Equations, Inequalities, and Functions

What GRE Algebra Actually Tests

GRE Algebra rewards fluent manipulation, not exotic techniques. The whole domain fits into four families: linear equations and systems, quadratics, inequalities (including absolute value), and functions / coordinate geometry. Off-syllabus material — logarithms, matrices, trigonometry, and complex numbers — never appears, so do not waste effort there.

Linear Equations and Systems

Solve a single linear equation by isolating the variable. Solve a system by elimination or substitution:

• Elimination — when coefficients line up to cancel. Add or subtract the equations.
• Substitution — when one variable is already isolated.

Answer-the-question shortcut: the GRE often asks for an expression like $x + y$ or $xy$, not the individual values. You can frequently get it directly. If $x + y = 7$ and $2x - y = 5$, adding gives $3x = 12$, so $x = 4$, $y = 3$, and $xy = 12$ — but if it had asked only for $x$, you would have stopped at $x = 4$.

Quadratic Equations

You need three tools:

Tool	Use it when
Factoring	The trinomial splits cleanly, e.g. $x^2 - 5x + 6 = (x - 2)(x - 3)$
Quadratic formula	$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ when factoring is messy
Special forms	Difference of squares $a^2 - b^2 = (a+b)(a-b)$; perfect square $a^2 \pm 2ab + b^2 = (a \pm b)^2$

Discriminant $b^2 - 4ac$ reveals the number of real solutions without solving: positive ⇒ two, zero ⇒ one (a repeated root), negative ⇒ none.

Inequalities and Absolute Value

Inequalities behave like equations with one rule that ruins careless work: multiplying or dividing by a negative number flips the inequality sign.

An absolute-value inequality splits into a compound statement:

$|ax + b| < c \;\Longleftrightarrow\; -c < ax + b < c,$ $|ax + b| > c \;\Longleftrightarrow\; ax + b > c \text{ or } ax + b < -c.$

For example, $|2x - 3| < 7$ becomes $-7 < 2x - 3 < 7$, i.e. $-4 < 2x < 10$, i.e. $-2 < x < 5$.

Functions

Function notation is just a rule: $f(x) = 2x^2 - 3x + 1$ means "square the input, and so on." Evaluate by substituting carefully — watch signs on negative inputs:

$f(-2) = 2(-2)^2 - 3(-2) + 1 = 2(4) + 6 + 1 = 15.$

Composition works inside-out: $g(f(x))$ means apply $f$ first, then $g$.

Coordinate-Geometry Basics

Quantity	Formula
Slope through $(x_1, y_1)$, $(x_2, y_2)$	$m = \dfrac{y_2 - y_1}{x_2 - x_1}$
Slope-intercept form	$y = mx + b$
Distance	$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Midpoint	$\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$
Parallel / perpendicular	equal slopes / slopes are negative reciprocals

To find a line through $(2, 1)$ and $(6, 9)$: slope $= \frac{9 - 1}{6 - 2} = 2$; then $1 = 2(2) + b$ gives $b = -3$, so $y = 2x - 3$.