Topics
4. Rational Functions and Equations
topic badge
United States of America
Grades 9-12

4.04 Adding and subtracting rational expressions

Lesson
Practice

Introduction

We will extend our understanding of adding and subtracting fractions from 5th grade to rational expressions containing variables. We will work up to expressions with trinomials and simplify complex fractions that contain rational expressions using tools practiced in lesson  4.02 Rational expressions  and lesson  4.03 Multiplying and dividing rational expressions  .

Adding and subtracting rational expressions

Exploration

  1. What are the similarities and differences between these two expressions and how we evaluate them? \text{Expression 1: } \ \dfrac{1}{7}-\dfrac{1}{14} \qquad \text{Expression 2: }\ \dfrac{1}{7}+\dfrac{3}{7}
  2. Create an expression in the form: \dfrac{⬚}{⬚} + \dfrac{⬚}{⬚} where each blank is filled with a unique, nonzero integer value.
  3. Rewrite your original expression into two new expressions by:
    • Multiplying one term by \dfrac{1}{x}, where x is a positive integer, resulting in: \dfrac{⬚}{⬚x} + \dfrac{⬚}{⬚} \text{\quad or \quad} \dfrac{⬚}{⬚} + \dfrac{⬚}{⬚x}
    • Adding x to the denominator of each fraction, resulting in: \dfrac{⬚}{x+ ⬚} + \dfrac{⬚}{x + ⬚}
  4. Work with a partner to determine how to add the fractions created in step 3.

The sum of two rational expressions will result in another rational expression, so rational expressions are closed under addition and subtraction. Recall that a common denominator is required in order to add or subtract fractions. The same is true for rational expressions A, B, and C:\dfrac{A}{B} + \dfrac{C}{B} = \dfrac{A+C}{B}

In order to add or subtract rational expressions which have different denominators, we will need to find a common multiple to rewrite the expressions so that they share a common denominator. Given \dfrac{A}{B} + \dfrac{C}{D} where A, B, C, and D are expressions, common multiple is BD, so we have:

\displaystyle \dfrac{A}{B} + \dfrac{C}{D}\displaystyle =\displaystyle \dfrac{A}{B} \cdot \dfrac{D}{D} + \dfrac{C}{D} \cdot \dfrac{B}{B}Multiplicative identity, since \dfrac{D}{D}= \dfrac{B}{B}=1
\displaystyle =\displaystyle \dfrac{AD}{BD} + \dfrac{CB}{DB}Definition of multiplying rational expressions
\displaystyle =\displaystyle \dfrac{AD}{BD} + \dfrac{BC}{BD}Commutative property of multiplication
\displaystyle =\displaystyle \dfrac{AD + BC}{BD}Adding fractions with a common denominator

We need to state restrictions on the variables so we do not get an expression with 0 in the denominator, leading to an undefined expression.

Examples

Example 1

Fully simplify the expression, justifying each step. State any restrictions on the variables.\frac{k - 4}{3 k} - \frac{k - 22}{3 k}

Worked Solution
Create a strategy

These two rational expressions have the same denominator, so we can subtract them by subtracting their numerators (being careful with the signs).

Any values for the variables that lead the denominators of the rational expressions to equal zero should be excluded.

Apply the idea

First, note that the denominators are both 3k. When 3k=0, the expressions are undefined, so we can exclude k=0.

\displaystyle \frac{k - 4}{3 k} - \frac{k - 22}{3 k}\displaystyle =\displaystyle \frac{\left(k - 4\right) - \left(k - 22\right)}{3 k}Rewrite as a single rational expression
\displaystyle =\displaystyle \frac{k - 4 - k + 22}{3 k}Distribute the subtraction and remove parentheses
\displaystyle =\displaystyle \frac{18}{3 k}Combine like terms in the numerator
\displaystyle =\displaystyle \frac{6}{k}Simplify using a common factor of 3

Since k=0 would also lead the denominator of the simplified expression to equal zero, we state that the solution is \dfrac{6}{k}, k \neq 0.

Reflect and check

Notice that even though neither of the original rational expressions had any common factors to simplify subtracting the expressions resulted in a common factor of 3.

When adding or subtracting rational expressions, make sure to check for any common factors after the addition or subtraction.

Example 2

Fully simplify the rational expressions, justifying each step. State any restrictions on the variables.

a

\frac{5m}{2p^5} + \frac{4}{p^2m^2}

Worked Solution
Create a strategy

Determine the restrictions on the variables.

These two rational expressions do not have the same denominator, so we will first want to rewrite them to have a common denominator before we add them.

In this case, the denominator will need to have factors of 2, p^5, p^2, and m^2. The smallest expression which does this is 2p^5m^2.

Apply the idea

The expression is undefined when 2p^5=0 and p^2m^2=0, so when p=0 or m=0, the expressions are undefined. We will exclude both values of p and m.

\displaystyle \frac{5m}{2p^5} + \frac{4}{p^2m^2}\displaystyle =\displaystyle \frac{5m}{2p^5} \cdot \frac{m^2}{m^2} + \frac{4}{p^2m^2} \cdot \frac{2p^3}{2p^3}Multiply each rational expression to create a common denominator
\displaystyle =\displaystyle \frac{5m^3}{2p^5m^2} + \frac{8p^3}{2p^5m^2}Exponent product property
\displaystyle =\displaystyle \frac{5m^3 + 8p^3}{2p^5m^2}Rewrite as a single rational expression

The simplified expression is \dfrac{5m^3 + 8p^3}{2p^5m^2}, p \neq 0 and m \neq 0.

Reflect and check

In the original rational expressions, one denominator had a factor of p^5 while the other had a factor of p^2.

Notice that p^2 is already a factor of p^5, however. So the final expression only needed to have a factor of p^5 and not p^7.

We can compare this to adding fractions such as \dfrac{1}{2} and \dfrac{1}{8}. In this case, the LCM will only be 8 and not 16, since 2 is already a factor of 8:\frac{1}{2} + \frac{1}{8} = \frac{4}{8} + \frac{1}{8} = \frac{5}{8}

Although any common multiple will work, using the least common multiple will reduce the amount of simplification required after performing the addition.

b

\frac{y - 2}{6} + \frac{y + 3}{y + 9}

Worked Solution
Create a strategy

These two rational expressions do not have the same denominator, so we will first want to rewrite them to have a common denominator before we add them.

In this case, the denominator will need to have factors of 6 and y + 9, and so the least common denominator will be 6\left(y + 9\right).

Apply the idea

Exclude the value of y that leads the denominator to equal zero, so y+9=0 \to y = -9 will be excluded.

\displaystyle \frac{y - 2}{6} + \frac{y + 3}{y + 9}\displaystyle =\displaystyle \frac{y - 2}{6} \cdot \frac{y + 9}{y + 9} + \frac{y + 3}{y + 9} \cdot \frac{6}{6}Multiply each rational expression to create a common denominator
\displaystyle =\displaystyle \frac{y^2 + 7y - 18}{6\left(y + 9\right)} + \frac{6y + 18}{6\left(y + 9\right)}Distribute multiplication in each numerator
\displaystyle =\displaystyle \frac{y^2 + 7y - 18 + 6y + 18}{6\left(y + 9\right)}Rewrite as a single rational expression
\displaystyle =\displaystyle \frac{y^2 + 13y}{6\left(y + 9\right)}Combine like terms in the numerator
\displaystyle =\displaystyle \frac{y\left(y + 13\right)}{6\left(y + 9\right)}Write numerator in factored form

The fully simplified expression is \dfrac{y\left(y + 13\right)}{6\left(y + 9\right)}, y \neq -9.

Reflect and check

We would determine any other values of y that would make the denominator of the simplified rational expression equal to zero after evaluating the addition, but notice that when y=-9, the expression in its simplified form is also undefined.

Example 3

Fully simplify the expression, justifying each step. State any restrictions on the variables.\frac{2x + 5}{x^2 - 2x - 3} - \frac{x}{x^2 - 6x + 9}

Worked Solution
Create a strategy

These two rational expressions do not have the same denominator, so we want to rewrite them to have a common denominator before we subtract them. In order to do that, we will first factor each denominator. Before rewriting the expressions with a common denominator, we will determine the values for which the expressions are undefined.

Apply the idea

Factoring the denominators, we get\frac{2x + 5}{x^2 - 2x - 3} - \frac{x}{x^2 - 6x + 9} = \frac{2x + 5}{\left(x - 3\right)\left(x + 1\right)} - \frac{x}{\left(x - 3\right)^2}So the least common denominator will be \left(x - 3\right)^2\left(x + 1\right).

The values for which either rational expression will be undefined are when x=3 and x=-1.

We can now use this to subtract the two rational expressions:

\displaystyle \frac{2x + 5}{\left(x - 3\right)\left(x + 1\right)} - \frac{x}{\left(x - 3\right)^2}\displaystyle =\displaystyle \frac{2x + 5}{\left(x - 3\right)\left(x + 1\right)} \cdot \frac{x - 3}{x - 3} - \frac{x}{\left(x - 3\right)^2} \cdot \frac{x + 1}{x + 1}Create a common denominator
\displaystyle =\displaystyle \frac{2x^2 - x - 15}{\left(x - 3\right)^2\left(x + 1\right)} - \frac{x^2 + x}{\left(x - 3\right)^2\left(x + 1\right)}Distribute multiplication in each numerator
\displaystyle =\displaystyle \frac{2x^2 - x - 15 - \left(x^2 + x\right)}{\left(x - 3\right)^2\left(x + 1\right)}Rewrite as a single rational expression
\displaystyle =\displaystyle \frac{2x^2 - x - 15 - x^2 - x}{\left(x - 3\right)^2\left(x + 1\right)}Distribute the subtraction
\displaystyle =\displaystyle \frac{x^2 - 2x - 15}{\left(x - 3\right)^2\left(x + 1\right)}Combine like terms in the numerator
\displaystyle =\displaystyle \frac{\left(x - 5\right)\left(x + 3\right)}{\left(x - 3\right)^2\left(x + 1\right)}Factor the numerator

The fully simplified expression is \dfrac{\left(x - 5\right)\left(x + 3\right)}{\left(x - 3\right)^2\left(x + 1\right)}, x \neq -1 and x \neq 3.

Example 4

Fully simplify the expression, justifying each step. State any restrictions on the variables.

\dfrac{ \dfrac{4}{x+3} + 6}{ 2 + \dfrac{2}{x+3}}

Worked Solution
Create a strategy

We will find the common denominator of the numerator and denominator separately, then simplify the numerator and denominator. The common denominator for both the numerator and denominator is x+3.

Then, we can turn the division into a multiplication by taking the reciprocal of the second rational expression. We then want to identify common factors that can be simplified before performing the multiplication.

Apply the idea

The value of x that will lead to an undefined expression is x=-3.

\displaystyle \dfrac{ \dfrac{4}{x+3} + 6}{ 2 + \dfrac{2}{x+3}}\displaystyle =\displaystyle \dfrac{ \dfrac{4}{x+3} + 6 \cdot \dfrac{x+3}{x+3}}{ 2 \cdot \dfrac{x+3}{x+3} + \dfrac{2}{x+3}}Multiply each expression to create a common denominator
\displaystyle =\displaystyle \dfrac{ \dfrac{4}{x+3} + \dfrac{6x+18}{x+3}}{ \dfrac{2x+6}{x+3} + \dfrac{2}{x+3}}Distribute multiplication in each numerator of the rational expressions in the complex fraction
\displaystyle =\displaystyle \dfrac{ \dfrac{6x+22}{x+3}}{ \dfrac{2x+8}{x+3}}Rewrite as two rational expressions
\displaystyle =\displaystyle \dfrac{6x+22}{x+3} \cdot \dfrac{x+3}{2x+8}Rewrite division as multiplication
\displaystyle =\displaystyle \dfrac{2 \left(3x+11 \right)}{x+3} \cdot \dfrac{x+3}{2 \left( x+4 \right)}Factor a GCF
\displaystyle =\displaystyle \dfrac{ \left(3x+11 \right)}{1} \cdot \dfrac{1}{ \left( x+4 \right)}Simplify common factors
\displaystyle =\displaystyle \dfrac{3x+11}{x+4} Simplify the product

Note that the denominator of the simplified expression will have a different value of x for which the expression is undefined. We can see that x=-4 will lead to 0 in the denominator, so it will need to be excluded.

The fully simplified expression is \dfrac{3x+11}{x+4}, x \neq -3 and x \neq -4.

Idea summary

Prior to adding or subtracting rational expressions, do the following:

  • Determine restrictions on the variables that will lead to undefined expressions
  • If necessary, rewrite rational expressions to get a common denominator, using the multiplicative identity property: \dfrac{A}{A}=1 for any rational expression, A where A\neq 0

Outcomes

A.SSE.A.2

Use the structure of an expression to identify ways to rewrite it.

A.APR.D.7 (+)

Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions.

What is Mathspace

About Mathspace