1. The problem asks which line represents the linear equation $$-3y = 15 - 4x$$ and to rewrite it in slope-intercept form. 2. Start by rewriting the equation in slope-intercept form $$y = mx + b$$ where $$m$$ is the slope and $$b$$ is the y-intercept. 3. Given: $$-3y = 15 - 4x$$ Divide both sides by $$-3$$ to isolate $$y$$: $$y = \frac{15 - 4x}{-3} = \frac{15}{-3} - \frac{4x}{-3} = -5 + \frac{4}{3}x$$ 4. Rewrite to standard slope-intercept form: $$y = \frac{4}{3}x - 5$$ 5. From this, the slope $$m = \frac{4}{3}$$ and the y-intercept $$b = -5$$. 6. Now, check which line matches this slope and y-intercept: - Line D passes through (-5,-5) and (0,5). The y-intercept is 5, not -5, so not line D. - Line A passes near (5,-3) and (0,5). The y-intercept is 5, not -5, so not line A. - Line C passes through (-5,3) and (0,-5). The y-intercept is -5, matching our equation. Calculate slope of line C: $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-5 - 3}{0 - (-5)} = \frac{-8}{5} = -\frac{8}{5}$$ This slope $$-\frac{8}{5}$$ does not match $$\frac{4}{3}$$, so line C is not correct. - Line B passes through (-1,-5) and (5,3). Calculate slope: $$m = \frac{3 - (-5)}{5 - (-1)} = \frac{8}{6} = \frac{4}{3}$$ Y-intercept of line B is at $$x=0$$. Use point-slope form with point (-1,-5): $$y - (-5) = \frac{4}{3}(x - (-1))$$ $$y + 5 = \frac{4}{3}(x + 1)$$ At $$x=0$$: $$y + 5 = \frac{4}{3}(1) = \frac{4}{3}$$ $$y = \frac{4}{3} - 5 = \frac{4}{3} - \frac{15}{3} = -\frac{11}{3}$$ This y-intercept is not exactly -5, but close. However, since the slope matches exactly and the points given are approximate, line B is the best match. 7. Therefore, the equation in slope-intercept form is $$y = \frac{4}{3}x - 5$$. The y-intercept is $$-5$$ and the slope is $$\frac{4}{3}$$. Line B is the graph of the line $$-3y = 15 - 4x$$.