8.4 Building rational expressions and the lcd  (Page 2/4)

$\frac{3}{4}=\frac{?}{20}.$

The original denominator 4 was multiplied by 5 to yield 20. What arithmetic process will yield 5 using 4 and 20?

$\frac{9}{10}=\frac{?}{10y}.$

The original denominator 10 was multiplied by $y$ to yield $10y$ .

$\frac{-6xy}{2a^{3}b}=\frac{?}{16a^5{b}^3}.$

The original denominator $2a^{3}b$ was multiplied by $8a^2{b}^2$ to yield $16a^5{b}^3.$

$\frac{5ax}{{\left(a+1\right)}^2}=\frac{?}{4{\left(a+1\right)}^2\left(a-2\right)}.$

The original denominator ${\left(a+1\right)}^2$ was multiplied by $4\left(a-2\right)$ to yield $4{\left(a+1\right)}^2\left(a-2\right)$ .

$$\begin{array}{l}\begin{array}{lll}\frac{8}{3}=\frac{N}{15}.\hfill & \hfill & \begin{array}{l}\text{The\hspace{0.17em}original\hspace{0.17em}denominator\hspace{0.17em}is\hspace{0.17em}3\hspace{0.17em}and\hspace{0.17em}the\hspace{0.17em}new\hspace{0.17em}}\\ \text{denominato\hspace{0.17em}is\hspace{0.17em}15}\text{.\hspace{0.17em}Divide\hspace{0.17em}the\hspace{0.17em}original\hspace{0.17em}}\\ \text{denominator\hspace{0.17em}into\hspace{0.17em}the\hspace{0.17em}new\hspace{0.17em}denominator\hspace{0.17em}and\hspace{0.17em}}\\ \text{multiply\hspace{0.17em}the\hspace{0.17em}numerator\hspace{0.17em}8\hspace{0.17em}by\hspace{0.17em}this\hspace{0.17em}result}\text{.}\end{array}\hfill \\ \hfill & \hfill & 15÷3=5\hfill \\ \hfill & \hfill & \text{Then,}8\cdot 5=40.\text{So,}\hfill \\ \frac{8}{3}=\frac{40}{15}\text{and}N=40.\hfill & \hfill & \hfill \end{array}\\ \text{Check\hspace{0.17em}by\hspace{0.17em}reducing}\frac{40}{15}.\end{array}$$

$$\begin{array}{lll}\frac{2x}{5b^{2}y}=\frac{N}{20b^5{y}^4}.\hfill & \hfill & \begin{array}{l}\text{The\hspace{0.17em}original\hspace{0.17em}denominator\hspace{0.17em}is\hspace{0.17em}}5b^{2}y\text{\hspace{0.17em}and\hspace{0.17em}the\hspace{0.17em}new\hspace{0.17em}}\\ \text{denominator\hspace{0.17em}is\hspace{0.17em}}20b^5{y}^4\text{.\hspace{0.17em}Divide\hspace{0.17em}the\hspace{0.17em}original\hspace{0.17em}}\\ \text{denominator\hspace{0.17em}into\hspace{0.17em}the\hspace{0.17em}new\hspace{0.17em}denominator\hspace{0.17em}and\hspace{0.17em}}\\ \text{multiply\hspace{0.17em}the\hspace{0.17em}numerator\hspace{0.17em}}2x\text{\hspace{0.17em}by\hspace{0.17em}this\hspace{0.17em}result}\text{.}\end{array}\hfill \\ \hfill & \hfill & \frac{20b^5{y}^4}{5b^{2}y}=4b^3{y}^3\hfill \\ \hfill & \hfill & \text{So,}2x·4b^3{y}^3=8b^{3}x{y}^3.\text{Thus,}\hfill \\ \frac{2x}{5b^{2}y}=\frac{8b^{3}x{y}^3}{20b^5{y}^4}\text{and}N=8b^{3}x{y}^3.\hfill & \hfill & \hfill \end{array}$$

$$\begin{array}{l}\begin{array}{lll}\frac{-6a}{a+2}=\frac{N}{\left(a+2\right)\left(a-7\right)}.\hfill & \hfill & \text{The\hspace{0.17em}new\hspace{0.17em}denominator\hspace{0.17em}divided\hspace{0.17em}by\hspace{0.17em}the\hspace{0.17em}original\hspace{0.17em}denominator\hspace{0.17em}is\hspace{0.17em}}\hfill \\ \hfill & \hfill & \frac{\left(a+2\right)\left(a-7\right)}{a+2}=a-7\hfill \\ \hfill & \hfill & \text{Multiply\hspace{0.17em}}-6a\text{by}a-7.\hfill \\ \hfill & \hfill & -6a\left(a-7\right)=-6a^2+42a\hfill \end{array}\\ \frac{-6a}{a+2}=\frac{-6a^2+42a}{\left(a+2\right)\left(a-7\right)}\text{\hspace{0.17em}and\hspace{0.17em}}N=-6a^2+42a.\end{array}$$

$$\begin{array}{l}\begin{array}{lll}\frac{-3\left(a-1\right)}{a-4}=\frac{N}{a^2-16}.\hfill & \hfill & \text{The\hspace{0.17em}new\hspace{0.17em}denominator\hspace{0.17em}divided\hspace{0.17em}by\hspace{0.17em}the\hspace{0.17em}original\hspace{0.17em}denominator\hspace{0.17em}is}\hfill \\ \hfill & \hfill & \frac{a^2-16}{a-4}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{\left(a+4\right)\cancel{\left(a-4\right)}}{\cancel{a-4}}\hfill \\ \hfill & \hfill & =a+4\hfill \\ \hfill & \hfill & \text{Multiply\hspace{0.17em}}-3\left(a-1\right)\text{by}a+4.\hfill \\ \hfill & \hfill & \begin{array}{ccc}-3\left(a-1\right)\left(a+4\right)& =& -3\left(a^2+3a-4\right)\end{array}\hfill \\ \hfill & \hfill & \begin{array}{ccc}& =& -3a^2-9a+12\end{array}\hfill \end{array}\\ \frac{-3\left(a-1\right)}{a-4}=\frac{-3a^2-9a+12}{a^2-16}\text{and}N=-3a^2-9a+12\end{array}$$

$$\begin{array}{lllll}7x\hfill & =\hfill & \frac{N}{x^2{y}^3}.\hfill & \hfill & \text{Write\hspace{0.17em}}7x\text{\hspace{0.17em}as\hspace{0.17em}}\frac{7x}{1}.\hfill \\ \frac{7x}{1}\hfill & =\hfill & \frac{N}{x^2{y}^3}\hfill & \hfill & \begin{array}{l}\text{Now\hspace{0.17em}we\hspace{0.17em}can\hspace{0.17em}see\hspace{0.17em}clearly\hspace{0.17em}that\hspace{0.17em}the\hspace{0.17em}original\hspace{0.17em}denominator\hspace{0.17em}}\\ \text{1\hspace{0.17em}was\hspace{0.17em}multiplied\hspace{0.17em}by\hspace{0.17em}}{x}^2{y}^3.\text{\hspace{0.17em}We\hspace{0.17em}need\hspace{0.17em}to\hspace{0.17em}multiply\hspace{0.17em}the\hspace{0.17em}}\\ \text{numerator\hspace{0.17em}}7x\text{\hspace{0.17em}by\hspace{0.17em}}{x}^2{y}^3.\end{array}\hfill \\ 7x\hfill & =\hfill & \frac{7x·x^2{y}^3}{x^2{y}^3}\hfill & \hfill & \hfill \\ 7x\hfill & =\hfill & \frac{7x^3{y}^3}{x^2{y}^3}\text{\hspace{0.17em}and\hspace{0.17em}}N=7x^3{y}^3.\hfill & \hfill & \hfill \end{array}$$

$$\begin{array}{l}\begin{array}{lll}\frac{5x}{x+3}=\frac{5x^2-20x}{N}.\hfill & \hfill & \begin{array}{l}\text{The\hspace{0.17em}same\hspace{0.17em}process\hspace{0.17em}works\hspace{0.17em}in\hspace{0.17em}this\hspace{0.17em}case}.\text{\hspace{0.17em}Divide\hspace{0.17em}the\hspace{0.17em}original\hspace{0.17em}}\\ \text{numerator\hspace{0.17em}}5x\text{\hspace{0.17em}into\hspace{0.17em}the\hspace{0.17em}new\hspace{0.17em}numerator\hspace{0.17em}}5x^2-20x.\end{array}\hfill \\ \hfill & \hfill & \frac{5x^2-20x}{5x}=\frac{\cancel{5x}\left(x-4\right)}{\cancel{5x}}\hfill \\ \hfill & \hfill & =x-4\hfill \\ \hfill & \hfill & \text{Multiply\hspace{0.17em}the\hspace{0.17em}denominator\hspace{0.17em}by\hspace{0.17em}}x-4.\hfill \\ \hfill & \hfill & \left(x+3\right)\left(x-4\right)\hfill \end{array}\\ \frac{5x}{x+3}=\frac{5x^2-20}{\left(x+3\right)\left(x-4\right)}\text{and}N=5x^2-20.\end{array}$$

$$\begin{array}{l}\begin{array}{lll}\frac{4x}{3-x}=\frac{N}{x-3}.\hfill & \hfill & \begin{array}{l}\text{The\hspace{0.17em}two\hspace{0.17em}denominators\hspace{0.17em}have\hspace{0.17em}nearly\hspace{0.17em}the\hspace{0.17em}same\hspace{0.17em}terms};\text{\hspace{0.17em}each\hspace{0.17em}has\hspace{0.17em}}\\ \text{the\hspace{0.17em}opposite\hspace{0.17em}sign}.\text{\hspace{0.17em}Factor\hspace{0.17em}}-1\text{\hspace{0.17em}from\hspace{0.17em}the\hspace{0.17em}original\hspace{0.17em}denominator}.\end{array}\hfill \\ \hfill & \hfill & 3-x=-1\left(-3+x\right)\hfill \\ \hfill & \hfill & =-\left(x-3\right)\hfill \end{array}\\ \frac{4x}{3-x}=\frac{4x}{-\left(x-3\right)}=\frac{-4x}{x-3}\text{\hspace{0.17em}and\hspace{0.17em}}N=-4x.\end{array}$$

It is important to note that we factored $-1$ from the original denominator. We did not multiply it by $-1$ . Had we multiplied only the denominator by $-1$ we would have had to multiply the numerator by $-1$ also.