Simplify $\frac{12x^3y}{18xy^4}$ and state the restrictions.

$\frac{2x^2}{3y^3}$, with $x \ne 0$ and $y \ne 0$.

Simplify $\frac{5x + 15}{x^2 - 9}$ and state the restrictions.

$\frac{5}{x - 3}$, with $x \ne \pm 3$.

Simplify $\frac{x^2 - x - 12}{x^2 + 5x + 6}$ and state the restrictions.

$\frac{x - 4}{x + 2}$, with $x \ne -2, -3$.

Simplify $\frac{6 - 2x}{x^2 - 9}$ and state the restrictions.

$\frac{-2}{x + 3}$, with $x \ne \pm 3$.

Simplify $\frac{x^3 - 4x}{2x^2 + x - 6}$, state the restrictions, and check your answer at $x = 1$.

$\frac{x(x - 2)}{2x - 3}$, with $x \ne -2, \frac{3}{2}$.

Grade 11 · MCR3U · Functions

MCR3U: Simplifying Rational Expressions

A rational expression is a fraction with a polynomial on the top and the bottom, like $\frac{x^2 - 9}{x^2 + x - 6}$ . Simplifying one works exactly like reducing $\frac{12}{18}$ to $\frac{2}{3}$ : find the common factors and divide them out. The new part is saying which values of $x$ are not allowed.

How It Works

Factor everything first. Common factor, then difference of squares, trinomials, or grouping for four terms. You can only cancel a factor, something multiplied by the rest, never a term that is added on. In $\frac{x + 3}{x}$ nothing cancels.

State the restrictions before you cancel. Set every factor of the denominator not equal to $0$ , using the factored form before anything cancels. A cancelled factor still makes the original undefined. $\frac{x^2 - 9}{x^2 + x - 6} = \frac{(x - 3)(x + 3)}{(x + 3)(x - 2)}$ needs $x \ne -3$ and $x \ne 2$ , even though $(x + 3)$ is about to go.

Cancel common factors. That example simplifies to $\frac{x - 3}{x - 2}$ , with $x \ne -3, 2$ . The simplified form equals the original wherever the original is defined, which is why the restrictions travel with it.

Watch for opposites. $2 - x$ and $x - 2$ are not the same, but $2 - x = -(x - 2)$ , so together they cancel to $-1$ . Factor the $-1$ out first so the match is easy to see.

Leading coefficients. Factor a trinomial like $2x^2 + 5x - 3$ by decomposition. Find two numbers that multiply to $2(-3) = -6$ and add to $5$ , which are $6$ and $-1$ . Then $2x^2 + 6x - x - 3 = 2x(x + 3) - (x + 3) = (2x - 1)(x + 3)$ .

Check by substitution. Pick a value that is not restricted, such as $x = 1$ , and evaluate the original and your answer. If they differ, a factor went wrong.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Simplify $\frac{12x^3y}{18xy^4}$ and state the restrictions.
Show solution
1. Restrictions: $x \ne 0$ and $y \ne 0$ .
2. Divide the coefficients by $6$ and subtract exponents: $x^{3 - 1} = x^2$ and $y^{1 - 4} = y^{-3}$ , which belongs in the denominator.
Answer: $\frac{2x^2}{3y^3}$ , with $x \ne 0$ and $y \ne 0$ .
Easy
Simplify $\frac{5x + 15}{x^2 - 9}$ and state the restrictions.
Show solution
1. Factor: $\frac{5(x + 3)}{(x - 3)(x + 3)}$ .
2. Restrictions from the factored denominator: $x \ne 3$ and $x \ne -3$ .
3. Cancel $(x + 3)$ .
Answer: $\frac{5}{x - 3}$ , with $x \ne \pm 3$ .
Medium
Simplify $\frac{x^2 - x - 12}{x^2 + 5x + 6}$ and state the restrictions.
Show solution
1. Factor: $x^2 - x - 12 = (x - 4)(x + 3)$ and $x^2 + 5x + 6 = (x + 2)(x + 3)$ .
2. Restrictions: $x \ne -2$ and $x \ne -3$ .
3. Cancel $(x + 3)$ .
Answer: $\frac{x - 4}{x + 2}$ , with $x \ne -2, -3$ .
Medium
Simplify $\frac{6 - 2x}{x^2 - 9}$ and state the restrictions.
Show solution
1. Factor the top so it matches a factor below: $6 - 2x = -2(x - 3)$ . The bottom is $(x - 3)(x + 3)$ .
2. Restrictions: $x \ne 3$ and $x \ne -3$ .
3. Cancel $(x - 3)$ , keeping the $-2$ .
Answer: $\frac{-2}{x + 3}$ , with $x \ne \pm 3$ .
Test level
Simplify $\frac{x^3 - 4x}{2x^2 + x - 6}$ , state the restrictions, and check your answer at $x = 1$ .
Show solution
1. Top: common factor, then difference of squares: $x(x^2 - 4) = x(x - 2)(x + 2)$ .
2. Bottom by decomposition: two numbers that multiply to $2(-6) = -12$ and add to $1$ are $4$ and $-3$ . So $2x^2 + 4x - 3x - 6 = 2x(x + 2) - 3(x + 2) = (2x - 3)(x + 2)$ .
3. Restrictions: $x \ne -2$ and $x \ne \frac{3}{2}$ .
4. Cancel $(x + 2)$ to get $\frac{x(x - 2)}{2x - 3}$ .
5. Check at $x = 1$ : the original is $\frac{1 - 4}{2 + 1 - 6} = \frac{-3}{-3} = 1$ , and the answer is $\frac{1(-1)}{-1} = 1$ .
Answer: $\frac{x(x - 2)}{2x - 3}$ , with $x \ne -2, \frac{3}{2}$ .

Learn It on Martian Lab

Martian Lab teaches this in the Introduction to Rational Expressions lesson of its MCR3U Rational Expressions unit. Watch it, then take the Rational Expressions practice tests to see if it stuck.

Watch the Introduction to Rational Expressions lesson free

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Free Rational Expressions practice test, with solutions

More Rational Expressions topics: Operations with Rational Expressions

More MCR3U topics: Inverse Functions · Transformations of Functions · Sine Law and the Ambiguous Case · Domain, Range and Function Notation · Max/Min and Vertex Form · Linear-Quadratic Systems · Quadratic Word Problems · Simplifying Radicals · Exponent Laws and Rational Exponents · Graphing Exponential Functions · Exponential Growth and Decay · Special Angles and the CAST Rule · Cosine Law and 3D Problems · Transformations of Sine and Cosine · Sinusoidal Word Problems · Arithmetic Sequences and Series · Geometric Sequences and Series · Pascal's Triangle and the Binomial Theorem · Simple and Compound Interest · Annuities: Future and Present Value

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