# Chapter 2 - Graphs and Functions - 2.5 Equations of Lines and Linear Models - 2.5 Exercises - Page 243: 38

slope = $-\frac{2}{3}$ y-intercept: $(0, \frac{16}{3})$ Refer to the graph below.

#### Work Step by Step

Solve for $y$: $2x+3y=16 \\2x+3y-2x=16-2x \\3y=-2x+16 \\\frac{3y}{3}=\frac{-2x+16}{3} \\y=-\frac{2}{3}x+\frac{16}{3}$ This means that the given equation is equivalent to $y=-\frac{2}{3}x+\frac{16}{3}$. RECALL: The slope-intercept form of a line's equation is $y=mx+b$ where $m$=slope and $(0, b)$ is the line's y-intercept. Thus, the equation $y=-\frac{2}{3}x+\frac{16}{3}$ has a slope of $-\frac{2}{3}$ and a y-intercept of $(0, \frac{16}{3})$. To graph the equation, perform the following steps: (1) Plot the y-intercept $(0, \frac{16}{3})$. (2) Use the slope to plot a second point. Since the slope is $-\frac{2}{3}$, from $(0, \frac{16}{3})$, move 2 units down (the rise) and 3 units to the right (the run) to reach the point $(3, \frac{10}{3})$. Plot $(3. \frac{10}{3})$. (3) Connect the points using a straight line. (Refer to the graph in the answer part above)

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