1. Stating the problem: We need to find the slope of each line given by the equations (a) through (p). 2. Recall: The slope-intercept form of a line is $y = mx + b$ where $m$ is the slope. 3. For each equation, rewrite it (if needed) into the form $y = mx + b$ and identify $m$. (a) $y = 3x - 4$ \- slope $m = 3$ (b) $y = 4x - 3$ \- slope $m = 4$ (c) $y = -\frac{1}{2}x + 5$ \- slope $m = -\frac{1}{2}$ (d) $y = 8 - 7x = -7x + 8$ \- slope $m = -7$ (e) $y = 5x$ \- slope $m = 5$ (f) $y = 5$ \- This is a horizontal line; slope $m = 0$ (g) $2y = 6x - 10 \Rightarrow y = 3x - 5$ \- slope $m = 3$ (h) $2y = 10x - 6 \Rightarrow y = 5x - 3$ \- slope $m = 5$ (i) $3y = -12x + 6 \Rightarrow y = -4x + 2$ \- slope $m = -4$ (j) $3y = 12 - 2x \Rightarrow y = 4 - \frac{2}{3}x = -\frac{2}{3}x + 4$ \- slope $m = -\frac{2}{3}$ (k) $y + x = 21 \Rightarrow y = 21 - x = -x + 21$ \- slope $m = -1$ (l) $2x = 12 - y \Rightarrow y = 12 - 2x = -2x + 12$ \- slope $m = -2$ (m) $\frac{1}{5}y = x - 3 \Rightarrow y = 5x - 15$ \- slope $m = 5$ (n) $\frac{1}{3}y = 2x - 6 \Rightarrow y = 6x - 18$ \- slope $m = 6$ (o) $\frac{1}{1}y = 7 - x \Rightarrow y = -x + 7$ \- slope $m = -1$ (p) $\frac{1}{4}y + 2x = 1 \Rightarrow \frac{1}{4}y = 1 - 2x \Rightarrow y = 4 - 8x = -8x + 4$ \- slope $m = -8$ Final answer: Slopes are (a)3, (b)4, (c)-1/2, (d)-7, (e)5, (f)0, (g)3, (h)5, (i)-4, (j)-2/3, (k)-1, (l)-2, (m)5, (n)6, (o)-1, (p)-8.