3.1 Algebraic Expressions, Equations & Inequalities

3.1 Algebraic Expressions, Equations & Inequalities

Algebra on the GRE Quantitative Reasoning section tests your ability to manipulate expressions, solve single-variable and multi-variable equations, work with inequalities, and evaluate absolute value statements. Unlike advanced calculus or abstract algebra, GRE algebra emphasizes structural recognition, speed, and precision. Master algebraic operations to avoid long calculations and solve problems efficiently.

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1. Simplifying & Factoring Algebraic Expressions

An algebraic expression is a mathematical phrase combining variables, constants, and operators without an equals sign (e.g., $3x^2 - 5x + 7$). Simplifying expressions requires combining like terms, expanding products using the distributive property, and factoring out common terms.

The Distributive Property and Greatest Common Factor (GCF)

The distributive property states that $a(b + c) = ab + ac$. Reversing this process is factoring out the Greatest Common Factor (GCF): $ab + ac = a(b + c)$

Core Algebraic Identities

Three special polynomial products appear frequently on the GRE. Recognizing these identities instantly allows you to factor or expand expressions in seconds:

1. Difference of Squares: $a^2 - b^2 = (a + b)(a - b)$
2. Square of a Sum (Perfect Square Trinomial): $(a + b)^2 = a^2 + 2ab + b^2$
3. Square of a Difference (Perfect Square Trinomial): $(a - b)^2 = a^2 - 2ab + b^2$

GRE Strategy Tip: Never expand quadratic expressions unless necessary. When given a complex expression like $\frac{x^2 - 16}{x - 4}$, immediately factor the numerator using the difference of squares: $\frac{(x+4)(x-4)}{x-4} = x + 4$ (for $x eq 4$).

Worked Example 3.1.1: Rational Expression Simplification

Simplify the expression $\frac{4x^2 - 36}{2x + 6}$ for $x eq -3$.

Solution:

1. Factor the numerator: First factor out the GCF of 4: $4x^2 - 36 = 4(x^2 - 9)$.
2. Apply the difference of squares identity to $(x^2 - 9)$: $x^2 - 9 = (x + 3)(x - 3)$. So the numerator is $4(x + 3)(x - 3)$.
3. Factor the denominator GCF: $2x + 6 = 2(x + 3)$.
4. Cancel the common factor $(x + 3)$: $\frac{4(x + 3)(x - 3)}{2(x + 3)} = \frac{4(x - 3)}{2} = 2(x - 3) = 2x - 6$

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2. Solving Linear Equations in One Variable

A linear equation in one variable can be written in the form $ax + b = 0$ (where $a eq 0$). Solving a linear equation means isolating the variable $x$ on one side of the equals sign.

Systematic Solving Steps

1. Clear Denominators: Multiply every term on both sides by the Least Common Denominator (LCD).
2. Distribute & Combine Like Terms: Expand parentheses and combine like terms on each side independently.
3. Isolate Variable Terms: Add or subtract terms to move all variable terms to one side.
4. Isolate Constant Terms: Add or subtract constants to move all numerical terms to the opposite side.
5. Divide by the Coefficient: Divide both sides by the coefficient of the variable.

Special Cases: No Solution vs. Infinitely Many Solutions

• No Solution (Contradiction): If simplifying leads to a false statement such as $0 = 7$, no value of $x$ satisfies the equation.
• Infinitely Many Solutions (Identity): If simplifying leads to a true statement such as $4x + 3 = 4x + 3$ or $0 = 0$, any real number $x$ satisfies the equation.

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3. Linear Inequalities & The Sign-Flipping Rule

Linear inequalities use comparison symbols ($<, \le, >, \ge$). Solving inequalities follows the exact same steps as linear equations, with one vital exception.

CRITICAL RULE (Sign Flipping): Whenever you multiply or divide both sides of an inequality by a negative number, you MUST reverse (flip) the direction of the inequality sign.

Why Does the Inequality Sign Reverse?

Consider the true statement $3 < 5$.

• Multiply both sides by $+2$: $6 < 10$ (True).
• Multiply both sides by $-1$: $-3$ and $-5$. On a number line, $-3$ lies to the right of $-5$, so $-3 > -5$. Keeping the original sign would yield $-3 < -5$, which is false!

Worked Example 3.1.2: Solving a Linear Inequality

Solve for $x$: $-3(2x - 5) + 4 \ge 31$.

Solution:

1. Distribute $-3$: $-6x + 15 + 4 \ge 31 \implies -6x + 19 \ge 31$.
2. Subtract 19 from both sides: $-6x \ge 31 - 19 \implies -6x \ge 12$.
3. Divide both sides by $-6$ (and flip the $\ge$ sign to $\le$): $x \le \frac{12}{-6} \implies x \le -2$

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4. Absolute Value Equations and Inequalities

The absolute value of a real number $x$, denoted $|x|$, represents the distance between $x$ and $0$ on the number line. Because distance is non-negative, $|x| \ge 0$ for all real numbers $x$.

Absolute Value Equations: $|ax + b| = c$

If $c > 0$, the equation $|ax + b| = c$ splits into two distinct linear cases: $ax + b = c \quad \text{OR} \quad ax + b = -c$ (If $c < 0$, the equation has NO real solution, because absolute value can never be negative.)

Absolute Value Inequalities

Absolute value inequalities represent distance intervals around a center point:

Worked Example 3.1.3: Absolute Value Inequality

Find all real values of $x$ satisfying $|4x - 3| \le 9$.

Solution:

1. Set up the three-part compound inequality: $-9 \le 4x - 3 \le 9$.
2. Add 3 to all three parts: $-9 + 3 \le 4x \le 9 + 3 \implies -6 \le 4x \le 12$.
3. Divide all three parts by $+4$: $\frac{-6}{4} \le x \le \frac{12}{4} \implies -1.5 \le x \le 3$. In interval notation, $x \in [-1.5, 3]$.

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5. GRE Trap Alerts & Quantitative Comparison Tactics

• The Variable Division Trap: Never divide an equation or inequality by a variable unless you are certain that the variable cannot equal zero. For example, dividing $x^2 = 5x$ by $x$ yields $x = 5$, but loses the valid root $x = 0$! Always factor instead: $x^2 - 5x = 0 \implies x(x - 5) = 0 \implies x = 0 \text{ or } 5$.
• Quantitative Comparison Zero/Negative Check: When comparing expressions involving variables, test zero, negative numbers, and fractions. For instance, if $x^2 > x$, $x$ could be $2$ ($4 > 2$), but $x$ could also be $-2$ ($4 > -2$).