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To add or subtract two or more rational expressions they must have the same denominator .
Building rational expressions allows us to transform fractions into fractions with the same denominators (which we can then add or subtract). The most convenient new denominator is the least common denominator (LCD) of the given fractions.
In arithmetic, the least common denominator is the smallest (least) quantity that each of the given denominators will divide into without a remainder. For algebraic expressions, the LCD is the polynomial of least degree divisible by each denominator. Some examples are shown below.
$\frac{3}{4},\frac{1}{6},\frac{5}{12}.$
The LCD is 12 since 12 is the smallest number that 4, 6, and 12 will divide into without a remainder.
$\frac{1}{3},\frac{5}{6},\frac{5}{8},\frac{7}{12}.$
The LCD is 24 since 24 is the smallest number that 3, 6, 8, and 12 will divide into without a remainder.
$\frac{2}{x},\frac{3}{{x}^{2}}.$
The LCD is
${x}^{2}$ since
${x}^{2}$ is the smallest quantity that
$x$ and
${x}^{2}$ will divide into without a remainder.
$\frac{5a}{6{a}^{2}b},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{3a}{8a{b}^{3}}.$
The LCD is
$24{a}^{2}{b}^{3}$ since
$24{a}^{2}{b}^{3}$ is the smallest quantity that
$6{a}^{2}b$ and
$8a{b}^{3}$ will divide into without a remainder.
$\frac{2y}{y-6},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{4{y}^{2}}{{\left(y-6\right)}^{3}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{y}{y-1}.$
The LCD is
${\left(y-6\right)}^{3}\left(y-1\right)$ since
${\left(y-6\right)}^{3}\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}\left(y-1\right)$ is the smallest quantity that
$y-6,{\left(y-6\right)}^{3}$ and
$y-1$ will divide into without a remainder.
We’ll now propose and demonstrate a method for obtaining the LCD.
Method for Obtaining the LCD
Find the LCD.
Find the LCD.
$\frac{3}{{x}^{2}},\text{\hspace{0.17em}}\frac{4}{{x}^{5}},\frac{-6}{xy}$
${x}^{5}y$
$\frac{x+1}{x-4},\frac{x-7}{{\left(x-4\right)}^{2}},\frac{-6}{x+1}$
${\left(x-4\right)}^{2}\left(x+1\right)$
$\frac{2}{m-6},\frac{-5m}{{\left(m+1\right)}^{2}\left(m-2\right)},\frac{12{m}^{2}}{{\left(m-2\right)}^{3}\left(m-6\right)}$
$\left(m-6\right){\left(m+1\right)}^{2}{\left(m-2\right)}^{3}$
$\frac{1}{{x}^{2}-1},\text{\hspace{0.17em}}\frac{2}{{x}^{2}-2x-3},\text{\hspace{0.17em}}\frac{-3x}{{x}^{2}-6x+9}$
$\left(x+1\right)\left(x-1\right){\left(x-3\right)}^{2}$
$\frac{3}{4{y}^{2}-8y},\frac{8}{{y}^{2}-4y+4},\frac{10y-1}{3{y}^{3}-6{y}^{2}}$
$12{y}^{2}{\left(y-2\right)}^{2}$
Change the given rational expressions into rational expressions having the same denominator.
$$\begin{array}{lll}\frac{3}{{x}^{2}},\frac{4}{x}.\hfill & \hfill & \begin{array}{l}\text{The\hspace{0.17em}LCD,\hspace{0.17em}by\hspace{0.17em}inspection,\hspace{0.17em}is\hspace{0.17em}}{x}^{2}\text{.\hspace{0.17em}Rewrite\hspace{0.17em}each\hspace{0.17em}expression\hspace{0.17em}}\\ \text{with\hspace{0.17em}}{x}^{2}\text{\hspace{0.17em}as\hspace{0.17em}the\hspace{0.17em}new\hspace{0.17em}denominator}\text{.\hspace{0.17em}}\end{array}\hfill \\ \frac{}{{x}^{2}},\frac{}{{x}^{2}}\hfill & \hfill & \begin{array}{l}\text{Determine\hspace{0.17em}the\hspace{0.17em}numerators}\text{.\hspace{0.17em}In\hspace{0.17em}}\frac{3}{{x}^{2}}\text{,\hspace{0.17em}the\hspace{0.17em}denominator\hspace{0.17em}was\hspace{0.17em}not\hspace{0.17em}}\\ \text{changed\hspace{0.17em}so\hspace{0.17em}we\hspace{0.17em}need\hspace{0.17em}not\hspace{0.17em}change\hspace{0.17em}the\hspace{0.17em}numerator}\text{.\hspace{0.17em}}\end{array}\hfill \\ \frac{3}{{x}^{2}},\frac{}{{x}^{2}}\hfill & \hfill & \begin{array}{l}\text{In\hspace{0.17em}the\hspace{0.17em}second\hspace{0.17em}fraction,\hspace{0.17em}the\hspace{0.17em}original\hspace{0.17em}denominator\hspace{0.17em}was\hspace{0.17em}}x\text{.\hspace{0.17em}}\\ \text{We\hspace{0.17em}can\hspace{0.17em}see\hspace{0.17em}that\hspace{0.17em}}x\text{\hspace{0.17em}must\hspace{0.17em}be\hspace{0.17em}multiplied\hspace{0.17em}by\hspace{0.17em}}x\text{\hspace{0.17em}to\hspace{0.17em}build\hspace{0.17em}it\hspace{0.17em}to\hspace{0.17em}}{x}^{2}\text{.\hspace{0.17em}}\\ \text{So\hspace{0.17em}we\hspace{0.17em}must\hspace{0.17em}also\hspace{0.17em}multiply\hspace{0.17em}the\hspace{0.17em}numerator\hspace{0.17em}4\hspace{0.17em}by\hspace{0.17em}}x\text{.\hspace{0.17em}Thus,\hspace{0.17em}4}\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}x=4x\text{.\hspace{0.17em}}\end{array}\hfill \\ \frac{3}{{x}^{2}},\frac{4x}{{x}^{2}}\hfill & \hfill & \hfill \end{array}$$
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