Simplify (x^2 - x)/(x), stating the restriction. [2 credits]

NYSED Regents (style)-style practice: Part III (constructed response). Solve for x: (2)/(x) + (1)/(3) = (5)/(x). State any restrictions and check for extraneous solutions.

A 4-credit question: credit for clearing denominators, solving, and the restriction check. The restriction is x ≠ 0. Multiply every term by the common denominator 3x: 3x · (2)/(x) + 3x · (1)/(3) = 3x · (5)/(x), giving 6 + x = 15, so x = 9. Check: x = 9 does not violate x ≠ 0, and substituting confirms (2)/(9) + (1)/(3) = (2)/(9) + (3)/(9) = (5)/(9), which matches (5)/(9). So x = 9 is valid. Omitting the restriction or skipping the check costs credit.

New YorkMathsSyllabus dot point

How do you simplify and combine rational expressions, and how do you solve a rational equation while avoiding extraneous solutions?

Simplify rational expressions by factoring and cancelling (noting domain restrictions); add, subtract, multiply, and divide them; and solve rational equations, checking for extraneous solutions introduced by the denominators.

A NY Regents Algebra II answer on rational expressions: simplifying by factoring with domain restrictions, the four operations on rational expressions, solving rational equations, and rejecting extraneous solutions.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
Simplifying rational expressions
The four operations
Solving rational equations
Extraneous solutions
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What this topic is asking

The Regents Algebra II exam (the Arithmetic with Polynomials and Rational Expressions, A-APR, and Reasoning with Equations, A-REI, clusters) wants you to simplify rational expressions by factoring and cancelling (tracking the domain restrictions), perform the four operations on them, and solve rational equations while rejecting extraneous solutions that the denominators forbid. Domain awareness is the theme throughout.

Simplifying rational expressions

A rational expression is a fraction of polynomials. To simplify, factor the top and bottom, then cancel any factor common to both. The crucial rule: you may cancel factors (things multiplied) but never terms (things added).

\frac{x^2 - 4}{x^2 + 5x + 6} = \frac{(x - 2)(x + 2)}{(x + 2)(x + 3)} = \frac{x - 2}{x + 3}, \quad x \neq -2, -3.

The domain restrictions come from every value that made an original denominator zero, here $x \neq -2$ and $x \neq -3$ , and they remain in force even after a factor cancels.

The four operations

Multiplication and division are easiest: factor, then multiply across (flipping the second fraction for division) and cancel. Addition and subtraction require a common denominator first, just like numerical fractions.

Solving rational equations

To solve an equation with rational terms, clear the denominators by multiplying every term by the least common denominator, then solve the resulting polynomial equation.

Extraneous solutions

The reason rational equations demand a check is that multiplying by a variable expression can introduce a candidate that makes an original denominator zero, an extraneous solution that is not actually valid. If solving had produced $x = 2$ above, it would be rejected, because $x = 2$ makes the denominators zero. A clarifying point worth stressing is that you must state the restrictions before solving and test every candidate after: the Regents awards a specific credit for identifying restrictions and rejecting an extraneous root, and a solution that ignores this can be marked down even when the algebra is otherwise correct. Treat the restriction step as part of the solution, not an afterthought.

Try this

Q1. Simplify $\frac{x^2 - x}{x}$ , stating the restriction. [2 credits]

Cue. $\frac{x(x - 1)}{x} = x - 1$ , with $x \neq 0$ .

Q2. Solve $\frac{6}{x} = 2$ . [1 credit]

Cue. Multiply by $x$ : $6 = 2x$ , so $x = 3$ (and $x \neq 0$ is satisfied).

Exam-style practice questions

Practice questions written in the style of NYSED exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Regents (style)2 marksPart I (multiple choice). Which expression is equivalent to

\frac{x^2 - 9}{x^2 - x - 6}

for

x \neq 3, -2

? (1)

\frac{x - 3}{x - 2}

(2)

\frac{x + 3}{x + 2}

(3)

\frac{x - 3}{x + 2}

(4)

\frac{x + 3}{x - 2}

Show worked answer →

The correct answer is (2).

Factor both: $\frac{x^2 - 9}{x^2 - x - 6} = \frac{(x - 3)(x + 3)}{(x - 3)(x + 2)}$ . Cancel the common $(x - 3)$ factor (valid for $x \neq 3$ ): $\frac{x + 3}{x + 2}$ . The remaining restriction $x \neq -2$ keeps the denominator nonzero. Cancelling terms instead of factors (for example crossing out the $9$ and $6$ ) is the classic error.

Regents (style)4 marksPart III (constructed response). Solve for

x

\frac{2}{x} + \frac{1}{3} = \frac{5}{x}

. State any restrictions and check for extraneous solutions.

Show worked answer →

A 4-credit question: credit for clearing denominators, solving, and the restriction check.

The restriction is $x \neq 0$ . Multiply every term by the common denominator $3x$ : $3x \cdot \frac{2}{x} + 3x \cdot \frac{1}{3} = 3x \cdot \frac{5}{x}$ , giving $6 + x = 15$ , so $x = 9$ . Check: $x = 9$ does not violate $x \neq 0$ , and substituting confirms $\frac{2}{9} + \frac{1}{3} = \frac{2}{9} + \frac{3}{9} = \frac{5}{9}$ , which matches $\frac{5}{9}$ . So $x = 9$ is valid. Omitting the restriction or skipping the check costs credit.

Related dot points

Sources & how we know this

Educator Guide to the Regents Examination in Algebra II — NYSED (2025)
New York State Next Generation Mathematics Learning Standards — NYSED (2017)