4.3 Linear Inequalities, Compound Inequalities, and Absolute Value Inequalities

4.3 Linear Inequalities, Compound Inequalities, and Absolute Value Inequalities

Quick Summary: Inequalities on the GMAT Focus Edition demand relentless vigilance regarding positive and negative sign behavior. Multiplying or dividing by a negative number flips the inequality symbol, while multiplying by an unknown variable without case analysis is strictly forbidden. By combining sign analysis charts with geometric distance interpretations of absolute values, you can navigate even the trickiest inequality problems with speed and certainty.

Inequalities represent ranges of possible values rather than discrete points. On the GMAT, inequalities frequently serve as a testing ground for number properties, boundary conditions, and sign analysis. Because multiple values satisfy an inequality, questions often ask for minimum values, maximum values, or the total count of satisfying integers.

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Fundamental Rules of Inequality Manipulation

Manipulating inequalities mirrors the algebra of linear equations, with one crucial exception: operations involving negative numbers and variable signs.

1. Operations that Preserve Inequality Direction

• Adding or Subtracting any Real Number: For any real numbers $a, b,$ and $c$: $a < b \implies a + c < b + c \quad \text{and} \quad a - c < b - c$
• Multiplying or Dividing by a Strictly Positive Number ($c > 0$): $a < b \implies ac < bc \quad \text{and} \quad \frac{a}{c} < \frac{b}{c}$

2. Operations that Reverse (Flip) Inequality Direction

• Multiplying or Dividing by a Strictly Negative Number ($c < 0$): $a < b \implies ac > bc \quad \text{and} \quad \frac{a}{c} > \frac{b}{c}$ Example: $-3x < 15$. Dividing by $-3$ reverses the sign: $x > -5$.
• Taking Reciprocals of Quantities with the Same Sign: $\text{If } 0 < a < b, \quad \text{then } \frac{1}{a} > \frac{1}{b}. \qquad \text{If } a < b < 0, \quad \text{then } \frac{1}{a} > \frac{1}{b}.$ Caution: If $a$ and $b$ have opposite signs ($a < 0 < b$), taking reciprocals preserves the sign direction because $\frac{1}{a}$ remains negative and $\frac{1}{b}$ remains positive (e.g., $-2 < 3 \implies -\frac{1}{2} < \frac{1}{3}$).

3. The Cardinal Prohibition: Multiplying by an Unknown Variable

Never multiply or divide both sides of an inequality by an algebraic variable unless you know its sign with 100% certainty.

Consider the inequality $\frac{6}{x} < 2$:

• Fatal Error: Multiplying by $x$ to conclude $6 < 2x \implies x > 3$. This completely ignores all negative values of $x$! If $x = -2$, $\frac{6}{-2} = -3 < 2$, which is completely valid, yet excluded by $x > 3$.
• Correct Method: Subtract 2 to set the right-hand side to zero and use sign analysis: $\frac{6}{x} - 2 < 0 \implies \frac{6 - 2x}{x} < 0 \implies \frac{2(3 - x)}{x} < 0$ The quotient is negative when numerator and denominator have opposite signs: $x < 0$ or $x > 3$.

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Combining Multiple Inequalities: Addition vs. Subtraction

Can You Add Inequalities?

Yes, provided they point in the same direction. $\text{If } a < b \quad \text{and} \quad c < d, \quad \text{then } a + c < b + d.$ Proof: Since $a < b$, adding $c$ gives $a + c < b + c$. Since $c < d$, adding $b$ gives $b + c < b + d$. By transitivity, $a + c < b + d$.

Can You Subtract Inequalities?

NO. Never subtract inequalities directly.

• Counterexample: $10 < 20$ and $1 < 19$. Subtracting term-by-term yields $10 - 1 < 20 - 19 \implies 9 < 1$, which is completely false!
• How to Subtract Correctly: To find the range of $x - y$ given $a < x < b$ and $c < y < d$:
  1. Multiply the $y$-inequality by $-1$ and flip the signs: $-d < -y < -c$.
  2. Add the transformed inequality to the $x$-inequality: $(a < x < b) + (-d < -y < -c) \implies a - d < x - y < b - c$

Can You Multiply Inequalities?

Only if all four terms are strictly positive: $\text{If } 0 < a < b \quad \text{and} \quad 0 < c < d, \quad \text{then } ac < bd.$ If negative numbers are involved, multiplication is invalid (e.g., $-5 < -2$ and $-4 < -1$, but $(-5)(-4) = 20 > 2 = (-2)(-1)$).

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Compound Inequalities: Conjunctions and Disjunctions

A compound inequality combines two individual inequality statements:

1. Conjunction ("AND"): Expressed in compact chain notation as $a \le x \le b$. The variable $x$ must satisfy both $x \ge a$ and $x \le b$ simultaneously. The solution set is the intersection (overlap) of the two individual sets.
2. Disjunction ("OR"): Written as $x \le a \text{ or } x \ge b$. The variable $x$ satisfies the statement if it belongs to either condition. The solution set is the union of the two disjoint intervals.

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Non-Linear and Rational Inequalities: The Sign Chart Method

When dealing with quadratic inequalities (e.g., $x^2 - x - 12 > 0$) or rational inequalities (e.g., $\frac{x - 3}{x + 5} \le 0$), algebraic isolation fails because terms change sign across the number line. You must use the Sign Chart (Wavy Curve) Method.

The 4-Step Sign Chart Protocol

1. Set Right-Hand Side to Zero: Move all terms to the left-hand side so the expression is compared strictly to 0 ($f(x) > 0, < 0, \ge 0, \le 0$).
2. Factor Completely: Factor numerator and denominator into linear factors.
3. Identify Critical Transition Points: Find values where each factor equals zero. These points divide the real number line into test intervals.
4. Determine Signs across Intervals: Test a test value in each interval or use factor multiplicity rules (signs alternate across simple roots).

Worked Walkthrough: Rational Inequality

Solve $\frac{x - 2}{x + 4} \ge 1$.

• Step 1: Subtract 1 from both sides: $\frac{x - 2}{x + 4} - 1 \ge 0 \implies \frac{(x - 2) - (x + 4)}{x + 4} \ge 0 \implies \frac{-6}{x + 4} \ge 0$
• Step 2: Inspect the numerator and denominator. The numerator is $-6$ (always negative).
• Step 3: For the entire quotient to be non-negative ($\ge 0$), the denominator must also be negative (negative divided by negative equals positive): $x + 4 < 0 \implies x < -4$ (Note: $x + 4$ cannot equal 0 because division by zero is undefined). The solution set is $x < -4$.

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Absolute Value: Number Line Distance and Transformations

By formal definition, the absolute value of a real number $x$, denoted $|x|$, represents its magnitude without regard to sign:

$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$

The Geometric Interpretation: Distance on the Number Line

Geometrically, $|x - c|$ represents the distance between $x$ and $c$ on the real number line. Because distance is inherently non-negative, $|x - c| \ge 0$ for all real numbers.

Absolute Value Equations with Variables on Both Sides

When solving $|ax + b| = cx + d$, remember that the left-hand side is an absolute value and must be non-negative. This establishes a strict constraint:

$cx + d \ge 0$

To solve, split into two cases and check both answers against the constraint:

• Case 1: $ax + b = cx + d$
• Case 2: $ax + b = -(cx + d)$

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Step-by-Step Worked Problem Solving Examples

Example 1: Rational Inequality with Interval Sign Analysis

Problem: What is the complete solution set for the inequality $\frac{x + 4}{x - 2} \le 0$?

Step-by-Step Solution:

1. Identify Critical Points:
   • Numerator equals zero: $x + 4 = 0 \implies x = -4$.
   • Denominator equals zero: $x - 2 = 0 \implies x = 2$.
2. State Domain Restrictions: The denominator cannot equal zero, so $x \neq 2$ strictly.
3. Partition the Number Line into Three Intervals:
   • Interval 1: $(-\infty, -4)$
   • Interval 2: $[-4, 2)$
   • Interval 3: $(2, \infty)$
4. Evaluate Signs in Each Interval:
   • For $x = -5$ (Interval 1): $\frac{-5 + 4}{-5 - 2} = \frac{-1}{-7} = +\frac{1}{7} > 0$ (Fails).
   • For $x = 0$ (Interval 2): $\frac{0 + 4}{0 - 2} = \frac{4}{-2} = -2 \le 0$ (Satisfies).
   • At boundary $x = -4$: $\frac{0}{-6} = 0 \le 0$ (Satisfies; endpoint included).
   • For $x = 3$ (Interval 3): $\frac{3 + 4}{3 - 2} = \frac{7}{1} = 7 > 0$ (Fails).
5. Conclusion: The solution set is $-4 \le x < 2$.

Example 2: Counting Integer Solutions to an Absolute Value Inequality

Problem: How many integer values of $x$ satisfy $|2x - 5| \le 13$?

Step-by-Step Solution:

1. Apply the Bounded Interval Transformation: $-13 \le 2x - 5 \le 13$
2. Isolate $2x$ by Adding 5 to All Three Parts: $-13 + 5 \le 2x \le 13 + 5 \implies -8 \le 2x \le 18$
3. Divide All Parts by 2 (Preserving Sign Direction since $2 > 0$): $-4 \le x \le 9$
4. Count Inclusive Integers: $\text{Count} = \text{Last} - \text{First} + 1 = 9 - (-4) + 1 = 9 + 4 + 1 = 14$
5. There are 14 satisfying integers: ${-4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}$.

Example 3: Minimizing an Algebraic Difference

Problem: If $-3 \le a \le 5$ and $-7 \le b \le 2$, what is the minimum possible value of the expression $a - b$?

Step-by-Step Solution:

1. Deconstruct the Objective: To make the difference $a - b$ as small (as negative) as possible, you must make $a$ as small as possible and subtract the largest possible value of $b$: $\min(a - b) = \min(a) - \max(b)$
2. Identify Extrema from Given Intervals:
   • The minimum value of $a$ is $-3$.
   • The maximum value of $b$ is $2$.
3. Calculate the Minimum: $\min(a - b) = -3 - (2) = -5$
4. Verify Boundary Combinations:
   • If $a = -3, b = -7 \implies -3 - (-7) = 4$
   • If $a = 5, b = -7 \implies 5 - (-7) = 12$ (maximum possible value)
   • If $a = 5, b = 2 \implies 5 - 2 = 3$
   • If $a = -3, b = 2 \implies -3 - 2 = -5$ (confirmed minimum).

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High-Frequency GMAT Traps & Pacing Strategies

• Trap 1: Cross-Multiplying Fractions with Variables. In an inequality like $\frac{x}{x-2} > 3$, multiplying both sides by $(x - 2)$ without knowing if $x - 2$ is positive or negative is fatal. Always subtract 3 and use a sign chart.
• Trap 2: Directly Subtracting Aligned Inequalities. If $x > 5$ and $y > 2$, claiming $x - y > 3$ is false. (Let $x = 6, y = 10$; then $6 > 5$ and $10 > 2$, but $6 - 10 = -4$, which is not greater than 3!).
• Trap 3: Squaring Inequalities with Negative Terms. If $x < 3$, squaring both sides does NOT mean $x^2 < 9$. If $x = -5$, $-5 < 3$ is true, but $(-5)^2 = 25$, which is not less than 9.
• Trap 4: Including Undefined Denominators in Solution Sets. In $\frac{x-1}{x-4} \le 0$, the root $x = 1$ is included ($x \ge 1$), but $x = 4$ must be strictly excluded ($x < 4$) because division by zero is undefined.