 12.
Hollow ends exclude distances exactly half a unit.
What it asks. Find nearby positions with the two maximum-distance ends excluded.
Plan. Locate 12 minus and plus 0.5, then keep positions strictly between them.
Less than leaves out equality.
Half a unit is 0.5.
Write |x − 12| < 0.5.The distance must be smaller than the stated half unit.
The end positions are 12 − 0.5 = 11.5 and 12 + 0.5 = 12.5.These are exactly half a unit from the center.
Keep 11.5 < x < 12.5, or (11.5, 12.5).The two ends have distance exactly 0.5 and must be left out, while every position between them is nearer.
Answer
|x − 12| < 0.5
11.5 < x < 12.5
Interval: (11.5, 12.5)
Check 12 has distance zero and works; the ends each have distance 0.5 and fail a strict less-than comparison.
Rung 4Rung 4: at least five from twelve
Describe all numbers at least 5 from 12.
The two outward pieces include their boundary values.
What it asks. Find faraway positions in either direction, including distance exactly 5.
Plan. Locate the positions five each way from 12, then keep the outward sides.
At least allows equality.
Nearer positions in the middle fail.
Write |x − 12| ≥ 5.The phrase at least requires distance 5 or greater.
The two boundary positions are 12 − 5 = 7 and 12 + 5 = 17.These are exactly five units from the center.
Keep x ≤ 7 or x ≥ 17.Positions to the left of 7 or to the right of 17 are far enough. Positions between them are too near.
Answer
|x − 12| ≥ 5
x ≤ 7 or x ≥ 17
Interval: (−∞, 7] ∪ [17, ∞)
Check 7 and 17 have distance 5 and work; the center 12 has distance 0 and fails.
Rung 5Rung 5: a thermostat's strict departure
A thermostat switches on when temperature T is more than 3 degrees from 70. Describe the switching temperatures.
The thermostat activates in either strict outside region.
What it asks. Find when the temperature is too far below or above the center.
Plan. Write the strict distance comparison, find the boundary temperatures, and keep the outward regions.
T is the actual temperature; 70 is the center.
More than excludes a difference of exactly 3.
Write |T − 70| > 3.The distance from temperature T to 70 must exceed 3.
The two boundary temperatures are 70 − 3 = 67 and 70 + 3 = 73.They are exactly three degrees from the center.
Keep T < 67 or T > 73.Those temperatures are farther away, while the two boundary temperatures are excluded by more than.
Answer
|T − 70| > 3
T < 67 or T > 73
Interval: (−∞, 67) ∪ (73, ∞)
Check 66 and 74 each differ by 4 and switch it on. 67 and 73 differ by exactly 3 and do not.
Rung 6Rung 6: a poll's margin of error
A poll reports 52% with a margin of error of 3 percentage points. Let p be the percentage number, such as 52. Describe the permitted values.
The percentage-number interval includes its margin-of-error limits.
What it asks. Find percentages within the stated departure from the reported center.
Plan. Treat p as a number on a percent scale and walk three percentage points each way from 52.
Here p = 52 means 52%, rather than the decimal 0.52.
A margin of error includes its stated limits.
Write |p − 52| ≤ 3.The distance from the reported percentage number is at most three percentage points.
52 − 3 = 49 and 52 + 3 = 55.These give the lower and upper percentage-number limits.
Keep 49 ≤ p ≤ 55.Every percentage number between the ends is within the margin, including both ends.
Answer
|p − 52| ≤ 3
49 ≤ p ≤ 55
Interval for p: [49, 55]
Percentages: 49% through 55%
Check 49 and 55 each differ from 52 by three percentage points. The center 52 differs by zero, so it also falls within the permitted margin.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: The distance from −6 is |x − 6|.
That expression measures from positive 6. At x = −2 it gives 8, while −2 and −6 are only four units apart.
✓ Instead: Distance from −6 is |x − (−6)| = |x + 6|. At x = −2 this is |4| = 4.
✗ Not this: A 5% tolerance at 680 permits a difference of 680.
680 is the center; the difference is five hundredths of it.
✓ Instead: 680 × 5100 = 34, so |R − 680| ≤ 34.
Tips and tricks
Read |x − c| as the distance between x and c.
The textbook uses within with both interpretations. Its Try It answers include the ends, while its $200 discussion means strictly less. If wording is only within, state whether you include the ends; use a strict sign when it says less than or strictly.
A margin of error permits the same maximum departure on either side of the reported value.
Trap. The center and permitted distance have different jobs. In |R − 680| ≤ 34, 680 names the rating, while 34 says how far the measurement may depart from it.
