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CLEP Precalculus study materials
CLEP Precalculus study resources

CLEP Precalculus: Algebraic Expressions, Equations, and Inequalities

CLEP Precalculus · Work through your exam, one topic at a time.

All guide sections

Algebraic Expressions, Equations, and Inequalities

The Algebraic Expressions, Equations, and Inequalities domain checks whether you can rewrite expressions fluently and solve equations and inequalities exactly - polynomial, rational, radical, absolute-value, exponential, and logarithmic. CLEP Precalculus items here are usually short but reward precision: the exam often hinges on whether you kept every restriction on the variable.

Core rules

  • Exponent and radical rules: a^m · a^n = a^(m+n), (a^m)^n = a^(mn), a^(-n) = 1/a^n, and x^(1/n) is the nth root. Rationalize denominators by multiplying by the conjugate.
  • Factoring toolkit: common factor first, then difference of squares a^2 - b^2, sum or difference of cubes, trinomial factoring, and grouping.
  • Quadratic equations: factor, complete the square, or use the quadratic formula. The discriminant b^2 - 4ac decides the count of real roots - positive gives two, zero gives one, negative gives none.
  • Polynomial division: the Remainder Theorem says dividing by (x - k) leaves remainder p(k), so p(k) = 0 means (x - k) is a factor.
  • Absolute value: |A| = b needs b >= 0 and gives A = ±b; |A| < b is the conjunction -b < A < b; |A| > b is the disjunction A < -b or A > b.
  • Rational inequalities: never cross-multiply by an expression whose sign is unknown. Move everything to one side, combine into a single fraction, then build a sign chart from the zeros of numerator and denominator.

Worked example

Solve (x - 1)/(x + 3) >= 2. Bring everything left: (x - 1)/(x + 3) - 2 >= 0. Combine over the common denominator: (x - 1 - 2x - 6)/(x + 3) = (-x - 7)/(x + 3) >= 0. Multiply numerator and denominator by -1, which flips the inequality: (x + 7)/(x + 3) <= 0. Critical values are x = -7 (numerator zero, included) and x = -3 (denominator zero, excluded). Testing the three intervals shows the quotient is non-positive on [-7, -3). The solution is [-7, -3).

Traps that cost marks

  • Multiplying both sides by (x + 3) and reversing the inequality incorrectly, or silently losing the exclusion x = -3.
  • Choosing a closed endpoint where the denominator vanishes, or an open one where only the numerator vanishes.
  • Squaring both sides of a radical equation and forgetting to test for extraneous roots.
  • Reading |A| > b as a conjunction instead of a disjunction.
  • Discarding the negative root when a quadratic genuinely has two solutions.