ORIGINAL WORKED EXAMPLE
Distance across zero
Find the distance between −3.5 and 2.25.
- Distance is |2.25−(−3.5)|.
- This equals |5.75|=5.75.
- Subtracting magnitudes would miss the fact that the points are on opposite sides of zero.
Answer 5.75 units
GRE · Arithmetic · TOPIC LESSON
Read signed positions and distances, solve absolute-value conditions and use reliable bounds.
BY THE END, YOU CAN
BEFORE YOU START
Original educational lesson. Work through the examples before revealing the practice answers. This original lesson supports current GRE General Test study; it does not reproduce official questions, certify a writing score or predict test content.
LEARN THROUGH ORIGINAL PRACTICE
All teaching passages, writing prompts, examples and exercises in this lesson are original. No official ETS questions, prompts or explanations are reproduced. Current structure and observed public sample metadata are separate from this original teaching; no official score or question probability is implied.
01 / UNDERSTAND & APPLY
The real number line increases from left to right. Negative numbers farther from zero are smaller: −8<−3. Integers, fractions and irrational numbers all occupy positions on the same line. A number is not positive merely because its absolute value is large.
GRE number lines and coordinate/data graphs are drawn to scale; ordinary geometric figures need not be. Read the marked units and tick spacing rather than imposing a scale from a different figure. A point between two labeled ticks is not necessarily an integer.
02 / UNDERSTAND & APPLY
|x| is the distance from x to zero, so it is never negative. It equals x for x≥0 and −x for x<0. The expression |x−a| is the distance between x and a. Distance between a and b can be written |a−b| or |b−a|.
Absolute value does not distribute over addition: |−4+7|=3, while |−4|+|7|=11. The latter sum is always at least as large as the absolute value of the sum. For a product, |ab|=|a||b|.
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ORIGINAL WORKED EXAMPLE
Find the distance between −3.5 and 2.25.
Answer 5.75 units
03 / UNDERSTAND & APPLY
For c>0, |x−a|=c means x=a−c or x=a+c. For c=0, the only solution is x=a. For c<0, there are no real solutions because a distance cannot be negative. Isolate the absolute value before splitting into cases.
Check both solutions in the original equation, especially when other expressions or denominators are present. A condition such as x>0 may eliminate one location.
ORIGINAL WORKED EXAMPLE
Solve |2x−5|=7 for x>0.
Answer x=6
04 / UNDERSTAND & APPLY
For r>0, |x−a|<r means a−r<x<a+r: the point lies within r units of a. The condition |x−a|>r means x<a−r or x>a+r: the point is outside. Replace strict symbols with inclusive ones when boundary distance r is allowed.
An open circle excludes an endpoint; a closed circle includes it. In interval notation, parentheses exclude and brackets include endpoints. Infinity is never a finite included endpoint. Distinguish “and” for an interior interval from “or” for separated exterior rays.
ORIGINAL WORKED EXAMPLE
A measurement x satisfies |x−12|≤0.4. State the interval.
Answer [11.6, 12.4]
05 / UNDERSTAND & APPLY
To remove absolute values, split the line at points where their inside expressions change sign. Within each region the sign is known, so the formula becomes linear. A formula derived for a region is valid only there; reject algebraic candidates outside it.
For |x−2|+|x−8|, any x from 2 through 8 has total distance 6. Outside that interval the sum grows. The geometric explanation is that inside the segment the two distances partition its length; outside it, the extra distance is traversed twice.
ORIGINAL WORKED EXAMPLE
Solve |x−2|+|x−8|=10.
Answer x=0 or x=10
06 / UNDERSTAND & APPLY
An estimate is useful only at a precision suited to the question. To compare products of positive numbers, lower and upper bounds can establish a result without exact multiplication. If one quantity is certainly below 30 and another certainly above 30, the ordering is settled.
Rounding each factor independently does not automatically give a rigorous upper bound unless each replacement is upward and the quantities are nonnegative. When signs or reciprocals appear, revisit how inequalities change. Exact equality cannot be justified by equal rounded displays.
ORIGINAL WORKED EXAMPLE
QC: Quantity A: 4.97×6.02. Quantity B: 31.
Answer B
07 / UNDERSTAND & APPLY
Knowing |x| determines magnitude but often not sign. If |x|=4, x may be −4 or 4. A question involving x itself may therefore be indeterminate even when a question involving x² is fixed.
For QC, list boundary, zero, negative and fractional possibilities permitted by the prompt. A pair of different allowed relationships proves D. A single successful trial cannot prove a universal relationship.
08 / UNDERSTAND & APPLY
All examples and practice in this lesson are original. For quantitative comparison (QC), use A if Quantity A is greater, B if Quantity B is greater, C if they are equal, and D if the relationship cannot be determined from the information given. A, B or C must hold for every permitted case. To establish D, give two allowed cases producing different relationships; an inability to calculate is not itself evidence for D.
A single-answer question requires one choice. For multiple-answer questions, obey the stated selection count or select every correct option if instructed. Test each option separately; do not stop after finding one. Multiple-answer questions receive credit only when the complete correct set is selected, with no incorrect choices. Numeric entry requires the requested value, with units handled in the calculation; round only when the question directs you to. A single numeric-entry box accepts an integer or decimal; a pair of fraction boxes accepts an integer numerator and a nonzero integer denominator. Fractions need not be reduced, but reducing makes checks easier. Do not enter a percent sign when the requested number is a percentage. Preserve exact fractions until the last step.
A calculator can check arithmetic but cannot decide the domain, interpretation or denominator for you. Estimate sign and size first, enter parentheses carefully, and compare the result with the original conditions. Cancel factors or exploit structure when doing so is faster and more reliable than decimal entry.
09 / UNDERSTAND & APPLY
Use the QC choices defined earlier.
ORIGINAL PRACTICE · 1
2
The two absolute values are 7 and 5, so the result is 2. Absolute values do not make the entire final expression necessarily positive; evaluate each separately.
ORIGINAL PRACTICE · 2
A and C
x+3=5 gives 2; x+3=−5 gives −8. The center is −3, not +3, and each answer is 5 units from that center.
ORIGINAL PRACTICE · 3
C
Distance less than 3 from 4 gives the interior interval with excluded endpoints. E incorrectly includes distances exactly equal to 3.
ORIGINAL PRACTICE · 4
D
x=6 makes A greater, while x=−6 makes B greater. The magnitude condition leaves sign undetermined.
ORIGINAL PRACTICE · 5
−1
The midpoint is their average, (−7+5)/2=−1. Half the distance is 6, but that is not the coordinate of the midpoint.
ORIGINAL PRACTICE · 6
A, B and E
The solution is x≤−1 or x≥3. At −1 and 3 the distance is exactly 2 and is allowed. Values 0 and 2 are only 1 unit away.
ORIGINAL PRACTICE · 7
D
At x=−1/2, A=1/2>B=1/4; at x=−3/2, A=3/2<B=9/4. Both are permitted. Negativity alone does not determine the comparison.
ORIGINAL PRACTICE · 8
A
Reciprocation reverses order for positive numbers, giving 1/20.2<1/x<1/19.8. C writes the larger endpoint first and does not define the required increasing interval.
KEEP THE BIG IDEAS
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The teaching notes, passages, worked explanations and practice are original. Official references identify the assessment framework and skill scope; they do not endorse this lesson or license official question reproduction.
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