8.3: Obtain the Lowest Common Denominator
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)As with fractions in arithmetic, the least common denominator or LCD is the lowest common multiple (LCM) of the denominators. Since rational expressions are fractions with polynomials, we use the LCD to add and subtract rational expression with different denominators. In this section, we obtain LCDs of rational expressions. First, let’s take a look at the method in finding the LCM in arithmetic.
Obtain the LCM in Arithmetic
Find \(\text{LCM}(3,6,15)\).
Solution
Find the prime factorization of each number in your set.
\[\begin{aligned}3&=3 \\ 6&=2\cdot 3 \\ 15&=3\cdot 5\end{aligned}\]
Next, take one of each factor and, for repeated factors, take the highest exponent. Hence, the \(\text{LCM}(3, 6, 15) = 2\cdot 3\cdot 5 = 30\). Notice all factors of each number is in the LCM:
\[\begin{array}{c}\underset{6}{\underbrace{2\cdot 3}}\cdot 5 \\ 2\cdot\underset{15}{\underbrace{3\cdot 5}}\end{array}\nonumber\]
Obtain the LCM with Monomials
We use the same method as in Example 8.3.1 , but now with variables.
Find the \(LCM(4x^2y^5, 6x^4y^3z^6)\).
Solution
Find the prime factorization of each expression in your set.
\[\begin{aligned}4x^2y^5&=2^2x^2y^5 \\ 6x^4y^3z^6&=2\cdot 3\cdot x^4y^3z^6\end{aligned}\]
Next, take one of each factor and, for repeated factors, take the highest exponent. Hence, the \(\text{LCM}(4x^2y^5 , 6x^4y^3z^6 ) = 2^2\cdot 3\cdot x^4 \cdot y^5\cdot z^6 = 12x^4y^3 z^6\). Notice we take the highest exponent of repeated factors so that all factors are contained in the LCM.
Obtain the LCM with Polynomials
We use the same method, but now we factor using factoring techniques to obtain the LCM between polynomials. Recall, all factors are contained in the LCM.
Find the \(\text{LCM}(x^2 + 2x − 3,\: x^2 − x − 12)\).
Solution
Find the prime factorization of each expression in your set.
\[\begin{aligned}x^2+2x-3&=(x+3)(x-1) \\ x^2-x-12&=(x-4)(x+3)\end{aligned}\]
Next, take one of each factor and, for repeated factors, take the highest exponent. Hence, the \(\text{LCM}(x^2 + 2x − 3,\: x^2 − x − 12) = (x − 1)(x + 3)(x − 4)\). Notice all factors are contained in the LCM:
\[\begin{array}{c}\underset{x^2+2x-3}{\underbrace{(x-1)(x+3)}}(x-4) \\ (x-1)\underset{x^2-x-12}{\underbrace{(x+3)(x-4)}}\end{array}\nonumber\]
Find the \(\text{LCM}(x^2 − 10x + 25,\: x^2 − 14x + 45)\).
Solution
Find the prime factorization of each expression in your set.
\[\begin{aligned}x^2-10x+25&=(x-5)^2 \\ x^2-14x+45&=(x-5)(x-9)\end{aligned}\]
Next, take one of each factor and, for repeated factors, take the highest exponent. Hence, the \(\text{LCM}(x^2 − 10x + 25,\: x^2 − 14x + 45) = (x − 5)^2 (x − 9)\).
Once we obtain the LCM of polynomial expressions, then this LCM can be used as the LCD in given rational expressions. We can then rewrite each fraction with the LCD. Recall, the LCD is the LCM of all denominators in the expression.
Rewrite Fractions with the Lowest Common Denominator
Find the LCD between \(\dfrac{5a}{4b^3c}\) and \(\dfrac{3c}{6a^2b}\). Rewrite each fraction with the LCD.
Solution
If we need to obtain the LCD, then we can follow a series of steps.
Step 1. Find the LCD, i.e., the LCM between denominators. In this case, we need to find the \(\text{LCM}(4b^3c,\: 6a^2b)\). \[\begin{aligned}4b^3c&=2^2\cdot b^3c \\ 6a^2b&=2\cdot 3\cdot a^2b\end{aligned}\] We can see that the \(\text{LCM}(4b^3c,\: 6a^2b)=2^2\cdot 3\cdot a^2\cdot b^3\cdot c=12a^2b^3c\). This is the LCD.
Step 2. Next, we rewrite each fraction with the LCD. \[\begin{array}{rl}\dfrac{5a}{4b^3c}&\text{Multiply the numerator and denominator by }3a^2 \\ \dfrac{5a}{4b^3c}\cdot\color{blue}{\dfrac{3a^2}{3a^2}}&\color{black}{}\text{Notice we get }12a^2b^3c\text{ in the denominator} \\ \dfrac{15a^3}{12a^2b^3c}&\text{The denominator is the LCD }\checkmark \\ \\ \dfrac{3c}{6a^2b}&\text{Multiply the numerator and denominator by }2b^2c \\ \dfrac{3c}{6a^2b}\cdot\color{blue}{\dfrac{2b^2c}{2b^2c}}&\color{black}{}\text{Notice we get }12a^2b^3c\text{ in the denominator} \\ \dfrac{6b^2c^2}{12a^2b^3c}&\text{The denominator is the LCD }\checkmark\end{array}\nonumber\] Hence, \(\dfrac{5a}{4b^3c}\) and \(\dfrac{3c}{6a^2b}\) can be written in the equivalent form with the \(\text{LCD}=12a^2b^3c\) as \[\dfrac{15a^3}{12a^2b^3c}\quad\text{and}\quad\dfrac{6b^2c^2}{12a^2b^3c},\nonumber\] respectively.
Find the LCD between \(\dfrac{5x}{x^2-5x-6}\) and \(\dfrac{x-2}{x^2+4x+3}\). Rewrite each fraction with the LCD.
Solution
If we need to obtain the LCD, then we can follow a series of steps.
Step 1. Find the LCD, i.e., the LCM between denominators. In this case, we need to find the \(\text{LCM}(x^2 − 5x − 6,\: x^2 + 4x + 3)\). \[\begin{aligned}x^2-5x-6&=(x+1)(x-6) \\ x^2+4x+3&=(x+3)(x+1)\end{aligned}\] We can see that the \(\text{LCM}(x^2 − 5x − 6,\: x^2 + 4x + 3) = (x + 3)(x + 1)(x − 6)\). This is the LCD.
Step 2. Next, we rewrite each fraction with the LCD. \[\begin{array}{rl}\dfrac{5x}{x^2-5x-6}&\text{Factor the denominator} \\ \dfrac{5x}{(x+1)(x-6)}&\text{Multiply the numerator and denominator by }(x+3) \\ \dfrac{5x}{(x+1)(x-6)}\cdot\color{blue}{\dfrac{(x+3)}{(x+3)}}&\color{black}{}\text{Notice we get the LCD in the denominator} \\ \dfrac{5x(x+3)}{(x+1)(x-6)(x+3)}&\text{The denominator is the LCD }\checkmark \\ \\ \dfrac{x-2}{x^2+4x+3}&\text{Factor the denominator} \\ \dfrac{(x-2)}{(x+3)(x+1)}&\text{Multiply the numerator and denominator by }(x-6) \\ \dfrac{(x-2)}{(x+3)(x+1)}\cdot\color{blue}{\dfrac{(x-6)}{(x-6)}}&\color{black}{}\text{Notice we get the LCD in the denominator} \\ \dfrac{(x-2)(x-6)}{(x+3)(x+1)(x-6)}&\text{The denominator is the LCD }\checkmark\end{array}\nonumber\] Hence, \(\dfrac{5x}{x^2-5x-6}\) and \(\dfrac{x-2}{x^2+4x+3}\) can be written in the equivalent from with the \(\text{LCD}=(x+3)(x+1)(x-6)\) as \[\dfrac{5x(x+3)}{(x+1)(x-6)(x+3)}\quad\text{and}\quad\dfrac{(x-2)(x-6)}{(x+3)(x+1)(x-6)},\nonumber\] respectively.
When the Egyptians began working with fractions, they expressed all fractions as a sum of a unit fraction. Rather than \(\dfrac{4}{5}\), they would write the fraction as the sum, \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{20}\). An interesting problem with this system is this is not a unique representation of \(\dfrac{4}{5}\); \(\dfrac{4}{5}\) is also equal to the sum \(\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{10}\).
Obtain the Lowest Common Denominator Homework
Find the equivalent numerator.
\(\dfrac{3}{8}=\dfrac{?}{48}\)
\(\dfrac{a}{x}=\dfrac{?}{xy}\)
\(\dfrac{2}{3a^3b^2c}=\dfrac{?}{9a^5b^2c^4}\)
\(\dfrac{2}{x+4}=\dfrac{?}{x^2-16}\)
\(\dfrac{x-4}{x+2}=\dfrac{?}{x^2+5x+6}\)
\(\dfrac{a}{5}=\dfrac{?}{5a}\)
\(\dfrac{5}{2x^2}=\dfrac{?}{8x^3y}\)
\(\dfrac{4}{3a^5b^2c^4}=\dfrac{?}{9a^5b^2c^4}\)
\(\dfrac{x+1}{x-3}=\dfrac{?}{x^2-6x+9}\)
\(\dfrac{x-6}{x+3}=\dfrac{?}{x^2-2x-15}\)
Find the lowest common multiple.
\(2a^3,\: 6a^4b^2,\: 4a^3b^5\)
\(x^2-3x,\: x-3,\: x\)
\(x+2,\: x-4\)
\(x^2-25,\: x+5\)
\(x^2+3x+2,\: x^2+5x+6\)
\(5x^2y,\: 25x^3y^5z\)
\(4x-8,\: x-2,\: 4\)
\(x,\: x-7,\: x+1\)
\(x^2-9,\: x^2-6x+9\)
\(x^2-7x+10,\: x^2-2x-15,\: x^2+x-6\)
Find the LCD and rewrite each fraction with the LCD.
\(\dfrac{3a}{5b^2},\:\dfrac{2}{10a^3b}\)
\(\dfrac{x+2}{x-3},\:\dfrac{x-3}{x+2}\)
\(\dfrac{x}{x^2-16},\:\dfrac{3x}{x^2-8x+16}\)
\(\dfrac{x+1}{x^2-36},\:\dfrac{2x+3}{x^2+12x+36}\)
\(\dfrac{4x}{x^2-x-6},\:\dfrac{x+2}{x-3}\)
\(\dfrac{3x}{x-4},\:\dfrac{2}{x+2}\)
\(\dfrac{5}{x^2-6x},\:\dfrac{2}{x},\:\dfrac{-3}{x-6}\)
\(\dfrac{5x+1}{x^2-3x-10},\:\dfrac{4}{x-5}\)
\(\dfrac{3x+1}{x^2-x-12},\:\dfrac{2x}{x^2+4x+3}\)
\(\dfrac{3x}{x^2-6x+8},\:\dfrac{x-2}{x^2+x-20},\:\dfrac{5}{x^2+3x-10}\)