8. (04.05 mc) in which quadrants do solutions for the inequality ( y \frac{1}{5}x + 3 ) exist? (1 point) i and ii i, ii, and iii i, iii, and iv all four quadrants

Question

8. (04.05 mc) in which quadrants do solutions for the inequality ( y > \frac{1}{5}x + 3 ) exist? (1 point) i and ii i, ii, and iii i, iii, and iv all four quadrants

Accepted Answer

# Explanation: ## Step1: Analyze the linear inequality The inequality is $ y > \frac{1}{5}x + 3 $. First, consider the line $ y=\frac{1}{5}x + 3 $. The y - intercept of this line is $ 3 $ (when $ x = 0 $, $ y=3 $) and the slope is $ \frac{1}{5} $ (positive, so the line rises from left to right). ## Step2: Determine the regions - **Quadrant I**: $ x>0,y > 0 $. For positive $ x $, as $ x $ increases, $ \frac{1}{5}x+3 $ is positive, and $ y>\frac{1}{5}x + 3 $ will also give positive $ y $, so solutions exist here. - **Quadrant II**: $ x<0,y>0 $. When $ x = 0 $, $ y = 3>0 $. For negative $ x $, $ \frac{1}{5}x+3 $ is still positive (since $ \frac{1}{5}x $ is negative but $ 3 $ is large enough to keep the sum positive for small - magnitude negative $ x $), and $ y>\frac{1}{5}x + 3 $ will have positive $ y $, so solutions exist here. - **Quadrant III**: $ x<0,y<0 $. Let's check if there are solutions. Suppose $ x=-20 $, then $ \frac{1}{5}x + 3=\frac{-20}{5}+3=-4 + 3=-1 $. We need $ y>-1 $, but in Quadrant III, $ y<0 $. So $ y>-1 $ and $ y < 0 $ is possible (e.g., $ y=-0.5 $ when $ x = - 20 $, $ y=-0.5> - 1=\frac{1}{5}\times(-20)+3 $). Wait, no, let's re - evaluate. Wait, when $ x$ is a large negative number, say $ x=-100 $, $ \frac{1}{5}x+3=\frac{-100}{5}+3=-20 + 3=-17 $. Then $ y>-17 $. But in Quadrant III, $ y<0 $. So $ y$ values between $ - 17$ and $ 0 $ (for $ x=-100 $) satisfy $ y>\frac{1}{5}x + 3 $ and are in Quadrant III? Wait, no, we made a mistake. Wait, the line $ y=\frac{1}{5}x + 3 $ when $ x=-15 $, $ y=\frac{-15}{5}+3=-3 + 3 = 0 $. When $ x<-15 $, $ \frac{1}{5}x+3<0 $. For example, $ x=-20 $, $ \frac{1}{5}x + 3=-1 $. So $ y>-1 $. In Quadrant III, $ y<0 $, so $ - 1 2 $, but in Quadrant III, $ y<0 $, so $ y>2 $ and $ y < 0 $ is impossible. Wait, I made a mistake earlier. Let's use the y - intercept. The line $ y=\frac{1}{5}x + 3 $ crosses the y - axis at $ (0,3) $ and has a positive slope. So the region $ y>\frac{1}{5}x + 3 $ is above the line. The line is above the x - axis (since y - intercept is $ 3>0 $) and has a positive slope. So in Quadrant III ($ x<0,y<0 $), the line $ y=\frac{1}{5}x + 3 $ is above the x - axis (for $ x$ such that $ \frac{1}{5}x+3>0\Rightarrow x>-15 $). For $ x < - 15 $, $ \frac{1}{5}x+3<0 $. But in Quadrant III, $ y<0 $. So when $ x < - 15 $, we have $ y>\frac{1}{5}x + 3$ (which is negative) and $ y < 0 $. For example, $ x=-20 $, $ \frac{1}{5}x+3=-4 + 3=-1 $, $ y=-0.5>-1 $ and $ y=-0.5<0 $ (Quadrant III). Wait, but when $ x=-5 $, $ \frac{1}{5}x + 3=-1 + 3 = 2 $, $ y>2 $ (Quadrant II or I). So there are solutions in Quadrant III? Wait, no, let's check the quadrant signs again. Quadrant III: $ x<0,y<0 $. If $ x=-20 $, $ y=-0.5 $ (which is $ y<0 $ and $ x<0 $), so it's in Quadrant III and $ y=-0.5>\frac{1}{5}\times(-20)+3=-1 $. So solutions exist in Quadrant III? Wait, no, I think I messed up. Wait, the correct way: The inequality $ y>\frac{1}{5}x + 3 $. Let's find where the line is. The line has a positive slope and y - intercept $ 3 $. So the region above the line: - In Quadrant I: $ x>0,y>0 $, as $ x$ increases, $ y=\frac{1}{5}x + 3 $ is positive, so $ y>\frac{1}{5}x + 3 $ gives positive $ y $, so solutions. - In Quadrant II: $ x<0,y>0 $, since at $ x = 0 $, $ y = 3 $, and for negative $ x $, $ \frac{1}{5}x+3 $ is still positive (because $ \frac{1}{5}x $ is negative but $ 3 $ is large enough to keep the sum positive for $ x$ in $ (-15,0) $), and $ y>\frac{1}{5}x + 3 $ will have positive $ y $, so solutions. - In Quadrant III: $ x<0,y<0 $. Let's take $ x=-10 $, $ \frac{1}{5}x+3=-2 + 3 = 1 $, $ y>1 $ (but $ y<0 $ in Quadrant III, so no solution here). Wait, I see my mistake earlier. When $ x=-10 $, $ \frac{1}{5}x + 3 = 1 $, so $ y>1 $, but in Quadrant III, $ y<0 $, so no solution. When $ x=-20 $, $ \frac{1}{5}x+3=-4 + 3=-1 $, $ y>-1 $. In Quadrant III, $ y<0 $, so $ - 1-1=\frac{1}{5}\times(-20)+3 $. So there are solutions in Quadrant III? Wait, no, the key is the y - intercept. The line $ y=\frac{1}{5}x + 3 $ is above the x - axis (since y - intercept is 3). The region above the line: - For Quadrant IV: $ x>0,y<0 $. Let's check. When $ x = 0 $, $ y = 3 $. For $ x>0 $, $ \frac{1}{5}x+3>0 $, so $ y>\frac{1}{5}x + 3 $ implies $ y>0 $, so no solutions in Quadrant IV (since in Quadrant IV, $ y<0 $). Wait, let's start over. The line $ y=\frac{1}{5}x + 3 $: - Crosses the y - axis at $ (0,3) $ (Quadrant I/II boundary, but $ y = 3>0 $, so in Quadrant I/II area). - Has a positive slope, so it goes from Quadrant II (x negative, y positive) to Quadrant I (x positive, y positive). - The inequality $ y>\frac{1}{5}x + 3 $ is the region above the line. - Quadrant I: $ x>0,y>0 $. Since the line is in Quadrant I (y - intercept 3, positive slope), above the line in Quadrant I: solutions exist. - Quadrant II: $ x<0,y>0 $. The line is in Quadrant II (x negative, y positive) for $ x\in(-15,0) $ (since when $ x=-15 $, $ y=\frac{-15}{5}+3 = 0 $), and above the line here (y positive) : solutions exist. - Quadrant III: $ x<0,y<0 $. For $ x < - 15 $, the line $ y=\frac{1}{5}x + 3 $ is below the x - axis (y negative). The region above the line here would be $ y>\frac{1}{5}x + 3 $ (y negative but greater than $ \frac{1}{5}x + 3 $). For example, $ x=-20 $, $ \frac{1}{5}x+3=-1 $, $ y=-0.5>-1 $ and $ y=-0.5<0 $ (Quadrant III). So solutions exist here. - Quadrant IV: $ x>0,y<0 $. The line $ y=\frac{1}{5}x + 3 $ for $ x>0 $ has $ y=\frac{1}{5}x + 3>0 $, so $ y>\frac{1}{5}x + 3 $ implies $ y>0 $, but in Quadrant IV, $ y<0 $, so no solutions here. Wait, but the options are: I and II; I, II and III; I, III and IV; All four quadrants. Wait, let's take test points: - **Quadrant I**: $ (x = 5,y = 5) $. $ \frac{1}{5}\times5+3=1 + 3 = 4 $, $ 5>4 $, so $ (5,5) $ is a solution. - **Quadrant II**: $ (x=-5,y = 4) $. $ \frac{1}{5}\times(-5)+3=-1 + 3 = 2 $, $ 4>2 $, so $ (-5,4) $ is a solution. - **Quadrant III**: $ (x=-20,y=-0.5) $. $ \frac{1}{5}\times(-20)+3=-4 + 3=-1 $, $ - 0.5>-1 $, so $ (-20,-0.5) $ is a solution. - **Quadrant IV**: $ (x = 5,y=-1) $. $ \frac{1}{5}\times5+3=4 $, $ - 1>4 $ is false, so no solution. So the quadrants with solutions are I, II, and III. # Answer: I, II, and III (the option "I, II, and III")

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QUESTION IMAGE

Question

Explanation:

Step1: Analyze the linear inequality

Step2: Determine the regions

Answer: