Tutorial: Rational expressions
Adaptive game version
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
$1$ 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 $\color{slateblue}{Q = 1}.$
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
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
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.
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:
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
$\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 $\color{#6968d0}{(x-1)}$.
Addition and subtraction: General case:
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
$\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: | $\dfrac{\color{#026fc1}{P}}{\color{#026fc1}{Q}}$ $\dfrac{\color{#c1026f}{R}}{\color{#c1026f}{S}}$ |
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 0.5 in Finite Mathematics and Applied Calculus.
or move ahead to the next tutorial by pressing "Next tutorial" on the sidebar.
Copyright © 2021 Stefan Waner and Steven R. Costenoble