mild 1. tell whether the ordered pair is a solution to the system of linear inequalities: (-5, 2) $ y x + 3 $ tell whether the ordered pair is a solution of the system of linear inequalities: 2. (-4, 3) 3. (-3, -1) 4. (-2, 0) 5. (1, 0.5) (graph: a coordinate plane with x from -5 to 1 and y from -4 to 4, showing two lines, one solid and one dashed, with a shaded region.) 6. graph the system. identify a solution if possible. $ y \frac{1}{2}x + 2 $. (graph: a coordinate plane with a marked error, showing two lines, one solid and one dashed.)

Question

mild 1. tell whether the ordered pair is a solution to the system of linear inequalities: (-5, 2) $ y < 4 $ $ y > x + 3 $ tell whether the ordered pair is a solution of the system of linear inequalities: 2. (-4, 3) 3. (-3, -1) 4. (-2, 0) 5. (1, 0.5) (graph: a coordinate plane with x from -5 to 1 and y from -4 to 4, showing two lines, one solid and one dashed, with a shaded region.) 6. graph the system. identify a solution if possible. $ y < x - 1 $ $ y \geq x + 1 $ (graph: a blank coordinate plane grid.) 7. write a system of linear inequalities represented by the graph. (graph: a coordinate plane with x from -2 to 4 and y from -2 to 2, showing two intersecting lines.) 8. describe and correct the error in graphing the system $ y \leq 3x + 4 $ and $ y > \frac{1}{2}x + 2 $. (graph: a coordinate plane with a marked error, showing two lines, one solid and one dashed.)

Accepted Answer

### Problem 1: # Explanation: ## Step1: Check $ y < 4 $ Substitute $ x = -5, y = 2 $ into $ y < 4 $. $ 2 < 4 $ is true. ## Step2: Check $ y > x + 3 $ Substitute $ x = -5, y = 2 $ into $ y > x + 3 $. $ 2 > -5 + 3 $ → $ 2 > -2 $ is true. # Answer: The ordered pair $(-5, 2)$ is a solution. ### Problem 2: (Assuming the system from Problem 1: $ y < 4 $, $ y > x + 3 $) # Explanation: ## Step1: Check $ y < 4 $ Substitute $ x = -4, y = 3 $ into $ y < 4 $. $ 3 < 4 $ is true. ## Step2: Check $ y > x + 3 $ Substitute $ x = -4, y = 3 $ into $ y > x + 3 $. $ 3 > -4 + 3 $ → $ 3 > -1 $ is true. # Answer: The ordered pair $(-4, 3)$ is a solution. ### Problem 3: (Assuming the system from Problem 1: $ y < 4 $, $ y > x + 3 $) # Explanation: ## Step1: Check $ y < 4 $ Substitute $ x = -3, y = -1 $ into $ y < 4 $. $ -1 < 4 $ is true. ## Step2: Check $ y > x + 3 $ Substitute $ x = -3, y = -1 $ into $ y > x + 3 $. $ -1 > -3 + 3 $ → $ -1 > 0 $ is false. # Answer: The ordered pair $(-3, -1)$ is not a solution. ### Problem 4: (Assuming the system from Problem 1: $ y < 4 $, $ y > x + 3 $) # Explanation: ## Step1: Check $ y < 4 $ Substitute $ x = -2, y = 0 $ into $ y < 4 $. $ 0 < 4 $ is true. ## Step2: Check $ y > x + 3 $ Substitute $ x = -2, y = 0 $ into $ y > x + 3 $. $ 0 > -2 + 3 $ → $ 0 > 1 $ is false. # Answer: The ordered pair $(-2, 0)$ is not a solution. ### Problem 5: (Assuming the system from Problem 1: $ y < 4 $, $ y > x + 3 $) # Explanation: ## Step1: Check $ y < 4 $ Substitute $ x = 1, y = 0.5 $ into $ y < 4 $. $ 0.5 < 4 $ is true. ## Step2: Check $ y > x + 3 $ Substitute $ x = 1, y = 0.5 $ into $ y > x + 3 $. $ 0.5 > 1 + 3 $ → $ 0.5 > 4 $ is false. # Answer: The ordered pair $(1, 0.5)$ is not a solution. ### Problem 6: # Explanation: The system is $ y < x - 1 $ (dashed line, shade below) and $ y \geq x + 1 $ (solid line, shade above). The lines $ y = x - 1 $ and $ y = x + 1 $ are parallel (same slope $ 1 $). The region below $ y = x - 1 $ and above $ y = x + 1 $ has no overlap (since $ x - 1 < x + 1 $ for all $ x $, so $ y < x - 1 $ implies $ y < x + 1 $, conflicting with $ y \geq x + 1 $). # Answer: The system has no solution. ### Problem 7: (To determine the system from the graph, assume two lines: one with positive slope, one with negative slope. Let's find equations. For the positive - slope line: passes through (0, -1) and (4, 1), slope $ m=\frac{1 - (-1)}{4 - 0}=\frac{2}{4}=\frac{1}{2} $, equation $ y=\frac{1}{2}x - 1 $ (dashed? Or solid? From graph arrows, maybe $ y < \frac{1}{2}x - 1 $ or $ y > \frac{1}{2}x - 1 $). For the negative - slope line: passes through (0, 2) and (-1, 0), slope $ m=\frac{0 - 2}{-1 - 0}=2 $, equation $ y=-2x + 2 $ (dashed? Or solid? Maybe $ y > -2x + 2 $ or $ y < -2x + 2 $). A possible system: $ y < \frac{1}{2}x - 1 $ and $ y > -2x + 2 $ (depends on shading, but generally, find the two linear inequalities by identifying the lines and their inequality directions). # Answer: A possible system is $ y < \frac{1}{2}x - 1 $ and $ y > -2x + 2 $ (exact system depends on graph shading details, but this is a common interpretation). ### Problem 8: # Explanation: - **Error Description**: For $ y\leq3x + 4 $, the line $ y = 3x+4 $ should be solid (since $ \leq $ includes equality), and for $ y>\frac{1}{2}x + 2 $, the line $ y=\frac{1}{2}x + 2 $ should be dashed (since $ > $ does not include equality). Also, check shading: $ y\leq3x + 4 $ shades below the line, $ y>\frac{1}{2}x + 2 $ shades above the line. If the graph has incorrect line styles (solid/dashed) or incorrect shading, that's the error. - **Correction**: Draw $ y = 3x + 4 $ as a solid line and shade below it. Draw $ y=\frac{1}{2}x + 2 $ as a dashed line and shade above it. # Answer: Error: Incorrect line styles (e.g., $ y=\frac{1}{2}x + 2 $ might be solid or $ y = 3x + 4 $ might be dashed) or incorrect shading. Correction: Make $ y = 3x + 4 $ solid (shade below), $ y=\frac{1}{2}x + 2 $ dashed (shade above).

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QUESTION IMAGE

Question

Explanation:

Problem 1:

Step1: Check \( y < 4 \)

Step2: Check \( y > x + 3 \)

Step1: Check \( y < 4 \)

Step2: Check \( y > x + 3 \)

Step1: Check \( y < 4 \)

Step2: Check \( y > x + 3 \)

Answer:

Problem 2: