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Tutorial: Rational expressions

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(This topic is also in Section 0.4 in Finite Mathematics and Applied Calculus)

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Note For this tutorial, it is assumed that you know how to multiply algebraic expressions using the distributive law. If you feel you need to review this, go back to %%multFacttut.

What is a rational expression?

Rational expression

A rational expression is an algebraic expression of the form $\dfrac{P}{Q},$ where $P$ and $Q$ are simpler expressions (usually polynomials), and the denominator $Q$ is not zero.
Examples
$\dfrac{1}{x-1}$ \t is rational with $\color{slateblue}{P = 1,\ \ Q = x-1}$ \\ \t   \\ $\dfrac{x^2+3x+1}{x^2+3}\ \ $ \t is rational with $\color{slateblue}{P = x^2+3x+1,\ \ Q = x^2+3}$ \\ \t  
Note As with numbers, we can think of expressions with no denominator (like whole numbers or polynomials) as rational expressions by dividing them by 1:
$x$ can be thought of as $\dfrac{x}{1},$ which is rational with $\color{slateblue}{P = x},$ and $\color{slateblue}{Q = 1}.$
$1$ can be thought of as $\dfrac{1}{1},$ which is rational with $\color{slateblue}{P = 1},$ and $\color{slateblue}{Q = 1}.$
$x^2y-2xy^2+1$ can be thought of as $\dfrac{x^2y-2xy^2+1}{1},$ which is rational with $\color{slateblue}{P = x^2y-2xy^2+1},$ and $\color{slateblue}{Q = 1}.$
Some for you to do

Algebra of rational expressions

The rules for manipulating rational expressions are the same as the rules for manipulating fractions. Let's look at these rules one-by-one:

The cancellation rule: Simplifying a rational expression

Cancellation rule

If $R$ is any nonzero expression that is a factor of both the numerator and denominator, then you can cancel it to simplify the rational expression:
$\dfrac{P\color{indianred}{R}}{Q\color{indianred}{R}} = \dfrac{P}{Q} \qquad \quad \ \ $ Cancel the R.
Caution $R$ must be a factor and not a summand; For instance,
$\dfrac{P+\color{indianred}{R}}{Q+\color{indianred}{R}}$ ≠ $\dfrac{P}{R} \qquad$ You cannot cancel a summand.

Suggested video for this topic: Video by Pat Kopf
Examples
$\dfrac{(x^2+3x+1)\color{indianred}{(x-1)}}{(x^2+3)\color{indianred}{(x-1)}}$ \t ${}= \dfrac{x^2+3x+1}{x^2+3}$ \t Cancel the factor $\color{#6968d0}{x-1}$. \\ $\dfrac{\color{indianred}{x^2}}{\color{indianred}{x^2}(x-1)}$ \t ${}= \dfrac{1}{x-1}$ \t Cancel the factor $\color{#6968d0}{x^2}$. \\ \t   \\ $\dfrac{\color{indianred}{(x^2-1)}(x^2y-2xy^2+1)}{\color{indianred}{(x^2-1)}}$ \t ${}= \dfrac{x^2y-2xy^2+1}{1}$ \t Cancel the factor $\color{#6968d0}{x^2-1}$. \\ \t ${}= x^2y-2xy^2+1$ \\ \t   \\ $\dfrac{6x^2}{10x^4}$ \t ${}= \dfrac{3 \cdot \color{indianred}{2 \cdot x \cdot x}}{5 \cdot \color{indianred}{2 \cdot x \cdot x} \cdot x \cdot x}$ \t Factor. \\ \t ${}= \dfrac{3}{5x^2}$ \t Cancel the factor $\color{#6968d0}{2x^2}$. \\ \t   \\ $\dfrac{x^3+x}{x^2+x}$ \t ${}= \dfrac{\color{indianred}{x}(x^2+1)}{\color{indianred}{x}(x+1)}$ \t Factor. \\ \t ${}= \dfrac{x^2+1}{x+1}$ \t Cancel the factor $\color{#6968d0}{x}$. \\ \t   \\ $\dfrac{x^2+3x+2}{x+1}$ \t ${}= \dfrac{\color{indianred}{(x+1)}(x+2)}{\color{indianred}{x+1}}$ \t Factor. \\ \t ${}= x+2$ \t Cancel the factor $\color{#6968d0}{x+1}$.

Multiplying and dividing rational expressions

Multiplying rational expressions

As is the case with ordinary fractions, we multiply two rational expressions by simply multiplying their numerators and denominators:
$\dfrac{\color{#026fc1}{P}}{\color{#026fc1}{Q}} \times \dfrac{\color{#c1026f}{R}}{\color{#c1026f}{S}} = \dfrac{\color{#026fc1}{P}\color{#c1026f}{R}}{\color{#026fc1}{Q}\color{#c1026f}{S}} \qquad $ \t Product of numerators over product of denominators
Note Before actually calculating the products on the top and bottom, you should first simplify by factoring and cancelling as above if possible (see the examples that follow).

Suggested video for this topic: Video by Professor Leonard
Examples
\\ $\dfrac{\color{#026fc1}{x+1}}{\color{#026fc1}{x}} \times \dfrac{\color{#c1026f}{x-1}}{\color{#c1026f}{2x+1}} $ \t ${}= \dfrac{\color{#026fc1}{(x+1)}\color{#c1026f}{(x-1)}}{\color{#026fc1}{x}\color{#c1026f}{(2x+1)}}$ \t Multiply top and bottom. \\ \t ${}= \dfrac{x^2-1}{2x^2+1}$ \t Calculate the products. \\ \t   \\ $\color{#026fc1}{2} \times \dfrac{\color{#c1026f}{4x}}{\color{#c1026f}{x-1}} $ \t ${}= \dfrac{\color{#026fc1}{2}}{\color{#026fc1}{1}} \times \dfrac{\color{#c1026f}{4x}}{\color{#c1026f}{x-1}} $ \t Convert to a rational expression. \\ \t ${}= \dfrac{\color{#026fc1}{2}\color{#c1026f}{(4x)}}{\color{#026fc1}{(1)}\color{#c1026f}{(x-1)}}$ \t Multiply top and bottom. \\ \t ${}= \dfrac{8x}{x-1}$ \t Calculate the products. \\ \t   \\ $\dfrac{\color{#026fc1}{4x}}{\color{#026fc1}{x-1}} \times \color{#c1026f}{(x^2+1)} $ \t ${}= \dfrac{\color{#026fc1}{4x}}{\color{#026fc1}{x-1}} \times \dfrac{\color{#c1026f}{x^2+1}}{\color{#c1026f}{1}} $ \t Convert to a rational expression. \\ \t ${}= \dfrac{\color{#026fc1}{4x}\color{#c1026f}{(x^2+1)}}{\color{#026fc1}{(x-1)}\color{#c1026f}{(1)}}$ \t Multiply top and bottom. \\ \t ${}= \dfrac{4x^3+4x}{x-1}$ \t Calculate the products. \\ \t   \\ $\dfrac{\color{#026fc1}{x-4}}{\color{#026fc1}{6x}} \times \dfrac{\color{#c1026f}{4x^3}}{\color{#c1026f}{2x+1}} $ \t ${}= \dfrac{\color{#026fc1}{(x-4)}\color{#c1026f}{4x^3}}{\color{#026fc1}{6x}\color{#c1026f}{(2x+1)}}$ \t Multiply top and bottom. \\ \t ${}=\dfrac{(x-4)2x}{3(2x+1)}$ \t Simplify: Cancel $\color{#6968d0}{2x}$. \\ \t ${}= \dfrac{2x^2-8x}{6x+3}$ \t Calculate the products. \\ \t   \\ $\dfrac{\color{#026fc1}{x-1}}{\color{#026fc1}{x}} \times \dfrac{\color{#c1026f}{2x^2-x}}{\color{#c1026f}{x^2-2x-1}} $ \t ${}= \dfrac{\color{#026fc1}{(x-1)}\color{#c1026f}{(2x^2-x)}}{\color{#026fc1}{x}\color{#c1026f}{(x^2-2x-1)}}$ \t Multiply top and bottom. \\ \t ${}=\dfrac{(x-1)(x)(2x-1)}{x(x-1)(x-1)}$ \t Factor. \\ \t ${}=\dfrac{2x-1}{x-1}$ \t Simplify: Cancel the $\color{#6968d0}{x}$ %%and $\color{#6968d0}{(x-1)}$.

Dividing rational expressions

As with ordinary fractions, division by a rational expression means multiplication by its reciprocal:
$\dfrac{\left(\dfrac{\color{#026fc1}{P}}{\color{#026fc1}{Q}}\right)}{\left(\dfrac{\color{#c1026f}{R}}{\color{#c1026f}{S}}\right)} = \dfrac{\color{#026fc1}{P}}{\color{#026fc1}{Q}} \times \dfrac{\color{#c1026f}{S}}{\color{#c1026f}{R}} \qquad$ \t
Flip the denominator and multiply.
\\ $\qquad = \dfrac{\color{#026fc1}{P}\color{#c1026f}{S}}{\color{#026fc1}{Q}\color{#c1026f}{R}} \qquad \quad$ \t Calculate the product.
Note As with products, before actually calculating the products on the top and bottom, you should first simplify by factoring and cancelling if possible.

Suggested video for this topic: Video by Pat Kopf
Examples
$\dfrac{\left(\dfrac{\color{#026fc1}{x+1}}{\color{#026fc1}{x}}\right)}{\left(\dfrac{\color{#c1026f}{x-1}}{\color{#c1026f}{2x+1}}\right)} $
\t ${}=\dfrac{\color{#026fc1}{x+1}}{\color{#026fc1}{x}} \times \dfrac{\color{#c1026f}{2x+1}}{\color{#c1026f}{x-1}}$ \t Flip the denominator and multiply. \\ \t ${}= \dfrac{\color{#026fc1}{(x+1)}\color{#c1026f}{(2x+1)}}{\color{#026fc1}{x}\color{#c1026f}{(x-1)}}$ \t Calculate the product. \\ \t ${}= \dfrac{2x^2+3x+1}{x^2-x}$ \\ \t   \\
$\dfrac{1}{\left(\dfrac{\color{#c1026f}{2x-1}}{\color{#c1026f}{x^3}}\right)}$
\t ${}= 1 \times \dfrac{\color{#c1026f}{x^3}}{\color{#c1026f}{2x-1}}$ \t
Flip the denominator and multiply.
\\ \t ${}= \dfrac{x^3}{2x-1}$ \\ \t   \\
 
\\
$\dfrac{\left(\dfrac{\color{#026fc1}{4x}}{\color{#026fc1}{x-1}}\right)}{\color{#c1026f}{x^2+1}} $
\t ${}= \dfrac{\color{#026fc1}{4x}}{\color{#026fc1}{x-1}} \times \dfrac{\color{#c1026f}{1}}{\color{#c1026f}{x^2+1}} $ \t Flip the denominator and multiply. \\ \t ${}= \dfrac{\color{#026fc1}{4x}\color{#c1026f}{(1)}}{\color{#026fc1}{(x-1)}\color{#c1026f}{(x^2+1)}}$ \t Multiply top and bottom. \\ \t ${}= \dfrac{4x}{x^3-x^2+x+1}$ \t Calculate the products. \\ \t   \\ $\dfrac{\color{#c1026f}{x^2+1}}{\left(\dfrac{\color{#026fc1}{4x}}{\color{#026fc1}{x-1}}\right)} $ \t ${}= \color{#c1026f}{(x^2+1)} \times \dfrac{\color{#026fc1}{x-1}}{\color{#026fc1}{4x}}$ \t Flip the denominator and multiply. \\ \t ${}= \dfrac{\color{#026fc1}{(x^2+1)}\color{#c1026f}{(x-1)}}{\color{#026fc1}{(1)}\color{#c1026f}{(4x)}}$ \t Multiply top and bottom. \\ \t ${}= \dfrac{x^3-x^2+x+1}{4x}$ \t Calculate the products.

Adding and subtracting rational expressions

Like the other rules we have seen, the rules for addition and subtraction of rational expressions are the same as for ordinary fractions. We start with the case in which the expressions we are adding or subtracting have the same denominator.
Addition and subtraction with common denominator:
$\dfrac{\color{#026fc1}{P}}{\color{#c1026f}{Q}} + \dfrac{\color{#026fc1}{R}}{\color{#c1026f}{Q}} = \dfrac{\color{#026fc1}{P} + \color{#026fc1}{R}}{\color{#c1026f}{Q}} \qquad $ \t Sum of numerators divided by common denominator \\   \\ $\dfrac{\color{#026fc1}{P}}{\color{#c1026f}{Q}} - \dfrac{\color{#026fc1}{R}}{\color{#c1026f}{Q}} = \dfrac{\color{#026fc1}{P} - \color{#026fc1}{R}}{\color{#c1026f}{Q}} \qquad $ \t Difference of numerators divided by common denominator
Notes
1. As with ordinary fractions, this formula works only when the two expressions have the same denominator.
2. When adding or subtracting expressions with the same denominator, do not factor or cancel before starting as you would with products or quotients; just leave them as they are until after doing the addition or subtractions.

Suggested video for this topic: Video by Sue Woolley
Examples
\\ $\dfrac{\color{#026fc1}{y}}{\color{#c1026f}{xy+1}} + \dfrac{\color{#026fc1}{x-1}}{\color{#c1026f}{xy+1}} $ \t ${}= \dfrac{\color{#026fc1}{y + x - 1}}{\color{#c1026f}{xy+1}}$ \t Add the numerators. \\ \t   \\ $\dfrac{\color{#026fc1}{x^2+1}}{\color{#c1026f}{x-1}} - \dfrac{\color{#026fc1}{2x}}{\color{#c1026f}{x-1}} $ \t ${}= \dfrac{\color{#026fc1}{x^2-2x+1}}{\color{#c1026f}{x-1}}$ \t Subtract the numerators. \\ \t ${}= \dfrac{(x-1)^2}{x-1}$ \t Factor. \\ \t ${}= x-1$ \t Simplify: Cancel the $\color{#6968d0}{(x-1)}$.

When the denominators are different we can use the following general rule, which works for all cases (including common denominators) although it does not always yield the most simple form of the answer:
Addition and subtraction: General case:
$\dfrac{\color{#026fc1}{P}}{\color{#026fc1}{Q}} + \dfrac{\color{#c1026f}{R}}{\color{#c1026f}{S}} = \dfrac{\color{#026fc1}{P}\color{#c1026f}{S} + \color{#026fc1}{Q}\color{#c1026f}{R}}{\color{#026fc1}{Q}\color{#c1026f}{S}} \qquad \quad \ \ $ Cross multiply to get the numerator: $\dfrac{\color{#026fc1}{P}}{\color{#026fc1}{Q}}$ $\dfrac{\color{#c1026f}{R}}{\color{#c1026f}{S}}$
Multiply across to get the denominator: $\dfrac{\color{#026fc1}{P}}{\color{#026fc1}{Q}}$ $\dfrac{\color{#c1026f}{R}}{\color{#c1026f}{S}}$
   
$\dfrac{\color{#026fc1}{P}}{\color{#026fc1}{Q}} - \dfrac{\color{#c1026f}{R}}{\color{#c1026f}{S}} = \dfrac{\color{#026fc1}{P}\color{#c1026f}{S} - \color{#026fc1}{Q}\color{#c1026f}{R}}{\color{#026fc1}{Q}\color{#c1026f}{S}} \qquad \quad \ \ $ Cross multiply to get the numerator: $\dfrac{\color{#026fc1}{P}}{\color{#026fc1}{Q}}$ $\dfrac{\color{#c1026f}{R}}{\color{#c1026f}{S}}$
Multiply across to get the denominator: $\dfrac{\color{#026fc1}{P}}{\color{#026fc1}{Q}}$ $\dfrac{\color{#c1026f}{R}}{\color{#c1026f}{S}}$
Notes
1. This formula works for ordinary fractions as well, and also when the two expressions have the same denominator (although cancellation is necessary to simplify the answer in that case).
2. When adding or subtracting expressions with different denominators, it helps to factor and/or cancel before starting as you would with products or quotients. This makes it easier to simplify the final answer.
3. If the denominators are the same, it is better to use the rule for addition and subtraction with common denominator; otherwise you will need to do additional work to simplify the answer.

Suggested video for this topic: Video by Pedro Thenumberbender
Examples
$\dfrac{\color{#026fc1}{3}}{\color{#026fc1}{2x+1}} + \dfrac{\color{#c1026f}{4}}{\color{#c1026f}{x-5}} $ \t ${}= \dfrac{\color{#026fc1}{3}\color{#c1026f}{(x-5)} + \color{#026fc1}{(2x+1)}\color{#c1026f}{4}}{\color{#026fc1}{(2x+1)}\color{#c1026f}{(x-5)}}$ \t Cross multiply for the numerator and multiply across for the denominator. \\ \t ${}= \dfrac{11x - 9}{(2x+1)(x-5)}$ \t Calculate the numerator. \\   \\ $\dfrac{\color{#026fc1}{2x}}{\color{#026fc1}{y-1}} - \dfrac{\color{#c1026f}{y+1}}{\color{#c1026f}{x}} $ \t ${}= \dfrac{\color{#026fc1}{2x}\color{#c1026f}{(x)} - \color{#026fc1}{(y-1)}\color{#c1026f}{(y+1)}}{\color{#026fc1}{(y-1)}\color{#c1026f}{(x)}}$ \t Cross multiply for the numerator and multiply across for the denominator. \\ \t ${}= \dfrac{2x^2-y^2+1}{(2x+1)(x-5)}$ \t Calculate the numerator. \\   \\ $\dfrac{\color{#026fc1}{5}}{\color{#026fc1}{2x}} - \dfrac{\color{#c1026f}{3}}{\color{#c1026f}{2(x+5)}} $ \t ${}= \dfrac{\color{#026fc1}{5}\cdot \color{#c1026f}{2(x+5)} - \color{#026fc1}{(2x)}\color{#c1026f}{3}}{\color{#026fc1}{2x}\cdot \color{#c1026f}{2(x+5)}}$ \t Cross multiply for the numerator and multiply across for the denominator. \\ \t ${}= \dfrac{4x + 50}{4x(x+5)}$ \t Calculate the numerator. \\ \t ${}= \dfrac{2(2x + 25)}{4x(x+5)}$ \t Factor the numerator. \\ \t ${}= \dfrac{2x + 25}{2x(x+5)}$ \t Simplify: Cancel the 2.

Putting it all together

One last quiz in which you will need to combine the above rules:

Now try the exercises in Section 0.4 in Finite Mathematics and Applied Calculus. or move ahead to the next tutorial by pressing "Next tutorial" on the sidebar.
Last Updated: April 2022
Copyright © 2021
Stefan Waner and Steven R. Costenoble

 

 

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