8.5: Compound Rational Expressions
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Compound fractions have fractions in either the numerator, or denominator, or usually both, i.e., fractions over fractions. These expressions are simplified by one of two ways.
- Simplify the numerator and denominator first, then divide as usual using techniques from this section.
- Simplify by multiplying each term in the expression by the lowest common denominator. Then simplify as usual using techniques from the previous section.
Let’s take a look at a few examples demonstrating both methods. We will start with a problem from arithmetic and then move on to algebraic expressions.
Simplify \(\dfrac{\dfrac{2}{3}-\dfrac{1}{4}}{\dfrac{5}{6}+\dfrac{1}{2}}\).
Solution
We simplify the numerator and denominator first, then divide as usual using techniques from this section.
\[\begin{array}{rl}\dfrac{\dfrac{2}{3}-\dfrac{1}{4}}{\dfrac{5}{6}+\dfrac{1}{2}}&\text{Rewrite each numerator and denominator in terms of the LCD} \\ \dfrac{\dfrac{8}{12}-\dfrac{3}{12}}{\dfrac{5}{6}+\dfrac{3}{6}}&\text{Simplify each numerator and denominator} \\ \dfrac{\dfrac{5}{12}}{\dfrac{8}{6}}&\text{Rewrite with the division sign} \\ \dfrac{5}{12}\div\dfrac{8}{6}&\text{Rewrite with multiplication by reciprocating the second fraction} \\ \dfrac{5}{12}\cdot \dfrac{6}{8}&\text{Multiply across numerators and denominators} \\ \dfrac{30}{96}&\text{Reduce by a factor of }6 \\ \dfrac{5}{16}&\text{Result}\end{array}\nonumber\]
Thus, the compound fraction reduces to \(\dfrac{5}{16}\).
Simplify \(\dfrac{1-\dfrac{1}{x^2}}{1-\dfrac{1}{x}}\).
Solution
We simplify by multiplying each term in the expression by the lowest common denominator. Then simplify as usual using techniques from the previous section.
\[\begin{array}{rl}\dfrac{1-\dfrac{1}{x^2}}{1-\dfrac{1}{x}}&\text{Multiply each term by the LCD}=x^2 \\ \dfrac{1\cdot\color{blue}{x^2}\color{black}{}-\dfrac{1}{x^2}\cdot\color{blue}{x^2}}{1\color{blue}{x^2}\color{black}{}-\dfrac{1}{x}\cdot\color{blue}{x^2}}&\color{black}{\text{Simplify each term}} \\ \dfrac{x^2-1}{x^2-x}&\text{Factor the numerator and denominator} \\ \dfrac{(x+1)(x-1)}{x(x-1)}&\text{Reduce by a factor of }(x-1) \\ \dfrac{(x+1)\color{blue}{\cancel{(x-1)}}}{x\color{blue}{\cancel{(x-1)}}}&\color{black}{\text{Rewrite}} \\ \dfrac{x+1}{x}&\text{Result} \end{array}\nonumber\]
Thus, the compound fraction reduces to \(\dfrac{x+1}{x}\).
As best practice, we use the second method, where we multiply each term in the expression by the lowest common denominator because this technique reduced the fraction to one denominator and one numerator. The first method kept the compound fraction until we rewrote is as two expressions with division. Moving forward, students should always apply the second method, multiplying each term in the expression by the lowest common denominator.
Simplify \(\dfrac{\dfrac{3}{x+4}-2}{5+\dfrac{2}{x+4}}\)
Solution
We simplify by multiplying each term in the expression by the lowest common denominator. Then simplify as usual using techniques from the previous section.
\[\begin{array}{rl}\dfrac{\dfrac{3}{x+4}-2}{5+\dfrac{2}{x+4}}&\text{Multiply each term by the LCD}=(x+4) \\ \dfrac{\dfrac{3}{x+4}\cdot\color{blue}{(x+4)}\color{black}{}-2\cdot\color{blue}{(x+4)}}{5\cdot\color{blue}{(x+4)}\color{black}{}+\dfrac{2}{x+4}\cdot\color{blue}{(x+4)}}&\color{black}{\text{Simplify each term}} \\ \dfrac{3-2(x+4)}{5(x+4)+2}&\text{Simplify} \\ \dfrac{3-2x-8}{5x+20+2}&\text{Combine like terms} \\ \dfrac{-2x-5}{5x+22}&\text{Result}\end{array}\nonumber\]
Thus, the compound fraction reduces to \(\dfrac{-2x-5}{5x+22}\).
Sophie Germain is one of the most famous women in mathematics. Many prime numbers, which are important to finding an LCD, carry her name. Germain primes are prime numbers where one more than double the prime number is also prime. For example, \(3\) is prime and so is \(2\cdot 3 + 1 = 7\). The largest known Germain prime (at the time of printing) is \(183027 \cdot 2^{265440} − 1\) which has \(79,911\) digits.
Compound Rational Expressions Homework
Simplify.
\(\dfrac{1+\dfrac{1}{x}}{1-\dfrac{1}{x^2}}\)
\(\dfrac{a-2}{\dfrac{4}{a}-a}\)
\(\dfrac{\dfrac{1}{a^2}-\dfrac{1}{a}}{\dfrac{1}{a^2}+\dfrac{1}{a}}\)
\(\dfrac{2-\dfrac{4}{x+2}}{5-\dfrac{10}{x+2}}\)
\(\dfrac{\dfrac{3}{2a-3}+2}{\dfrac{-6}{2a-3}-4}\)
\(\dfrac{\dfrac{x}{x+1}-\dfrac{1}{x}}{\dfrac{x}{x+1}+\dfrac{1}{x}}\)
\(\dfrac{\dfrac{3}{x}}{\dfrac{9}{x^2}}\)
\(\dfrac{\dfrac{a^2-b^2}{4a^2b}}{\dfrac{a+b}{16ab^2}}\)
\(\dfrac{1-\dfrac{3}{x}-\dfrac{10}{x^2}}{1+\dfrac{11}{x}+\dfrac{18}{x^2}}\)
\(\dfrac{1-\dfrac{2x}{3x-4}}{x-\dfrac{32}{3x-4}}\)
\(\dfrac{x-1+\dfrac{2}{x-4}}{x+3+\dfrac{6}{x-4}}\)
\(\dfrac{\dfrac{1}{y^2}-1}{1+\dfrac{1}{y}}\)
\(\dfrac{\dfrac{25}{a}-a}{5+a}\)
\(\dfrac{\dfrac{1}{b}+\dfrac{1}{2}}{\dfrac{4}{b^2-1}}\)
\(\dfrac{4+\dfrac{12}{2x-3}}{5+\dfrac{15}{2x-3}}\)
\(\dfrac{\dfrac{-5}{b-5}-3}{\dfrac{10}{b-5}+6}\)
\(\dfrac{\dfrac{2a}{a-1}-\dfrac{3}{a}}{\dfrac{-6}{a-1}-4}\)
\(\dfrac{\dfrac{x}{3x-2}}{\dfrac{x}{9x^2-4}}\)
\(\dfrac{1-\dfrac{1}{x}-\dfrac{6}{x^2}}{1-\dfrac{4}{x}+\dfrac{3}{x^2}}\)
\(\dfrac{\dfrac{15}{x^2}-\dfrac{2}{x}-1}{\dfrac{4}{x^2}-\dfrac{5}{x}+4}\)
\(\dfrac{1-\dfrac{12}{3x+10}}{x-\dfrac{8}{3x+10}}\)
\(\dfrac{x-5-\dfrac{18}{x+2}}{x+7+\dfrac{6}{x+2}}\)
\(\dfrac{x-4+\dfrac{9}{2x+3}}{x+3-\dfrac{5}{2x+3}}\)
\(\dfrac{\dfrac{2}{b}-\dfrac{5}{b+3}}{\dfrac{3}{b}+\dfrac{3}{b+3}}\)
\(\dfrac{\dfrac{2}{b^2}-\dfrac{5}{ab}-\dfrac{3}{a^2}}{\dfrac{2}{b^2}+\dfrac{7}{ab}+\dfrac{3}{a^2}}\)
\(\dfrac{\dfrac{y}{y+2}-\dfrac{y}{y-2}}{\dfrac{y}{y+2}+\dfrac{y}{y-2}}\)
\(\dfrac{\dfrac{1}{a}-\dfrac{3}{a-2}}{\dfrac{2}{a}+\dfrac{5}{a-2}}\)
\(\dfrac{\dfrac{1}{y^2}-\dfrac{1}{xy}-\dfrac{2}{x^2}}{\dfrac{1}{y^2}-\dfrac{3}{xy}+\dfrac{2}{x^2}}\)
\(\dfrac{\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}}{\dfrac{x-1}{x+1}+\dfrac{x+1}{x-1}}\)
\(\dfrac{\dfrac{x+1}{x-1}-\dfrac{1-x}{1+x}}{\dfrac{1}{(x+1)^2}+\dfrac{1}{(x-1)^2}}\)