# GRE Inequalities and Absolute Values: Examples | topin

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## Record the domain and the meaning of each sign

[ETS](https://www.ets.org/gre/test-takers/general-test/prepare/content/quantitative-reasoning.html) includes linear and quadratic inequalities, absolute value and number-line reasoning in GRE Quant’s scope. Its conventions treat numbers as real unless the problem supplies a narrower restriction. If the question says x is an integer, your answer can be a finite list or count. If it says only x is a number, decimals and fractions inside the solution interval also remain possible.

Strict inequalities, < and >, exclude the boundary. Non-strict inequalities, ≤ and ≥, include it when the expression is defined there. “Between” needs the exact accompanying wording; do not assume both endpoints are allowed. Our first line of scratch work is the domain and any forbidden values, such as x ≠ 2 for a denominator x − 2.

A solution is the complete set of values that satisfy every condition, not one number that works. That is exactly what a [select-one-or-more question](/gre/quant/multiple-choice) tests when it asks which values “could be” x: solve the interval once, then check every choice against it. In Quantitative Comparison the same idea runs the other way. One counterexample disproves a claimed relationship, but one value that works proves nothing about the whole interval. The [GRE Quant section guide](/gre/quant) sets out all five question formats.

## Keep the direction when the operation permits it

Adding or subtracting the same quantity preserves an inequality. Multiplying or dividing both sides by a positive number preserves its direction; a negative number reverses it. If the multiplier is a variable with unknown sign, first split the sign cases or use a method that avoids that operation. Assuming the sign is a common way to lose half the valid possibilities.

Squaring also needs care. For nonnegative quantities, squaring preserves order. With arbitrary real quantities, it can change the relationship because a large negative value has a large square. From −3 < 2, squaring gives 9 > 4\. This simple counterexample explains why blindly squaring an inequality is not the same as performing an always-valid reversible algebra step.

__Operations and their conditions__
| Operation                       | Effect                               | Required check                                |
| ------------------------------- | ------------------------------------ | --------------------------------------------- |
| Add the same expression         | Direction unchanged                  | Expressions are defined                       |
| Multiply by a positive constant | Direction unchanged                  | Constant is greater than zero                 |
| Divide by a negative constant   | Reverse the sign                     | Constant is nonzero                           |
| Multiply by unknown x           | Need sign cases                      | x may be positive, negative or zero           |
| Square both sides               | Safe comparison under suitable signs | Do not assume arbitrary reals are nonnegative |

## Original problem: solve an interval and count integers

Original question: find all integers x satisfying −2 < 3 − 2x ≤ 9\. Split the chain. From −2 < 3 − 2x, subtract three to get −5 < −2x. Dividing by negative two reverses the sign and gives x < 2.5\. From 3 − 2x ≤ 9, subtract three and divide by negative two to obtain x ≥ −3.

The real solution is −3 ≤ x < 2.5\. The integer solutions are −3, −2, −1, 0, 1 and 2, a total of six. The upper endpoint 2.5 is excluded even before applying the integer restriction. The lower endpoint −3 is included: substitution makes the middle expression nine, which satisfies the non-strict right-hand inequality.

Test a boundary, an inside value and an outside value. At x = 2, the middle expression is −1, valid. At x = 3, it is −3, invalid. These checks target sign reversal and endpoint mistakes. If a question asks for the sum rather than the count of integer solutions, continue to that requested output instead of stopping at six.

A compact original integer check is 1/2 < x ≤ 7/2\. The integers are one, two and three. Rounding the lower boundary to one and then treating the new inequality as strict would incorrectly discard one. Keep the original exact interval until the integer selection is complete; an approximate boundary can change the permitted set.

## Original problem: read absolute value as distance

Original question: solve |2x − 5| < 7 for real x. The expression inside the absolute value must lie between −7 and 7: −7 < 2x − 5 < 7\. Add five, giving −2 < 2x < 12, then divide by two to get −1 < x < 6\. Both endpoints are excluded because the original inequality was strict.

A distance view gives the same result: |2x − 5| = 2|x − 2.5|, so x must lie less than 3.5 units from 2.5\. That creates one central interval. By contrast, |2x − 5| > 7 asks for distances greater than 3.5, giving x < −1 or x > 6\. The change from “less than” to “greater than” changes one interval into two outside regions.

If the threshold is negative, pause before splitting. Since an absolute value is nonnegative, |x − 4| < −2 has no solutions, while |x − 4| > −2 is true for every real x. At threshold zero, equality or strictness matters: |x − 4| ≤ 0 gives only x = 4, whereas |x − 4| < 0 is impossible.

## Original problem: solve absolute value on both sides

Original question: solve |x − 2| = x + 4\. The right side must be nonnegative, so x ≥ −4\. For x ≥ 2, the equation becomes x − 2 = x + 4, which is impossible. For x < 2, it becomes 2 − x = x + 4, giving x = −1\. This value meets both the branch condition and the right-side condition.

Substitution confirms |−1 − 2| = 3 and −1 + 4 = 3\. Checking the original equation is essential because a candidate derived in one branch must belong to that branch. Solving both algebraic lines and retaining every numerical result without checking domains can produce an extra answer. The branch inequalities belong to the calculation, not to optional commentary afterwards.

For |x − 2| = 6, the right side is already a nonnegative constant and the two branches simply give x − 2 = 6 or x − 2 = −6, hence x = 8 or −4\. Do not assume two solutions in every absolute-value equation: an equation can have two, one, no solutions or, under different structures, an entire set.

## Original problem: quadratic and rational inequalities

Original question: solve (x − 1)(x + 3) ≤ 0\. The critical points are −3 and 1\. For x below −3, both factors are negative and their product is positive. Between −3 and 1, one factor is negative and the other positive, giving a negative product. Above 1, both are positive. Include the zeros, so the solution is −3 ≤ x ≤ 1.

Now solve (x − 1)/(x + 3) ≤ 0\. The same sign regions occur, but x = −3 is undefined and must be excluded. The solution becomes −3 < x ≤ 1\. This difference is why multiplying through by x + 3 without its sign and zero conditions is unsafe. Mark denominator zeros as forbidden before using any sign chart.

This interval reasoning connects with [divisibility and remainders](/articles/gre-divisibility-remainders), where a range must be intersected with a pattern of integers. Once the algebra gives an interval, apply any separate integer, positive or divisibility condition. Do not build that restriction into the algebra by assuming every possible x is a convenient positive whole number.

## Use counterexamples and endpoints as targeted checks

Our practice routine is to solve a chain inequality, an inside-distance condition, an outside-distance condition, a case equation and a rational sign chart. For each, identify the smallest set of checks likely to catch your mistake. A denominator zero checks domain handling. A negative substitution checks an assumed sign. An endpoint checks strictness. This is more focused than redoing every arithmetic step.

Use the [Quantitative Comparison guide](/gre/quant/quantitative-comparison) when the output is a relationship rather than an interval. Try permitted negative numbers, zero and fractions where relevant, but do not claim a universal result from a handful of examples. After concept practice, use [topin’s free GRE mock](/gre/practice-test), marked on the official scale, to check transfer under mixed-topic conditions.

If the model is settled, most examples here need little calculator work. ETS advises using the calculator for tedious computations rather than automatically. Keep exact boundary fractions until the question gives a rounding instruction. In review, record “sign reversal”, “case condition”, “undefined endpoint” or “wrong domain”. These labels identify the repair more precisely than calling every missed inequality a careless error.

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## FAQs

When do I reverse an inequality sign?

Reverse it when multiplying or dividing both sides by a negative quantity. If the quantity’s sign is unknown, split into sign cases rather than choosing a direction without justification.

How do I solve an absolute value less than a positive number?

For |A| < k with k > 0, solve −k < A < k. This represents an inside interval. For |A| > k, use A < −k or A > k, giving outside regions.

Can I square both sides of any inequality?

No. Squaring preserves comparisons for nonnegative quantities but can change the relationship for arbitrary real values. Establish the relevant signs or use cases before taking a step that changes the solution set.

Are denominator zeros included with a less-than-or-equal sign?

No. An expression is undefined at a denominator zero, regardless of a non-strict inequality. Mark those values as excluded before completing the sign chart.

How do I count integer solutions from an interval?

First solve the real interval with the correct endpoints, then list or count the integers it contains. Keep strictness and any additional restrictions. The answer might request their count, sum or a particular extreme value.

## Sources (checked 5 October 2026)

* [ETS: Quantitative Reasoning overview and calculator guidance (checked 5 October 2026)](https://www.ets.org/gre/test-takers/general-test/prepare/content/quantitative-reasoning.html)
* [ETS: GRE Math Review (checked 5 October 2026)](https://www.ets.org/content/dam/ets-org/pdfs/gre/gre-math-review.pdf)
* [ETS: GRE Mathematical Conventions (checked 5 October 2026)](https://www.ets.org/content/dam/ets-org/pdfs/gre/gre-math-conventions.pdf)
* [ETS: Guidelines specific to the on-screen calculator (checked 5 October 2026)](https://www.ets.org/content/dam/ets-org/pdfs/gre/on-screen-calculator-guidelines.pdf)

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