Linear Equations Inequality C6252A

1. **State the problem:** Solve and graph the equation $-x + 4y = 8$. 2. **Rewrite the equation in slope-intercept form:** $$-x + 4y = 8$$ Add $x$ to both sides: $$4y = x + 8$$ Divide both sides by 4: $$y = \frac{\cancel{4}x}{\cancel{4}4} + \frac{8}{4} = \frac{1}{4}x + 2$$ 3. **Find points to graph:** - When $x=0$, $y=2$ (point $(0,2)$) - When $x=4$, $y=\frac{1}{4} \times 4 + 2 = 1 + 2 = 3$ (point $(4,3)$) --- 1. **State the problem:** Graph the line $y = -\frac{5}{3}x + 3$. 2. **Find points:** - When $x=0$, $y=3$ (point $(0,3)$) - When $x=3$, $y = -\frac{5}{3} \times 3 + 3 = -5 + 3 = -2$ (point $(3,-2)$) --- 1. **State the problem:** Graph the inequality $y < 3x + 2$. 2. **Use the boundary line:** $y = 3x + 2$ 3. **Find points:** - When $x=0$, $y=2$ (point $(0,2)$) - When $x=1$, $y=3(1) + 2 = 5$ (point $(1,5)$) 4. **Graphing rules:** - Use a dashed line for the boundary because the inequality is strict ($<$). - Shade the region below the line because $y$ is less than $3x + 2$. --- **Summary:** - The first equation is correctly converted to slope-intercept form and points are correctly calculated. - The second equation points are correctly calculated. - The inequality is correctly graphed with a dashed line and shading below. Therefore, the solutions and graphs are correct.