4.2 Rational expressions: building rational expressions and the  (Page 3/4)

$\frac{3}{4},\frac{1}{6},\frac{5}{12}.$

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.

$\frac{5a}{6a^{2}b},\frac{3a}{8a{b}^3}.$

The LCD is $24a^2{b}^3$ since $24a^2{b}^3$ is the smallest quantity that $6a^{2}b$ and $8a{b}^3$ will divide into without a remainder.

$\frac{2y}{y-6},\frac{4y^2}{{\left(y-6\right)}^3},\frac{y}{y-1}.$

The LCD is ${\left(y-6\right)}^3\left(y-1\right)$ since ${\left(y-6\right)}^3·\left(y-1\right)$ is the smallest quantity that $y-6,{\left(y-6\right)}^3$ and $y-1$ will divide into without a remainder.

• $\frac{1}{x},\frac{3}{x^3},\frac{2}{4y}$

1. The denominators are already factored.
2. Note that $x$ appears as $x$ and $x^3$ . Use only the $x$ with the higher exponent, $x^3$ . The term $4y$ appears, so we must also use $4y$ .
3. The LCD is $4x^{3}y$ .

• $\frac{5}{{\left(x-1\right)}^2},\frac{2x}{\left(x-1\right)\left(x-4\right)},\frac{-5x}{x^2-3x+2}$

1. Only the third denominator needs to be factored.
   
   $x^2-3x+2=\left(x-2\right)\left(x-1\right)$
   
   Now the three denominators are ${\left(x-1\right)}^2,\left(x-1\right)\left(x-4\right)$ , and $\left(x-2\right)\left(x-1\right).$
2. Note that $x-1$ appears as ${\left(x-1\right)}^2,x-1$ , and $x-1.$ Use only the $x-1$ with the highest exponent, ${\left(x-1\right)}^2$ . Also appearing are $x-4$ and $x-2.$
3. The LCD is ${\left(x-1\right)}^2\left(x-4\right)\left(x-2\right).$

• $\frac{-1}{6a^4},\frac{3}{4a^{3}b},\frac{1}{3a^3\left(b+5\right)}$

1. The denominators are already factored.
2. We can see that the LCD of the numbers 6, 4, and 3 is 12. We also need $a^4$ , $b$ , and $b+5$ .
3. The LCD is $12a^{4}b\left(b+5\right).$

• $\frac{9}{x},\frac{4}{8y}$

1. The denominators are already factored.
2. $x,8y.$
3. The LCD is $8xy$ .

$$\begin{array}{lll}\frac{3}{x^2},\frac{4}{x}.\hfill & \hfill & \begin{array}{l}\text{The\hspace{0.17em}LCD,\hspace{0.17em}by\hspace{0.17em}inspection,\hspace{0.17em}is\hspace{0.17em}}{x}^2\text{.\hspace{0.17em}Rewrite\hspace{0.17em}each\hspace{0.17em}expression\hspace{0.17em}}\\ \text{with\hspace{0.17em}}{x}^2\text{\hspace{0.17em}as\hspace{0.17em}the\hspace{0.17em}new\hspace{0.17em}denominator}\text{.\hspace{0.17em}}\end{array}\hfill \\ \frac{}{x^2},\frac{}{x^2}\hfill & \hfill & \begin{array}{l}\text{Determine\hspace{0.17em}the\hspace{0.17em}numerators}\text{.\hspace{0.17em}In\hspace{0.17em}}\frac{3}{x^2}\text{,\hspace{0.17em}the\hspace{0.17em}denominator\hspace{0.17em}was\hspace{0.17em}not\hspace{0.17em}}\\ \text{changed\hspace{0.17em}so\hspace{0.17em}we\hspace{0.17em}need\hspace{0.17em}not\hspace{0.17em}change\hspace{0.17em}the\hspace{0.17em}numerator}\text{.\hspace{0.17em}}\end{array}\hfill \\ \frac{3}{x^2},\frac{}{x^2}\hfill & \hfill & \begin{array}{l}\text{In\hspace{0.17em}the\hspace{0.17em}second\hspace{0.17em}fraction,\hspace{0.17em}the\hspace{0.17em}original\hspace{0.17em}denominator\hspace{0.17em}was\hspace{0.17em}}x\text{.\hspace{0.17em}}\\ \text{We\hspace{0.17em}can\hspace{0.17em}see\hspace{0.17em}that\hspace{0.17em}}x\text{\hspace{0.17em}must\hspace{0.17em}be\hspace{0.17em}multiplied\hspace{0.17em}by\hspace{0.17em}}x\text{\hspace{0.17em}to\hspace{0.17em}build\hspace{0.17em}it\hspace{0.17em}to\hspace{0.17em}}{x}^2\text{.\hspace{0.17em}}\\ \text{So\hspace{0.17em}we\hspace{0.17em}must\hspace{0.17em}also\hspace{0.17em}multiply\hspace{0.17em}the\hspace{0.17em}numerator\hspace{0.17em}4\hspace{0.17em}by\hspace{0.17em}}x\text{.\hspace{0.17em}Thus,\hspace{0.17em}4}·x=4x\text{.\hspace{0.17em}}\end{array}\hfill \\ \frac{3}{x^2},\frac{4x}{x^2}\hfill & \hfill & \hfill \end{array}$$