8.5 Adding and subtracting rational expressions

$$\begin{array}{lllll}\frac{1}{6}+\frac{3}{6}\hfill & \hfill & \hfill & \hfill & \text{The\hspace{0.17em}denominators\hspace{0.17em}are\hspace{0.17em}the\hspace{0.17em}same}\text{.\hspace{0.17em}Add\hspace{0.17em}the\hspace{0.17em}numerators}\text{.}\hfill \\ \frac{1}{6}+\frac{3}{6}\hfill & =\hfill & \frac{1+3}{6}=\frac{4}{6}\hfill & \hfill & \text{Reduce}\text{.}\hfill \\ \frac{1}{6}+\frac{3}{6}\hfill & =\hfill & \frac{2}{3}\hfill & \hfill & \hfill \end{array}$$

$$\begin{array}{ccccc}\frac{5}{x}+\frac{8}{x}\hfill & \hfill & \hfill & \hfill & \text{The\hspace{0.17em}denominators\hspace{0.17em}are\hspace{0.17em}the\hspace{0.17em}same}\text{.\hspace{0.17em}Add\hspace{0.17em}the\hspace{0.17em}numerators}\text{.}\hfill \\ \frac{5}{x}+\frac{8}{x}\hfill & =\hfill & \frac{5+8}{x}=\frac{13}{x}\hfill & \hfill & \hfill \end{array}$$

$$\begin{array}{ccc}\frac{2ab}{y^{2}w}-\frac{5b}{y^{2}w}\hfill & \hfill & \text{The\hspace{0.17em}denominators\hspace{0.17em}are\hspace{0.17em}the\hspace{0.17em}same}\text{.\hspace{0.17em}Subtract\hspace{0.17em}the\hspace{0.17em}numerators}\text{.}\hfill \\ \frac{2ab}{y^{2}w}-\frac{5b}{y^{2}w}=\frac{2ab-5b}{y^{2}w}\hfill & \hfill & \hfill \end{array}$$

$$\begin{array}{llll}\frac{3x^2+x+2}{x-7}+\frac{x^2-4x+1}{x-7}\hfill & \hfill & \text{The\hspace{0.17em}denominators\hspace{0.17em}are\hspace{0.17em}the\hspace{0.17em}same}\text{.\hspace{0.17em}Add\hspace{0.17em}the\hspace{0.17em}numerators}\text{.}\hfill & \hfill \\ \frac{3x^2+x+2}{x-7}+\frac{x^2-4x+1}{x-7}\hfill & =\hfill & \frac{3x^2+x+2+x^2-4x+1}{x-7}\hfill & \hfill \\ \hfill & =\hfill & \frac{4x^2-3x+3}{x-7}\hfill & \hfill \end{array}$$

$$\begin{array}{l}\begin{array}{llll}\frac{5y+3}{2y-5}-\frac{2y+4}{2y-5}\hfill & \hfill & \hfill & \begin{array}{l}\text{The\hspace{0.17em}denominators\hspace{0.17em}are\hspace{0.17em}the\hspace{0.17em}same}\text{.\hspace{0.17em}Subtract\hspace{0.17em}the\hspace{0.17em}numerators}\text{.\hspace{0.17em}}\\ \text{But\hspace{0.17em}}becareful\text{\hspace{0.17em}to\hspace{0.17em}subtract\hspace{0.17em}the\hspace{0.17em}}entire\text{\hspace{0.17em}numerator}\text{.\hspace{0.17em}Use\hspace{0.17em}parentheses!}\end{array}\hfill \\ \frac{5y+3}{2y-5}-\frac{2y+4}{2y-5}\hfill & =\hfill & \frac{5y+3-\left(2y+4\right)}{2y-5}\hfill & \hfill \\ \hfill & =\hfill & \frac{5y+3-2y-4}{2y-5}\hfill & \hfill \\ \hfill & =\hfill & \frac{3y-1}{2y-5}\hfill & \hfill \end{array}\\ \begin{array}{ccc}Note:& & \frac{5y+3}{2y-5}-\frac{2y+4}{\underset{\text{Observe\hspace{0.17em}this\hspace{0.17em}part}}{\underbrace{2y-5}}}.\end{array}\\ \text{The\hspace{0.17em}term}-\frac{2y+4}{2y-5}\text{\hspace{0.17em}could\hspace{0.17em}be\hspace{0.17em}written\hspace{0.17em}as}\\ +\frac{-\left(2y+4\right)}{2y-5}=\frac{-2y-4}{2y-5}\end{array}$$

A common mistake is to write

$$-\frac{2y+4}{2y-5}\text{\hspace{0.17em}as\hspace{0.17em}}\frac{-2y+4}{2y-5}$$
This is not correct, as the negative sign is not being applied to the entire numerator.