Operations on Rational Expressions
Learning Objectives
- Introduction to rational expressions
- Recognize and define a rational expression
- Determine the domain of a rational expression
- Simplify a rational expression
- Multiply and Divide Rational Expressions
- Add and Subtract Rational Expressions
- Identify the domain of a sum or difference of rational expressions
- Identify the least dcommon denominator of two rational expressions
- Add and subtract rational expressions using a greatest common denominator
Determine the domain of a rational expression
One sure way you can break math is to divide by zero. Consider the following rational expression evaluated at x = 2:Evaluate [latex]\frac{x}{x-2}[/latex] for [latex]x=2[/latex]
Substitute [latex]x=2[/latex]
[latex]\begin{array}{l}\frac{2}{2-2}\\\text{}\\=\frac{2}{0}\end{array}[/latex]
This means that for the expression [latex]\frac{x}{x-2}[/latex], x cannot be 2 because it will result in an undefined ratio. In general, finding values for a variable that will not result in division by zero is called finding the domain. Finding the domain of a rational expression or function will help you not break math.Domain of a rational expression or equation
The domain of a rational expression or equation is a collection of the values for the variable that will not result in an undefined mathematical operation such as division by zero. For a = any real number, we can notate the domain in the following way:x is all real numbers where [latex]x\neq{a}[/latex]
Example
Identify the domain of the expression. [latex] \frac{x+7}{{{x}^{2}}+8x-9}[/latex]Answer: Find any values for x that would make the denominator equal to 0 by setting the denominator equal to 0 and solving the equation.
[latex]x^{2}+8x-9=0[/latex]
Solve the equation by factoring. The solutions are the values that are excluded from the domain.[latex]\begin{array}{c}(x+9)(x-1)=0\\x=-9\,\,\,\text{or}\,\,\,x=1\end{array}[/latex]
Answer
The domain is all real numbers except [latex]−9[/latex] and [latex]1[/latex].Simplify Rational Expressions
Before we dive in to simplifying rational expressions, let's review the difference between a factor, a term, and an expression. This will hopefully help you avoid another way to break math when you are simplifying rational expressions. Factors are the building blocks of multiplication. They are the numbers that you can multiply together to produce another number: 2 and 10 are factors of 20, as are 4, 5, 1, 20. Terms are single numbers, or variables and numbers connected by multiplication. -4, 6x and [latex]x^2[/latex] are all terms. Expressions are groups of terms connected by addition and subtraction. [latex]2x^2-5[/latex] is an expression. This distinction is important when you are required to divide. Let's use an example to show why this is important. Simplify: [latex]\large\frac{2x^2}{12x}[/latex] The numerator and denominator of this fraction consist of factors. To simplify it, we can divide without being impeded by addition or subtraction. [latex-display]\begin{array}{cc}\large\frac{2x^2}{12x}\\=\large\frac{2\cdot{x}\cdot{x}}{2\cdot3\cdot2\cdot{x}}\\=\large\frac{\cancel{2}\cdot{\cancel{x}}\cdot{x}}{\cancel{2}\cdot3\cdot2\cdot{\cancel{x}}}\end{array}[/latex-display] We can do this because [latex]\frac{2}{2}=1\text{ and }\frac{x}{x}=1[/latex], so our expression simplifies to [latex]\large\frac{x}{6}[/latex] Compare that to the expression [latex]\large\frac{2x^2+x}{12-2x}[/latex], notice the denominator and numerator consist of two terms connected by addition and subtraction. We have to tip-toe around the addition and subtraction. When asked to simplify it is tempting to want to cancel out like terms as we did when we just had factors. But you can't do that, it will break math!Example
Simplify and state the domain for the expression. [latex] \frac{x+3}{{{x}^{2}}+12x+27}[/latex]Answer: To find the domain (and the excluded values), find the values for which the denominator is equal to 0. Factor the quadratic, and apply the zero product principle.
[latex]\begin{array}{c}x+3=0\,\,\,\,\,\,\text{or}\,\,\,\,\,\,x+9=0\\x=0-3\,\,\,\,\,\,\text{or}\,\,\,\,\,\,x=0-9\\x=-3\,\,\,\,\,\,\text{or}\,\,\,\,\,\,x=-9\\\\x=-3\,\,\,\,\,\,\text{or}\,\,\,\,\,\,x=-9\end{array}[/latex]
The domain is all real numbers except [latex]x=-3[/latex] or [latex]x=-9[/latex]. Factor the numerator and denominator. Identify the factors that are the same in the numerator and denominator, and simplify.[latex]\large\begin{array}{c}\frac{x+3}{x^{2}+12x+27}\\\\=\frac{x+3}{\left(x+3\right)\left(x+9\right)}\\\\\frac{\cancel{x+3}}{\cancel{\left(x+3\right)}\left(x+9\right)}\\\\\normalsize=1\cdot\large\frac{1}{x+9}\end{array}[/latex]
Answer
[latex-display] \frac{x+3}{{{x}^{2}}+12x+27}=\frac{1}{x+9}[/latex-display] The domain is all real numbers except [latex]−3[/latex] and [latex]−9[/latex].Example
Simplify and state the domain for the expression. [latex]\frac{x^{2}+10x+24}{x^{3}-x^{2}-20x}[/latex]Answer: To find the domain, determine the values for which the denominator is equal to 0.
[latex]\begin{array}{r}x^{3}-x^{2}-20x=0\\x\left(x^{2}-x-20\right)=0\\x\left(x-5\right)\left(x+4\right)=0\end{array}[/latex]
The domain is all real numbers except 0, 5, and −4. To simplify, factor the numerator and denominator of the rational expression. Identify the factors that are the same in the numerator and denominator, and simplify.[latex] \large\begin{array}{c}\frac{x^{2}+10x+24}{x^{3}-x^{2}-20x}\\\\=\frac{\left(x+4\right)\left(x+6\right)}{x\left(x-5\right)\left(x+4\right)}\\\\=\frac{\cancel{\left(x+4\right)}\left(x+6\right)}{x\left(x-5\right)\cancel{\left(x+4\right)}}\end{array}[/latex]
Simplify. It is acceptable to either leave the denominator in factored form or to distribute multiplication.[latex]\frac{x+6}{x\left(x-5\right)}\,\,\,\text{or}\,\,\,\frac{x+6}{x^{2}-5x}[/latex]
Answer
[latex-display] \frac{x+6}{x(x-5)}[/latex] or [latex] \frac{x+6}{{{x}^{2}}-5x}[/latex-display] The domain is all real numbers except 0, 5, and [latex]−4[/latex].Example
Simplify [latex]\frac{{x}^{2}-9}{{x}^{2}+4x+3}[/latex], state the domain.Answer: The special product in the numerator is a difference of squares. [latex-display]\begin{array}\frac{\left(x+3\right)\left(x - 3\right)}{\left(x+3\right)\left(x+1\right)}\hfill & \hfill & \hfill & \hfill & \text{Factor the numerator and the denominator}.\hfill \\ \frac{x - 3}{x+1}\hfill & \hfill & \hfill & \hfill & \text{Cancel common factor }\left(x+3\right).\hfill \end{array}[/latex-display] With the denominator factored it is easier to find the domain of the expression. Determine the values for which the denominator is equal to 0. [latex-display]\begin{array}{cc}\left(x+3\right)=0,\left(x+1\right)=0\\x\ne-3,\text{ AND }x\ne-1\end{array}[/latex-display]
Answer
[latex-display]\frac{{x}^{2}-9}{{x}^{2}+4x+3}=\frac{x - 3}{x+1}[/latex], Domain: [latex]x\ne-3,\text{ AND }x\ne-1[/latex-display]Steps for Simplifying a Rational Expression
To simplify a rational expression, follow these steps:- Determine the domain. The excluded values are those values for the variable that result in the expression having a denominator of 0.
- Factor the numerator and denominator.
- Find common factors for the numerator and denominator and simplify.
Multiply and Divide Rational Expressions
Just as you can multiply and divide fractions, you can multiply and divide rational expressions. In fact, you use the same processes for multiplying and dividing rational expressions as you use for multiplying and dividing numeric fractions. The process is the same even though the expressions look different!Multiply Rational Expressions
Remember that there are two ways to multiply numeric fractions. One way is to multiply the numerators and the denominators and then simplify the product, as shown here.[latex] \displaystyle \frac{4}{5}\cdot \frac{9}{8}=\frac{36}{40}=\frac{3\cdot 3\cdot 2\cdot 2}{5\cdot 2\cdot 2\cdot 2}=\frac{3\cdot 3\cdot \cancel{2}\cdot\cancel{2}}{5\cdot \cancel{2}\cdot\cancel{2}\cdot 2}=\frac{3\cdot 3}{5\cdot 2}\cdot 1=\frac{9}{10}[/latex]
A second way is to factor and simplify the fractions before performing the multiplication.[latex]\frac{4}{5}\cdot\frac{9}{8}=\frac{2\cdot2}{5}\cdot\frac{3\cdot3}{2\cdot2\cdot2}=\frac{\cancel{2}\cdot\cancel{2}\cdot3\cdot3}{\cancel{2}\cdot5\cdot\cancel{2}\cdot2}=1\cdot\frac{3\cdot3}{5\cdot2}=\frac{9}{10}[/latex]
Notice that both methods result in the same product. In some cases you may find it easier to multiply and then simplify, while in others it may make more sense to simplify fractions before multiplying. The same two approaches can be applied to rational expressions. Our first two examples apply both techniques to one expression. After that we will let you decide which works best for you.Example
Multiply.[latex] \displaystyle \frac{5{{a}^{2}}}{14}\cdot \frac{7}{10{{a}^{3}}}[/latex] State the product in simplest form.Answer: Multiply the numerators, and then multiply the denominators.
[latex]\frac{5a^{2}}{14}\cdot\frac{7}{10a^{3}}=\frac{35a^{2}}{140a^{3}}[/latex]
Simplify by finding common factors in the numerator and denominator. Simplify the common factors.[latex]\large\begin{array}{l}\frac{35a^{2}}{140a^{3}}=\frac{5\cdot7\cdot{a}^{2}}{5\cdot7\cdot2\cdot2\cdot{a}^{2}\cdot{a}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{\cancel{5}\cdot\cancel{7}\cdot\cancel{{a}^{2}}}{\cancel{5}\cdot\cancel{7}\cdot2\cdot2\cdot\cancel{{a}^{2}}\cdot{a}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\normalsize1\cdot\large\frac{1}{4a}\end{array}[/latex]
Simplify.[latex] \displaystyle \frac{1}{4a}[/latex]
Answer
[latex-display] \displaystyle \frac{5{{a}^{2}}}{14}\cdot \frac{7}{10{{a}^{3}}}=\frac{1}{4a}[/latex][latex] \displaystyle [/latex-display]Example
Multiply. [latex]\frac{5a^{2}}{14}\cdot\frac{7}{10a^{3}}[/latex] State the product in simplest form.Answer: Factor the numerators and denominators. Look for the greatest common factors.
[latex] \displaystyle \frac{5\cdot {{a}^{2}}}{7\cdot 2}\cdot \frac{7}{5\cdot 2\cdot {{a}^{2}}\cdot a}[/latex]
Simplify common factors, then multiply.[latex]\large\begin{array}{c}\frac{5\cdot {{a}^{2}}}{7\cdot 2}\cdot \frac{7}{5\cdot 2\cdot {{a}^{2}}\cdot a}\\\\=\frac{\cancel{5}\cdot\cancel{{a}^{2}}}{\cancel{7}\cdot 2}\cdot \frac{\cancel{7}}{\cancel{5}\cdot 2\cdot\cancel{{a}^{2}}\cdot a}\\\\=\frac{1\cdot1\cdot1}{2\cdot2\cdot{a}}=\frac{1}{4a}\end{array}[/latex]
Answer
[latex-display]\frac{5a^{2}}{14}\cdot\frac{7}{10a^{3}}=\frac{1}{4a}[/latex-display]Example
Multiply. [latex] \displaystyle \frac{{{a}^{2}}-a-2}{5a}\cdot \frac{10a}{a+1}\,\,,\,\,\,\,\,\,a\,\ne \,\,-1\,,\,\,0[/latex] State the product in simplest form.Answer:
Factor the numerators and denominators.
[latex]\frac{\left(a-2\right)\left(a+1\right)}{5\cdot{a}}\cdot\frac{5\cdot2\cdot{a}}{\left(a+1\right)}[/latex]
Simplify common factors:[latex]\large\begin{array}{c}\frac{\left(a-2\right)\cancel{\left(a+1\right)}}{\cancel{5}\cdot{\cancel{a}}}\cdot\frac{\cancel{5}\cdot2\cdot{\cancel{a}}}{\cancel{\left(a+1\right)}}\\\\=\frac{a-2}{1}\cdot\frac{2}{1}\end{array}[/latex]
Multiply simplified rational expressions. This expression can be left with the numerator in factored form or multiplied out.[latex]\begin{array}{c}\frac{\left(a-2\right)}{1}\cdot\frac{2}{1}\\\\=2\left(a-2\right)\end{array}[/latex]
Answer
[latex-display] \displaystyle \frac{{{a}^{2}}-a-2}{5a}\cdot \frac{10a}{a+1}=2a-4[/latex-display]Example
Multiply. [latex]\frac{a^{2}+4a+4}{2a^{2}-a-10}\cdot\frac{a+5}{a^{2}+2a},\,\,\,a\neq-2,0,\frac{5}{2}[/latex] State the product in simplest form.Answer: Factor the numerators and denominators.
[latex]\frac{\left(a+2\right)\left(a+2\right)}{\left(2a-5\right)\left(a+2\right)}\cdot\frac{a+5}{a\left(a+2\right)}[/latex]
Simplify common factors.[latex]\large\frac{\cancel{\left(a+2\right)}\cancel{\left(a+2\right)}}{\left(2a-5\right)\cancel{\left(a+2\right)}}\cdot\frac{a+5}{a\cancel{\left(a+2\right)}}[/latex]
Multiply simplified rational expressions. This expression can be left with the denominator in factored form or multiplied out.[latex]\frac{1}{\left(2a-5\right)}\cdot\frac{a+5}{a}=\frac{a+5}{a\left(2a-5\right)}[/latex]
Answer
[latex-display]\frac{a^{2}+4a+4}{2a^{2}-a-10}\cdot\frac{a+5}{a^{2}+2a}=\frac{a+5}{a\left(2a-5\right)}[/latex-display]Divide Rational Expressions
You've seen that you multiply rational expressions as you multiply numeric fractions. It should come as no surprise that you also divide rational expressions the same way you divide numeric fractions. Specifically, to divide rational expressions, keep the first rational expression, change the division sign to multiplication, and then take the reciprocal of the second rational expression. Let’s begin by recalling division of numerical fractions.[latex]\frac{2}{3}\div\frac{5}{9}=\frac{2}{3}\cdot\frac{9}{5}=\frac{18}{15}=\frac{6}{5}[/latex]
Use the same process to divide rational expressions. You can think of division as multiplication by the reciprocal, and then use what you know about multiplication to simplify.Example
State the domain, then divide. [latex]\frac{5x^{2}}{9}\div\frac{15x^{3}}{27}[/latex]Answer: State the Domain: Find excluded values. 9 and 27 can never equal 0. Because [latex]15x^{3}[/latex] becomes the denominator in the reciprocal of [latex] \displaystyle \frac{15{{x}^{3}}}{27}[/latex], you must find the values of x that would make [latex]15x^{3}[/latex] equal 0.
[latex]\begin{array}{c}15x^{3}=0\\x=0\,\text{is an excluded value}.\end{array}[/latex]
Divide: State the quotient in simplest form. Rewrite division as multiplication by the reciprocal.[latex]\frac{5x^{2}}{9}\cdot\frac{27}{15x^{3}}[/latex]
Factor the numerators and denominators.[latex]\frac{5\cdot{x}\cdot{x}}{3\cdot3}\cdot\frac{3\cdot3\cdot3}{5\cdot3\cdot{x}\cdot{x}\cdot{x}}[/latex]
Simplify common factors. Simplify.[latex]\large\begin{array}{c}\frac{\cancel{5}\cdot{\cancel{x}}\cdot{\cancel{x}}}{\cancel{3}\cdot\cancel{3}}\cdot\frac{\cancel{3}\cdot\cancel{3}\cdot\cancel{3}}{\cancel{5}\cdot\cancel{3}\cdot{\cancel{x}}\cdot{\cancel{x}}\cdot{x}}\\\\=\frac{1}{x}\end{array}[/latex]
Answer
[latex-display] \displaystyle \frac{5{{x}^{2}}}{9}\div \frac{15{{x}^{3}}}{27}=\frac{1}{x},x\ne 0[/latex-display]Example
Divide. [latex]\frac{3x^{2}}{x+2}\div\frac{6x^{4}}{\left(x^{2}+5x+6\right)}[/latex] State the quotient in simplest form, and express the domain of the expression.Answer: Determine the excluded values that make the denominators and the numerator of the divisor equal to 0.
[latex]\begin{array}{r}\left(x+2\right)=0\,\,\,\,\,\\x=-2\\\left({{x}^{2}}+5x+6 \right)=0\,\,\,\,\,\\\left(x+3\right)\left(x+2\right)=0\,\,\,\,\,\\x=-3\,\,\,\,\text{or}\,\,\,\,-2\\6x^{4}=0\,\,\,\,\,\\x=0\,\,\,\,\,\end{array}[/latex]
Domain is all real numbers except [latex]0[/latex], [latex]−2[/latex], and [latex]−3[/latex]. Rewrite division as multiplication by the reciprocal.[latex]\frac{3x^{2}}{x+2}\cdot\frac{\left(x^{2}+5x+6\right)}{6x^{4}}[/latex]
Factor the numerators and denominators.[latex]\frac{3\cdot{x}\cdot{x}}{x+2}\cdot\frac{\left(x+2\right)\left(x+3\right)}{2\cdot3\cdot{x}\cdot{x}\cdot{x}\cdot{x}}[/latex]
Simplify common factors[latex]\large\frac{\cancel{3}\cdot{\cancel{x}}\cdot{\cancel{x}}}{\cancel{x+2}}\cdot\frac{\cancel{\left(x+2\right)}\left(x+3\right)}{2\cdot\cancel{3}\cdot{\cancel{x}}\cdot{\cancel{x}}\cdot{x}\cdot{x}}[/latex]
Simplify.[latex]\frac{(x+3)}{2{{x}^{2}}}[/latex]
Answer
[latex] \displaystyle \frac{3{{x}^{2}}}{x+2}\div \frac{6{{x}^{4}}}{({{x}^{2}}+5x+6)}=\frac{x+3}{2{{x}^{2}}}[/latex]. The domain is all real numbers except 0, [latex]−2[/latex], and [latex]−3[/latex].Add and Subtract Rational Expressions
In beginning math, students usually learn how to add and subtract whole numbers before they are taught multiplication and division. However, with fractions and rational expressions, multiplication and division are sometimes taught first because these operations are easier to perform than addition and subtraction. Addition and subtraction of rational expressions are not as easy to perform as multiplication because, as with numeric fractions, the process involves finding common denominators.Adding Rational Expressions
To find the LCD of two rational expressions, we factor the expressions and multiply all of the distinct factors. For instance, given the rational expressions[latex]\large\frac{6}{\left(x+3\right)\left(x+4\right)},\text{ and }\frac{9x}{\left(x+4\right)\left(x+5\right)}[/latex]
The LCD would be [latex]\left(x+3\right)\left(x+4\right)\left(x+5\right)[/latex].
To find the LCD, we count the greatest number of times a factor appears in each denominator, and make sure it is represented in the LCD that many times. For example, in [latex]\large\frac{6}{\left(x+3\right)\left(x+4\right)}[/latex], [latex]\left(x+3\right)[/latex] is represented once and [latex]\left(x+4\right)[/latex] is represented once, so they both appear exactly once in the LCD. In [latex]\large\frac{9x}{\left(x+4\right)\left(x+5\right)}[/latex], [latex]\left(x+4\right)[/latex] appears once, and [latex]\left(x+5\right)[/latex] appears once. We have already accounted for [latex]\left(x+4\right)[/latex], so the LCD just needs one factor of [latex]\left(x+5\right)[/latex] to be complete. Once we find the LCD, we need to multiply each expression by the form of 1 that will change the denominator to the LCD. What do we mean by " the form of 1"? [latex]\frac{x+5}{x+5}=1[/latex] so multiplying an expression by it will not change it's value. For example, we would need to multiply the expression [latex]\large\frac{6}{\left(x+3\right)\left(x+4\right)}[/latex] by [latex]\frac{x+5}{x+5}[/latex] and the expression [latex]\frac{9x}{\left(x+4\right)\left(x+5\right)}[/latex] by [latex]\frac{x+3}{x+3}[/latex]. Hopefully this process will become clear after you practice it yourself. As you look through the examples on this page, try to identify the LCD before you look at the answers. Also, try figuring out which "form of 1" you will need to multiply each expression by so that it has the LCD.Example
Add the rational expressions: [latex]\frac{5}{x}+\frac{6}{y}[/latex], and define the domain. State the sum in simplest form.Answer: First, let's define the domain of each term. Since we have x and y in the denominators, we can say [latex]x\ne0 ,\text{ and }y\ne0[/latex]. Now we have to find the LCD. Since x appears once and y appears once, the LCD will be [latex]xy[/latex]. We then multiply each expression by the appropriate form of 1 to obtain [latex]xy[/latex] as the denominator for each fraction.
The domain is [latex]x\ne-4[/latex]
Answer
[latex-display] \displaystyle \frac{2{{x}^{2}}}{x+4}+\frac{8x}{x+4}=2x,x\ne -4[/latex], [latex]x\ne0 ,\text{ and }y\ne0[/latex-display]Analysis of the Solution
Multiplying by [latex]\frac{y}{y}[/latex] or [latex]\frac{x}{x}[/latex] does not change the value of the original expression because any number divided by itself is 1, and multiplying an expression by 1 gives the original expression. Here is one more example of adding rational expressions, but in this case, the expressions have denominators with multi-term polynomials. First, we will factor, then find the LCD. Note that [latex]x^2-4[/latex] is a difference of squares and can be factored using special products.Example
Simplify[latex]\frac{2{{x}^{2}}}{{{x}^{2}}-4}+\frac{x}{x-2}[/latex], and give the domain. State the result in simplest form.Answer: Find the least common multiple by factoring each denominator. Multiply each factor the maximum number of times it appears in a single factorization. Remember that x cannot be [latex]2[/latex] or [latex]-2[/latex] because the denominators would be 0. [latex]\left(x+2\right)[/latex] appears a maximum of one time, as does [latex]\left(x–2\right)[/latex]. This means the LCM is [latex]\left(x+2\right)\left(x–2\right)[/latex].
[latex]\begin{array}{l}x^{2}-4=\left(x+2\right)\left(x-2\right)\\\,\,x-2=x-2\\\,\,x+2=x+2\\\,\,\text{LCM}=\left(x+2\right)\left(x-2\right)\end{array}[/latex]
The LCM becomes the common denominator. Multiply each expression by the equivalent of 1 that will give it the common denominator.[latex]\begin{array}{r}\frac{2{{x}^{2}}}{{{x}^{2}}-4}=\frac{2{{x}^{2}}}{(x+2)(x-2)}\\\frac{x}{x-2}\cdot \frac{x+2}{x+2}=\frac{x(x+2)}{(x+2)(x-2)}\end{array}[/latex]
Rewrite the original problem with the common denominator. It makes sense to keep the denominator in factored form in order to check for common factors.[latex] \displaystyle \frac{2{{x}^{2}}}{(x+2)(x-2)}+\frac{x(x+2)}{(x+2)(x-2)}[/latex]
Combine the numerators.[latex] \begin{array}{c}\frac{2{{x}^{2}}+x(x+2)}{(x+2)(x-2)}\\\\\frac{2{{x}^{2}}+{{x}^{2}}+2x}{(x+2)(x-2)}\end{array}[/latex]
Check for simplest form. Since neither [latex]\left(x+2\right)[/latex] nor [latex]\left(x-2\right)[/latex] is a factor of [latex]3{{x}^{2}}+2x[/latex], this expression is in simplest form.[latex]\frac{3{{x}^{2}}+2x}{(x+2)(x-2)}[/latex]
Answer
[latex-display] \displaystyle \frac{2{{x}^{2}}}{{{x}^{2}}-4}+\frac{x}{x-2}=\frac{3{{x}^{2}}+2x}{(x+2)(x-2)}[/latex][latex] \displaystyle x\ne 2,-2[/latex-display]Subtracting Rational Expressions
To subtract rational expressions, follow the same process you use to add rational expressions. You will need to be careful with signs, though.Example
Subtract[latex]\frac{2}{t+1}-\frac{t-2}{{{t}^{2}}-t-2}[/latex], define the domain. State the difference in simplest form.Answer: Find the LCD of each expression. [latex]t+1[/latex] cannot be factored any further, but [latex]{{t}^{2}}-t-2[/latex] can be. Note that t cannot be [latex]-1[/latex] or [latex]2[/latex] because the denominators would be 0.
[latex]\begin{array}{c}t+1=t+1\\t^{2}-t-2=\left(t-2\right)\left(t+1\right)\end{array}[/latex]
Find the least common multiple. [latex]t+1[/latex] appears exactly once in both of the expressions, so it will appear once in the least common denominator. [latex]t–2[/latex] also appears once. This means that [latex]\left(t-2\right)\left(t+1\right)[/latex] is the least common multiple. In this case, it is easier to leave the common multiple in terms of the factors, so you will not multiply it out. Use the least common multiple for your new common denominator, it will be the LCD.[latex]\begin{array}{c}t+1=t+1\\t^{2}-t-2=\left(t-2\right)\left(t+1\right)\\\text{LCM}:\left(t+1\right)\left(t-1\right)\end{array}[/latex]
Compare each original denominator and the new common denominator. Now rewrite the rational expressions to each have the common denominator of [latex]\left(t+1\right)\left(t–2\right)[/latex]. You need to multiply [latex]t+1[/latex] by [latex]t–2[/latex] to get the LCD, so multiply the entire rational expression by [latex] \displaystyle \frac{t-2}{t-2}[/latex]. The second expression already has a denominator of [latex]\left(t+1\right)\left(t–2\right)[/latex], so you do not need to multiply it by anything.[latex] \begin{array}{c}\frac{2}{t+1}\cdot \frac{t-2}{t-2}=\frac{2(t-2)}{(t+1)(t-2)}\\\\\,\,\,\frac{t-2}{{{t}^{2}}-t-2}=\frac{t-2}{(t+1)(t-2)}\end{array}[/latex]
Then rewrite the subtraction problem with the common denominator.[latex] \frac{2\left(t-2\right)}{\left(t+1\right)\left(t-2\right)}-\frac{t-2}{\left(t+1\right)\left(t-2\right)}[/latex]
Subtract the numerators and simplify. Remember that parentheses need to be included around the second [latex]\left(t–2\right)[/latex] in the numerator because the whole quantity is subtracted. Otherwise you would be subtracting just the t.[latex] \begin{array}{c}\frac{2(t-2)-(t-2)}{(t+1)(t-2)}\\\\\frac{2t-4-t+2}{(t+1)(t-2)}\\\\\frac{t-2}{(t+1)(t-2)}\end{array}[/latex]
The numerator and denominator have a common factor of [latex]t–2[/latex], so the rational expression can be simplified.[latex]\large\begin{array}{c}\frac{\cancel{t-2}}{(t+1)\cancel{(t-2)}}\\\\=\frac{1}{t+1}\end{array}[/latex]
Answer
[latex-display] \displaystyle \frac{2}{t+1}-\frac{t-2}{{{t}^{2}}-t-2}=\frac{1}{t+1},t\ne -1,2[/latex-display]Example
Subtract the rational expressions: [latex]\frac{6}{{x}^{2}+4x+4}-\frac{2}{{x}^{2}-4}[/latex], and define the domain. State the difference in simplest form.Answer: Note that the denominator of the first expression is a perfect square trinomial, and the denominator of the second expression is a difference of squares so they can be factored using special products. [latex-display]\begin{array}{cc}\frac{6}{{\left(x+2\right)}^{2}}-\frac{2}{\left(x+2\right)\left(x - 2\right)}\hfill & \text{Factor}.\hfill \\ \frac{6}{{\left(x+2\right)}^{2}}\cdot \frac{x - 2}{x - 2}-\frac{2}{\left(x+2\right)\left(x - 2\right)}\cdot \frac{x+2}{x+2}\hfill & \text{Multiply each fraction to get LCD as denominator}.\hfill \\ \frac{6\left(x - 2\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}-\frac{2\left(x+2\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Multiply}.\hfill \\ \frac{6x - 12-\left(2x+4\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Apply distributive property}.\hfill \\ \frac{4x - 16}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Subtract}.\hfill \\ \frac{4\left(x - 4\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Simplify}.\hfill \end{array}[/latex-display]
The domain is [latex]x\ne-6[/latex]
Answer
[latex-display] \displaystyle \frac{4x+7}{x+6}-\frac{2x+8}{x+6}=\frac{2x-1}{x+6},\text{}x\ne-6[/latex-display]Analysis of the solution
In the last example, the LCD was [latex]\left(x+2\right)^2\left(x-2\right)[/latex]. The reason we need to include [latex]\left(x+2\right)[/latex] two times is because it appears two times in the expression [latex]\frac{6}{{x}^{2}+4x+4}[/latex]. The video that follows contains an example of subtracting rational expressions. https://www.youtube.com/watch?v=MMlNtCrkakI&feature=youtu.beSummary
An additional consideration for rational expressions is to determine what values are excluded from the domain. Since division by 0 is undefined, any values of the variables that result in a denominator of 0 must be excluded. Excluded values must be identified in the original equation, not from its factored form.Rational expressions are fractions containing polynomials. They can be simplified much like numeric fractions. To simplify a rational expression, first determine common factors of the numerator and denominator, and then remove them by rewriting them as expressions equal to 1. Rational expressions are multiplied and divided the same way as numeric fractions. To multiply, first find the greatest common factors of the numerator and denominator. Next, regroup the factors to make fractions equivalent to one. Then, multiply any remaining factors. To divide, first rewrite the division as multiplication by the reciprocal of the denominator. The steps are then the same as for multiplication. When expressing a product or quotient, it is important to state the excluded values. These are all values of a variable that would make a denominator equal zero at any step in the calculations.Licenses & Attributions
CC licensed content, Original
- Screenshot: Breaking Math. Provided by: Lumen Learning License: CC BY: Attribution.
- Revision and Adaptation. Provided by: Lumen Learning License: CC BY: Attribution.
- Simplify and Give the Domain of Rational Expressions. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Screenshot: Multiply and Divide. Provided by: Lumen Learning License: CC BY: Attribution.
- Screenshot: Reciprocal Architecture. Provided by: Lumen Learning License: CC BY: Attribution.
- Multiply Rational Expressions and Give the Domain. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Divide Rational Expressions and Give the Domain. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Screenshot: Caution. Provided by: Lumen Learning License: CC BY: Attribution.
- Image: What do they have in common?. Provided by: Lumen Learning License: CC BY: Attribution.
- Ex: Add Rational Expressions with Unlike Denominators. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Subtract Rational Expressions with Unlike Denominators and Give the Domain. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Subtract Rational Expressions with Unlike Denominators and Give the Domain. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
CC licensed content, Shared previously
- Unit 15: Rational Expressions, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education Located at: https://www.nroc.org/. License: CC BY: Attribution.
- College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et al.. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution. License terms: Download for free at : http://cnx.org/contents/[email protected]:1/Preface.
- Ex 1: Simplify a Complex Fraction (No Variables). Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
- Ex 2: Simplify a Complex Fraction (Variables). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Ex 3: Simplify a Complex Fraction (Variables). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.