8.5 Adding and subtracting rational expressions  (Page 2/2)

$$\begin{array}{l}\frac{3x^2+4x+5}{\left(x+6\right)\left(x-2\right)}+\frac{2x^2+x+6}{x^2+4x-12}-\frac{x^2-4x-6}{x^2+4x-12}\\ \\ \text{Factor\hspace{0.17em}the\hspace{0.17em}denominators\hspace{0.17em}to\hspace{0.17em}determine\hspace{0.17em}if\hspace{0.17em}they're\hspace{0.17em}the\hspace{0.17em}same}\text{.}\\ \\ \frac{3x^2+4x+5}{\left(x+6\right)\left(x-2\right)}+\frac{2x^2+x+6}{\left(x+6\right)\left(x-2\right)}-\frac{x^2-4x-6}{\left(x+6\right)\left(x-2\right)}\\ \\ \text{The\hspace{0.17em}denominators\hspace{0.17em}are\hspace{0.17em}the\hspace{0.17em}same}\text{.\hspace{0.17em}Combine\hspace{0.17em}the\hspace{0.17em}numerators\hspace{0.17em}being\hspace{0.17em}careful\hspace{0.17em}to\hspace{0.17em}note\hspace{0.17em}the\hspace{0.17em}negative\hspace{0.17em}sign}\text{.}\\ \\ \frac{3x^2+4x+5+2x^2+x+6-\left(x^2-4x+6\right)}{\left(x+6\right)\left(x-2\right)}\\ \\ \frac{3x^2+4x+5+2x^2+x+6-x^2+4x+6}{\left(x+6\right)\left(x-2\right)}\\ \\ \frac{4x^2+9x+17}{\left(x+6\right)\left(x-2\right)}\end{array}$$

$$\begin{array}{lll}\frac{4a}{3y}+\frac{2a}{9y^2}.\hfill & \hfill & \text{The\hspace{0.17em}denominators\hspace{0.17em}are\hspace{0.17em}}not\text{\hspace{0.17em}the\hspace{0.17em}same}\text{.}\text{Find}\text{the}\text{LCD}\text{.}\text{By}\text{inspection,}\text{the}\text{LCD}\text{is}9y^2.\hfill \\ \frac{}{9y^2}+\frac{2a}{9y^2}\hfill & \hfill & \begin{array}{l}\text{The}\text{denominator}\text{of}\text{the}\text{first}\text{rational}\text{expression}\text{has}\text{been}\text{multiplied}\text{by}3y,\hfill \\ \text{so}\text{the}\text{numerator}\text{must}\text{be}\text{multiplied}\text{by}3y.\hfill \\ 4a·3y=12ay\hfill \end{array}\hfill \\ \frac{12ay}{9y^2}+\frac{2a}{9y^2}\hfill & \hfill & \text{The}\text{denominators}\text{are}\text{now}\text{the}\text{same}\text{.\hspace{0.17em}Add}\text{the}\text{numerators}\text{.}\hfill \\ \frac{12ay+2a}{9y^2}\hfill & \hfill & \hfill \end{array}$$

$$\begin{array}{llll}\frac{3b}{b+2}+\frac{5b}{b-3}.\hfill & \hfill & \begin{array}{l}\text{The}\text{denominators}\text{are}not\text{the}\text{same}\text{.\hspace{0.17em}The}\text{LCD}\text{is}(b+2)(b-3).\hfill \\ \hfill \end{array}\hfill & \hfill \\ \frac{}{(b+2)(b-3)}+\frac{}{(b+2)(b-3)}\hfill & \hfill & \begin{array}{l}\text{The}\text{denominator}\text{of}\text{the}\text{first}\text{rational}\text{expression}\text{has}\text{been}\text{multiplied}\text{by}b-3,\hfill \\ \text{so}\text{the}\text{numerator}\text{must}\text{be}\text{multiplied}\text{by}b-3.3b(b-3)\hfill \end{array}\hfill & \hfill \\ \frac{3b(b-3)}{(b+2)(b-3)}+\frac{}{(b+2)(b-3)}\hfill & \hfill & \begin{array}{l}\text{The}\text{denominator}\text{of}\text{the}\text{second}\text{rational}\text{expression}\text{has}\text{been}\text{multiplied}\text{by}b+2,\hfill \\ \text{so}\text{the}\text{numerator}\text{must}\text{be}\text{multiplied}\text{by}b+2.5b(b+2)\hfill \end{array}\hfill & \hfill \\ \frac{3b(b-3)}{(b+2)(b-3)}+\frac{5b(b+2)}{(b+2)(b-3)}\hfill & \hfill & \begin{array}{l}\text{The}\text{denominators}\text{are}\text{now}\text{the}\text{same}\text{.\hspace{0.17em}Add}\text{the}\text{numerators}\text{.}\hfill \\ \hfill \end{array}\hfill & \hfill \\ \frac{3b(b-3)+5b(b+2)}{(b-3)(b+2)}\hfill & =\hfill & \frac{3b^2-9b+5b^2+10b}{(b-3)(b+2)}\hfill & \hfill \\ \hfill & =\hfill & \frac{8b^2+b}{(b-3)(b-2)}\hfill & \hfill \end{array}$$

$$\begin{array}{llll}\frac{x+3}{x-1}+\frac{x-2}{4x+4}.\hfill & \hfill & \hfill & \begin{array}{l}\text{The\hspace{0.17em}denominators\hspace{0.17em}are\hspace{0.17em}}not\text{\hspace{0.17em}the\hspace{0.17em}same}\text{.}\hfill \\ \text{Find\hspace{0.17em}the\hspace{0.17em}LCD}\text{.}\hfill \end{array}\hfill \\ \frac{x+3}{x-1}+\frac{x-2}{4\left(x+1\right)}\hfill & \hfill & \hfill & \text{The\hspace{0.17em}LCD\hspace{0.17em}is\hspace{0.17em}}\left(x+1\right)\left(x-1\right)\hfill \\ \frac{}{4\left(x+1\right)\left(x-1\right)}+\frac{}{4\left(x+1\right)\left(x-1\right)}\hfill & \hfill & \hfill & \begin{array}{l}\text{The\hspace{0.17em}denominator\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}first\hspace{0.17em}rational\hspace{0.17em}expression\hspace{0.17em}has\hspace{0.17em}been\hspace{0.17em}multiplied\hspace{0.17em}by\hspace{0.17em}}4\left(x+1\right)\text{\hspace{0.17em}so\hspace{0.17em}}\hfill \\ \text{the\hspace{0.17em}numerator\hspace{0.17em}must\hspace{0.17em}be\hspace{0.17em}multiplied\hspace{0.17em}by\hspace{0.17em}}4\left(x+1\right).4\left(x+3\right)\left(x+1\right)\hfill \end{array}\hfill \\ \frac{4\left(x+3\right)\left(x+1\right)}{4\left(x+1\right)\left(x-1\right)}+\frac{}{4\left(x+1\right)\left(x-1\right)}\hfill & \hfill & \hfill & \begin{array}{l}\text{The\hspace{0.17em}denominator\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}second\hspace{0.17em}rational\hspace{0.17em}expression\hspace{0.17em}has\hspace{0.17em}been\hspace{0.17em}multiplied\hspace{0.17em}by\hspace{0.17em}}x-1\hfill \\ \text{so\hspace{0.17em}the\hspace{0.17em}numerator\hspace{0.17em}must\hspace{0.17em}be\hspace{0.17em}multiplied\hspace{0.17em}by\hspace{0.17em}}x-1.\left(x-1\right)\left(x-2\right)\hfill \end{array}\hfill \\ \frac{4\left(x+3\right)\left(x+1\right)}{4\left(x+1\right)\left(x-1\right)}+\frac{\left(x-1\right)\left(x-2\right)}{4\left(x+1\right)\left(x-1\right)}\hfill & \hfill & \hfill & \begin{array}{l}\text{The\hspace{0.17em}denominators\hspace{0.17em}are\hspace{0.17em}now\hspace{0.17em}the\hspace{0.17em}same.}\hfill \\ \text{Add\hspace{0.17em}the\hspace{0.17em}numerators.}\hfill \end{array}\hfill \\ \frac{4\left(x+3\right)\left(x+1\right)+\left(x-1\right)\left(x-2\right)}{4\left(x+1\right)\left(x-1\right)}\hfill & \hfill & \hfill & \hfill \\ \frac{4\left(x^2+4x+3\right)+x^2-3x+2}{4\left(x+1\right)\left(x-1\right)}\hfill & \hfill & \hfill & \hfill \\ \frac{4x^2+16x+12+x^2-3x+2}{4\left(x+1\right)\left(x-1\right)}\hfill & =\hfill & \frac{5x^2+13x+14}{4\left(x+1\right)\left(x-1\right)}\hfill & \hfill \end{array}$$

$$\begin{array}{lll}\frac{x+5}{x^2-7x+12}+\frac{3x-1}{x^2-2x-3}\hfill & \hfill & \text{Determine}\text{the}\text{LCD}\text{.}\hfill \\ \frac{x+5}{(x-4)(x-3)}+\frac{3x-1}{(x-3)(x+1)}\hfill & \hfill & \text{The}\text{LCD}\text{is}(x-4)(x-3)(x+1).\hfill \\ \frac{}{(x-4)(x-3)(x+1)}+\frac{}{(x-4)(x-3)(x+1)}\hfill & \hfill & \text{The}\text{first}\text{numerator}\text{must}\text{be}\text{multiplied}\text{by}x+1\text{and}\text{the}\text{second}\text{by}x-4.\hfill \\ \frac{(x+5)(x+1)}{(x-4)(x-3)(x+1)}+\frac{(3x-1)(x-4)}{(x-4)(x-3)(x+1)}\hfill & \hfill & \text{The}\text{denominators}\text{are}\text{now}\text{the}\text{same}\text{.}\text{Add}\text{the}\text{numerators}\text{.}\hfill \\ \frac{(x+5)(x+1)+(3x-1)(x-4)}{(x-4)(x-3)(x+1)}\hfill & \hfill & \hfill \\ \frac{x^2+6x+5+3x^2-13x+4}{(x-4)(x-3)(x+1)}\hfill & \hfill & \hfill \\ \frac{4x^2-7x+9}{(x-4)(x-3)(x+1)}\hfill & \hfill & \hfill \end{array}$$

$$\begin{array}{lll}\frac{a+4}{a^2+5a+6}-\frac{a-4}{a^2-5a-24}\hfill & \hfill & \text{Determine\hspace{0.17em}the\hspace{0.17em}LCD.}\hfill \\ \frac{a+4}{\left(a+3\right)\left(a+2\right)}-\frac{a-4}{\left(a+3\right)\left(a-8\right)}\hfill & \hfill & \text{The\hspace{0.17em}LCD\hspace{0.17em}is\hspace{0.17em}}\left(a+3\right)\left(a+2\right)\left(a-8\right).\hfill \\ \frac{}{\left(a+3\right)\left(a+2\right)\left(a-8\right)}-\frac{}{\left(a+3\right)\left(a+2\right)\left(a-8\right)}\hfill & \hfill & \text{The\hspace{0.17em}first\hspace{0.17em}numerator\hspace{0.17em}must\hspace{0.17em}be\hspace{0.17em}multiplied\hspace{0.17em}by\hspace{0.17em}}a-8\text{\hspace{0.17em}and\hspace{0.17em}the\hspace{0.17em}second\hspace{0.17em}by\hspace{0.17em}}a+2.\hfill \\ \frac{\left(a+4\right)\left(a-8\right)}{\left(a+3\right)\left(a+2\right)\left(a-8\right)}-\frac{\left(a-4\right)\left(a+2\right)}{\left(a+3\right)\left(a+2\right)\left(a-8\right)}\hfill & \hfill & \text{The\hspace{0.17em}denominators\hspace{0.17em}are\hspace{0.17em}now\hspace{0.17em}the\hspace{0.17em}same}\text{.\hspace{0.17em}Subtract\hspace{0.17em}the\hspace{0.17em}numerators}\text{.\hspace{0.17em}}\hfill \\ \frac{\left(a+4\right)\left(a-8\right)-\left(a-4\right)\left(a+2\right)}{\left(a+3\right)\left(a+2\right)\left(a-8\right)}\hfill & \hfill & \hfill \\ \frac{a^2-4a-32-\left(a^2-2a-8\right)}{\left(a+3\right)\left(a+2\right)\left(a-8\right)}\hfill & \hfill & \hfill \\ \frac{a^2-4a-32-a^2+2a+8}{\left(a+3\right)\left(a+2\right)\left(a-8\right)}\hfill & \hfill & \hfill \\ \frac{-2a-24}{\left(a+3\right)\left(a+2\right)\left(a-8\right)}\hfill & \hfill & \text{Factor\hspace{0.17em}}-2\text{\hspace{0.17em}from\hspace{0.17em}the\hspace{0.17em}numerator}\text{.\hspace{0.17em}}\hfill \\ \frac{-2\left(a+12\right)}{\left(a+3\right)\left(a+2\right)\left(a-8\right)}\hfill & \hfill & \hfill \end{array}$$

$$\begin{array}{lllll}\frac{3x}{7-x}+\frac{5x}{x-7}.\hfill & \hfill & \hfill & \hfill & \begin{array}{l}\text{The\hspace{0.17em}denominators\hspace{0.17em}are}nearly\text{the\hspace{0.17em}same.}\text{\hspace{0.17em}They\hspace{0.17em}differ\hspace{0.17em}only\hspace{0.17em}in\hspace{0.17em}sign.}\hfill \\ \text{Our\hspace{0.17em}technique\hspace{0.17em}is\hspace{0.17em}to\hspace{0.17em}factor\hspace{0.17em}}-1\text{\hspace{0.17em}from\hspace{0.17em}one\hspace{0.17em}of\hspace{0.17em}them.}\hfill \end{array}\hfill \\ \frac{3x}{7-x}=\frac{3x}{-\left(x-7\right)}\hfill & =\hfill & \frac{-3x}{x-7}\hfill & \hfill & \text{Factor\hspace{0.17em}}-1\text{\hspace{0.17em}from\hspace{0.17em}the\hspace{0.17em}first\hspace{0.17em}term}\text{.\hspace{0.17em}}\hfill \\ \frac{3x}{7-x}+\frac{5x}{x-7}\hfill & =\hfill & \frac{-3x}{x-7}+\frac{5x}{x-7}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{-3x+5x}{x-7}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{2x}{x-7}\hfill & \hfill & \hfill \end{array}$$

$$\begin{array}{lllll}3+\frac{7}{x-1}.\hfill & \hfill & \hfill & \hfill & \text{Rewrite\hspace{0.17em}the\hspace{0.17em}expression.}\hfill \\ \frac{3}{1}+\frac{7}{x-1}\hfill & \hfill & \hfill & \hfill & \text{The\hspace{0.17em}LCD\hspace{0.17em}is\hspace{0.17em}}x-1.\hfill \\ \frac{3\left(x-1\right)}{x-1}+\frac{7}{x-1}\hfill & =\hfill & \frac{3x-3}{x-1}+\frac{7}{x-1}\hfill & =\hfill & \frac{3x-3+7}{x-1}\hfill \\ \hfill & \hfill & \hfill & =\hfill & \frac{3x+4}{x-1}\hfill \end{array}$$

$$\begin{array}{llll}3y+4-\frac{y^2-y+3}{y-6}.\hfill & \hfill & \hfill & \text{Rewrite\hspace{0.17em}the\hspace{0.17em}expression.}\hfill \\ \frac{3y+4}{1}-\frac{y^2-y+3}{y-6}\hfill & \hfill & \hfill & \text{The\hspace{0.17em}LCD\hspace{0.17em}is\hspace{0.17em}}y-6.\hfill \\ \frac{\left(3y+4\right)\left(y-6\right)}{y-6}-\frac{y^2-y+3}{y-6}\hfill & =\hfill & \frac{\left(3y+4\right)\left(y-6\right)-\left(y^2-y+3\right)}{y-6}\hfill & \hfill \\ \hfill & =\hfill & \frac{3y^2-14y-24-y^2+y-3}{y-6}\hfill & \hfill \\ \hfill & =\hfill & \frac{2y^2-13y-27}{y-6}\hfill & \hfill \end{array}$$