ORIGINAL WORKED EXAMPLE
Apply the idea
Solve −3x+5≥17.
- Subtract 5: −3x≥12.
- Divide by −3 and reverse the inequality.
- Check the boundary: x=−4 gives 17; x=−5 gives 20.
Answer x≤−4.
GMAT · Algebra and constraints · TOPIC LESSON
Inequalities, absolute value and solution restrictions
Original GMAT Problem Solving instruction: reason from constraints, calculate efficiently without a calculator, and check the requested quantity.
BY THE END, YOU CAN
- Explain and apply inequality reversal on negative multiplication.
- Explain and apply interval intersections.
- Explain and apply compound inequalities.
- Explain and apply sign cases.
- Explain and apply absolute value as distance.
- Explain and apply absolute-value equations and inequalities.
- Explain and apply integer restrictions.
- Explain and apply denominator restrictions.
BEFORE YOU START
- Linear equations, signed arithmetic and integer notation.
Original educational lesson. Work through the examples before revealing the practice answers. This original lesson supports current GMAT study through an editorial skill map. It does not reproduce official questions, predict exam content or guarantee a score.
LEARN THROUGH ORIGINAL PRACTICE
Original GMAT-style worked examples and practice
All passages, examples and exercises here are original. No official GMAC questions, choices or explanations are reproduced. This independent preparation resource is not affiliated with or endorsed by GMAC. GMAT is a trademark of the Graduate Management Admission Council.
01 / UNDERSTAND & APPLY
Why a negative multiplier reverses order
An inequality compares positions on a number line. Adding the same real quantity to both sides preserves order. Multiplication by a positive number preserves order; multiplication by a negative number reverses it because the number line is reflected. Division follows the same sign rule.
If the sign of a variable multiplier is unknown, do not multiply through without cases. For example, x>2 does not permit multiplying by y and keeping the same direction unless y>0 is known. When y=0, strict inequality becomes equality after multiplication.
ORIGINAL PRACTICE · 1
Solve 7−2x<1.
Reveal answer & reasoning
x>3.
Subtract 7 to get −2x<−6, then divide by −2 and reverse the sign. Testing x=4 gives −1<1; the reversed alternative x<3 fails for x=0.
02 / UNDERSTAND & APPLY
All simultaneous restrictions must overlap
A value satisfying two simultaneous inequalities lies in their intersection. For x>−2 and x≤5, the allowed interval is −2<x≤5. Open endpoints exclude equality; closed endpoints include it. In an “or” statement, take the union instead, allowing values satisfying either condition.
An intersection can be empty, a single value or an interval. Do not select the wider interval simply because it contains more values. Every retained value must satisfy every required restriction.
ORIGINAL WORKED EXAMPLE
Apply the idea
Solve x≥−1 and x<4, then impose x>2.
- The first two restrictions give −1≤x<4.
- Intersect this with x>2.
- Only values strictly above 2 and below 4 survive.
Answer 2<x<4.
ORIGINAL PRACTICE · 1
What values satisfy x≤3 and x≥3?
Reveal answer & reasoning
x=3 only.
Both inclusive bounds meet at 3. An empty interval would result if either condition were strict at the same boundary; inclusive endpoints matter.
03 / UNDERSTAND & APPLY
Operate on all parts of a chained inequality
A chain such as −5<2x+1≤9 states two conditions joined by “and.” Apply the same operation to all three parts. If dividing by a negative number, reverse both comparisons and then rewrite the chain in increasing order.
A chain with incompatible endpoints has no solution. Keep strictness attached to the correct boundary. Separate the two inequalities if a single chained transformation becomes difficult to track.
ORIGINAL WORKED EXAMPLE
Apply the idea
Solve −5<2x+1≤9.
- Subtract 1 throughout: −6<2x≤8.
- Divide by positive 2.
- The left endpoint remains excluded and the right included.
Answer −3<x≤4.
ORIGINAL PRACTICE · 1
Solve 2≤−x<7.
Reveal answer & reasoning
−7<x≤−2.
Multiply all parts by −1, reversing both signs: −2≥x>−7. Rewrite in increasing order. The included endpoint is −2, not −7.
04 / UNDERSTAND & APPLY
Factor and test signs between critical values
A product is positive when its factors have the same sign and negative when they have opposite signs. Zeros of factors divide the number line into intervals where signs are constant. Test one point in each interval or reason from each factor’s sign.
For rational inequalities, denominator zeros also divide the line, but they are always excluded. A repeated even power does not change sign across its zero. A sign chart must reflect multiplicity, not assume every critical point flips the sign.
ORIGINAL WORKED EXAMPLE
Apply the idea
Solve (x−2)(x+3)<0.
- The zeros are −3 and 2.
- For x<−3 both factors are negative, so the product is positive.
- For −3<x<2 the signs differ; for x>2 both are positive. Strict inequality excludes both zeros.
Answer −3<x<2.
ORIGINAL PRACTICE · 1
Solve (x−1)²>0.
Reveal answer & reasoning
All real x except 1.
A square is positive unless its base is zero. It is positive on both sides of 1, so a rule that automatically alternates signs would incorrectly exclude half the line.
05 / UNDERSTAND & APPLY
Absolute value measures distance
The absolute value |x−a| is the distance from x to a on the number line. Distances are nonnegative. The equation |x−a|=r has two solutions a−r and a+r when r>0, one when r=0, and none when r<0.
This interpretation prevents sign mistakes. |x−a| is not always x−a: it equals x−a for x≥a and a−x for x<a. A negative input to absolute value does not produce a negative distance.
ORIGINAL WORKED EXAMPLE
Apply the idea
Solve |x−4|=7.
- x is 7 units from 4.
- The two positions are 4−7 and 4+7.
- Substitution gives distance 7 in both cases.
Answer x=−3 or x=11.
ORIGINAL PRACTICE · 1
How many real solutions satisfy |2x+1|=−3?
Reveal answer & reasoning
None.
The left side is always nonnegative, so it cannot equal −3. Splitting into 2x+1=−3 and 2x+1=3 would incorrectly ignore the impossibility of a negative absolute-value output.
06 / UNDERSTAND & APPLY
Inside means and; outside means or
For r>0, |x−a|<r means a−r<x<a+r: the points are inside a distance band. In contrast, |x−a|>r means x<a−r or x>a+r: the points are outside. Non-strict versions include the boundaries.
Check the right side first. An absolute value is always at least zero, so |u|≥−2 holds for every allowed u, while |u|<−2 has no solution. Blindly applying a two-branch formula can produce false intervals.
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ORIGINAL WORKED EXAMPLE
Apply the idea
Solve |2x−3|≤5.
- Write −5≤2x−3≤5.
- Add 3 throughout: −2≤2x≤8.
- Divide by 2.
Answer −1≤x≤4.
ORIGINAL PRACTICE · 1
Solve |x+2|>3.
Reveal answer & reasoning
x<−5 or x>1.
The centre is −2 and the distance must exceed 3, so values lie beyond −5 or 1. Joining these branches with “and” would require x to be simultaneously below −5 and above 1.
07 / UNDERSTAND & APPLY
Continuous solutions and allowed counts are different
First solve the inequality over the reals, then retain only permitted integers. An interval can contain many real values but no integers. When counting integer endpoints, list a short interval or use greatest and least allowed integers carefully.
For consecutive integers from L to U inclusive, the count is U−L+1. Strict inequalities may require moving an endpoint inward by 1 when that endpoint is an integer. A count of objects must also be nonnegative unless the problem explicitly defines signed counts.
ORIGINAL WORKED EXAMPLE
Apply the idea
How many integers satisfy −2<x≤5?
- The allowed integers are −1,0,1,2,3,4,5.
- The least is −1 and greatest is 5.
- Count 5−(−1)+1.
Answer 7.
ORIGINAL PRACTICE · 1
If n is an integer and 3<2n<9, list all possible n.
Reveal answer & reasoning
2, 3, 4.
Divide by 2 to get 1.5<n<4.5, then keep integers. Rounding both endpoints outward would incorrectly include 1 or 5.
08 / UNDERSTAND & APPLY
A denominator’s sign controls inequality operations
For a rational inequality, identify all excluded denominator zeros first. To avoid multiplying by an unknown-sign denominator, put all terms over one denominator and analyze signs in intervals. Alternatively, split into denominator-positive and denominator-negative cases explicitly.
Cancellation may simplify an expression but does not remove the original excluded points. A positive squared denominator permits multiplication without reversing order, provided it is nonzero; this can be a useful alternative when the algebra stays simple.
ORIGINAL WORKED EXAMPLE
Apply the idea
Solve (x−1)/(x+2)≥0.
- Exclude x=−2. Critical values are −2 and 1.
- For x<−2 both numerator and denominator are negative, giving a positive quotient. Between −2 and 1 the quotient is negative.
- For x≥1 the quotient is nonnegative; x=1 is allowed because the denominator is 3.
Answer x<−2 or x≥1.
ORIGINAL PRACTICE · 1
Solve 1/x>0 for real x.
Reveal answer & reasoning
x>0.
The denominator excludes zero. A positive numerator divided by x is positive exactly when x is positive. Multiplying by x without a sign case might erase the essential distinction between positive and negative x.
KEEP THE BIG IDEAS
Quick revision
- Reverse inequalities when multiplying or dividing by a negative value.
- Use intersection for and, union for or.
- Interpret absolute value as distance and check whether its target is possible.
- Exclude denominator zeros and impose integer restrictions at the end.
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