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5. Rational Functions
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Algebra 2

5.05 Add and subtract rational expressions

Lesson
Practice

Add and subtract rational expressions

Exploration

  1. What are the similarities and differences between these two expressions and how we evaluate them? \text{Expression 1: } \ \frac{1}{7}-\frac{1}{14} \qquad \text{Expression 2: }\ \frac{1}{7}+\frac{3}{7}

  2. Create an expression in the form: \frac{⬚}{⬚} + \frac{⬚}{⬚} 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: \frac{⬚}{⬚x} + \frac{⬚}{⬚} \text{\quad or \quad} \frac{⬚}{⬚} + \frac{⬚}{⬚x}

    • Adding x to the denominator of each fraction, resulting in: \frac{⬚}{x+ ⬚} + \frac{⬚}{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. 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:\frac{A}{B} + \frac{C}{B} = \frac{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 B\cdot D, so we have:

\displaystyle \frac{A}{B} + \frac{C}{D}\displaystyle =\displaystyle \frac{A}{B} \cdot \frac{D}{D} + \frac{C}{D} \cdot \frac{B}{B}Multiplicative identity, since \dfrac{D}{D}= \dfrac{B}{B}=1
\displaystyle =\displaystyle \frac{AD}{BD} + \frac{CB}{DB}Multiply the fractions
\displaystyle =\displaystyle \frac{AD}{BD} + \frac{BC}{BD}Commutative property of multiplication
\displaystyle =\displaystyle \frac{AD + BC}{BD}Add the 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. Write 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}Distributive property
\displaystyle =\displaystyle \frac{18}{3 k}Combine like terms in the numerator
\displaystyle =\displaystyle \frac{6}{k}Divide out 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

When subtracting rational expressions, we have to be careful to distribute the negative sign to each term in the numerator of the second expression. Notice that we used parentheses when combining the rational expressions.

Without parentheses, you might incorrectly simplify the numerator to k - 4 - k - 22, which leads a very different answer.

\displaystyle \frac{k - 4}{3 k} - \frac{k - 22}{3 k}\displaystyle =\displaystyle \frac{k - 4 - k - 22}{3 k}
\displaystyle =\displaystyle \frac{-26}{3 k}

By using parentheses, we can ensure each term is subtracted correctly.

Example 2

Determine whether the two expressions are equivalent, justifying your answer.

Expression 1:\frac{3x}{\left(x-2\right)} + \frac{\left(-2x+1\right)}{5x}

Expression 2:\frac{3x}{\left(x-2\right)} - \frac{\left(2x+1\right)}{5x}

Worked Solution
Create a strategy

To determine if these expressions are equivalent, we can fully simplify each expression by adding or subtracting the terms, then compare the resulting fractions.

Both terms in the given expressions have uncommon denominators, so we first have to find a common denominator before we can add or subtract. Because the denominators in the given expressions are the same, each will need to have factors of x-2 and 5x, so the least common denominator will be 5x\left(x-2\right).

Apply the idea

First, we will exclude the values of x that lead the denominator to equal zero. By setting the factors equal to 0 and solving, we find x\neq 2 and x\neq0.

Let's start with our first expression:

\displaystyle \frac{3x}{(x-2)} + \frac{\left(-2x+1\right)}{5x}\displaystyle =\displaystyle \frac{3x}{\left(x-2\right)} \cdot \frac{5x}{5x} + \frac{\left(-2x+1\right)}{5x} \cdot \frac{\left(x-2\right)}{\left(x-2\right)}Create a common denominator
\displaystyle =\displaystyle \frac{15x^{2} }{5x\left(x-2\right)} + \frac{-2x^{2}+4x+x -2}{5x\left(x-2\right)}Distributive property
\displaystyle =\displaystyle \frac{15x^{2} }{5x\left(x-2\right)} + \frac{-2x^{2}+5x -2}{5x\left(x-2\right)}Combine like terms
\displaystyle =\displaystyle \frac{15x^{2} -2x^{2} +5x -2}{5x\left(x-2\right)}Rewrite as a single fraction
\displaystyle =\displaystyle \frac{13x^{2}+5x-2}{5x\left(x-2\right)}Combine like terms

The numerator is not factorable, so this expression is fully simplified.

Now, we need to do the same process with the second expression:

\displaystyle \frac{3x}{\left(x-2\right)} - \frac{\left(2x+1\right)}{5x}\displaystyle =\displaystyle \frac{3x}{\left(x-2\right)} \cdot \frac{5x}{5x} - \frac{\left(2x+1\right)}{5x} \cdot \frac{\left(x-2\right)}{\left(x-2\right)}Create a common denominator
\displaystyle =\displaystyle \frac{15x^{2} }{5x\left(x-2\right)} - \frac{2x^{2}-4x+x -2}{5x\left(x-2\right)}Distributive property
\displaystyle =\displaystyle \frac{15x^{2} }{5x\left(x-2\right)} - \frac{2x^{2}-3x -2}{5x\left(x-2\right)}Combine like terms
\displaystyle =\displaystyle \frac{15x^{2} -\left(2x^{2} -3x -2\right)}{5x\left(x-2\right)}Rewrite as a single fraction
\displaystyle =\displaystyle \frac{15x^{2} -2x^{2} +3x +2}{5x\left(x-2\right)}Distribute the negative
\displaystyle =\displaystyle \frac{13x^{2}+3x+2}{5x\left(x-2\right)}Combine like terms in the numerator

The numerator is not factorable, so this expression is fully simplified.

Now compare the simplified expressions.

Simplified expression 1:\frac{13x^{2}+5x-2}{5x\left(x-2\right)} Simplified expression 2:\frac{13x^{2}+3x +2}{5x\left(x-2\right)}

The numerators of the expressions are different, so the two expressions are not equivalent.

Reflect and check

You may have noticed that the original expressions look very similar, with one being addition and one subtraction. Recall that subtracting a positive value is equivalent to adding a negative value.

If we apply that concept to this example, the second expression could have been written as the addition of the negative fraction, but we must be sure to distribute the negative to the entire numerator.

Expression 2:\frac{3x}{\left(x-2\right)} - \frac{\left(2x+1\right)}{5x}

\frac{3x}{\left(x-2\right)} + \frac{-\left(2x+1\right)}{5x}

\frac{3x}{\left(x-2\right)} + \frac{\left(-2x-1\right)}{5x}

This is another way to see that Expression 2 is not equivalent to Expression 1:\frac{3x}{\left(x-2\right)} + \frac{\left(-2x+1\right)}{5x}

Example 3

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

a

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

Apply the idea

The expression is undefined when 2p^{5}=0 and p^{2}m^{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^{2}m^{2}}\displaystyle =\displaystyle \frac{5m}{2p^{5}} \cdot \frac{m^{2}}{m^{2}} + \frac{4}{p^{2}m^{2}} \cdot \frac{2p^{3}}{2p^{3}}Create a common denominator of 2p^{5}m^{2}
\displaystyle =\displaystyle \frac{5m^{3}}{2p^{5}m^{2}} + \frac{8p^{3}}{2p^{5}m^{2}}Product rule of exponents
\displaystyle =\displaystyle \frac{5m^{3} + 8p^{3}}{2p^{5}m^{2}}Rewrite as a single fraction

The simplified expression is \dfrac{5m^{3} + 8p^{3}}{2p^{5}m^{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}Create a common denominator of 6\left(y + 9\right)
\displaystyle =\displaystyle \frac{y^{2} + 7y - 18}{6\left(y + 9\right)} + \frac{6y + 18}{6\left(y + 9\right)}Distributive property
\displaystyle =\displaystyle \frac{y^{2} + 7y - 18 + 6y + 18}{6\left(y + 9\right)}Rewrite as a single fraction
\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)}Factor the numerator

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

Reflect and check

We should 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 4

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)}Distributive property
\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 fraction
\displaystyle =\displaystyle \frac{2x^{2} - x - 15 - x^{2} - x}{\left(x - 3\right)^{2}\left(x + 1\right)}Distributive property
\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 5

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

\frac{ \frac{4}{x+3} + 6}{ 2 + \frac{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 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.

Creating a common denominator for the numerator:

\displaystyle \frac{4}{x+3} + 6\displaystyle =\displaystyle \frac{4}{x+3} + 6\cdot \frac{x+3}{x+3}Multiply by \dfrac{x+3}{x+3}
\displaystyle =\displaystyle \frac{4}{x+3} + \frac{6x+18}{x+3}Distributive property
\displaystyle =\displaystyle \frac{6x+22}{x+3}Rewrite as a single term

Creating a common denominator for the denominator:

\displaystyle 2 + \frac{2}{x+3}\displaystyle =\displaystyle 2\cdot \dfrac{x+3}{x+3} + \frac{2}{x+3}Multiply by \dfrac{x+3}{x+3}
\displaystyle =\displaystyle \frac{2x+6}{x+3} + \frac{2}{x+3}Distributive property
\displaystyle =\displaystyle \frac{2x+8}{x+3}Rewrite as a single term

Now, we can continue simplifying the expression:

\displaystyle \frac{ \frac{4}{x+3} + 6}{ 2 + \frac{2}{x+3}}\displaystyle =\displaystyle \frac{ \frac{6x+22}{x+3}}{ \frac{2x+8}{x+3}}Rewrite with the simplified numerator and denominator
\displaystyle =\displaystyle \frac{6x+22}{x+3} \cdot \frac{x+3}{2x+8}Rewrite as multiplication by the reciprocal
\displaystyle =\displaystyle \frac{2 \left(3x+11 \right)}{x+3} \cdot \frac{x+3}{2 \left( x+4 \right)}Factor out the GCFs
\displaystyle =\displaystyle \frac{\cancel{2} \left(3x+11 \right)}{\cancel{x+3}} \cdot \frac{\cancel{x+3}}{\cancel{2} \left( x+4 \right)}Divide out common factors
\displaystyle =\displaystyle \frac{3x+11}{x+4} Simplify

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

A2.EO.1

The student will perform operations on and simplify rational expressions.

A2.EO.1a

Add, subtract, multiply, or divide rational algebraic expressions, simplifying the result.

A2.EO.1b

Justify and determine equivalent rational algebraic expressions with monomial and binomial factors. Algebraic expressions should be limited to linear and quadratic expressions.

A2.EO.1c

Recognize a complex algebraic fraction and simplify it as a product or quotient of simple algebraic fractions.

A2.EO.1d

Represent and demonstrate equivalence of rational expressions written in different forms.

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