Question

a. sketch the graph of the following inequality. ( y > -2x + 5 )
b. circle those ordered pairs that are solutions to the inequality.
i. ( (6,0) ) ii. ( (-3, -1) ) iii. ( (4, -3) ) iv. ( (7,5) ) v. ( (0, 4) ) vi. ( (0, 5) )

Explanation:

Part a: Sketching the Graph of \( y > -2x + 5 \)

Step 1: Identify the Boundary Line

The inequality is \( y > -2x + 5 \). First, consider the boundary line \( y = -2x + 5 \). Since the inequality is "greater than" (not "greater than or equal to"), the boundary line will be dashed (to indicate that points on the line are not included in the solution set).

Step 2: Find Two Points on the Boundary Line

To graph the line \( y = -2x + 5 \), we can find two points:

• When \( x = 0 \), \( y = -2(0) + 5 = 5 \). So, the point is \( (0, 5) \).
• When \( x = 2 \), \( y = -2(2) + 5 = 1 \). So, the point is \( (2, 1) \).

Plot these two points and draw a dashed line through them.

Step 3: Determine the Region to Shade

To determine which side of the line to shade, we can test a point not on the line. A common test point is the origin \( (0, 0) \), but let's check if it satisfies the inequality:
Substitute \( x = 0 \) and \( y = 0 \) into \( y > -2x + 5 \):
\( 0 > -2(0) + 5 \)
\( 0 > 5 \)
This is false. So, the origin is not in the solution region. Therefore, we shade the region above the dashed line \( y = -2x + 5 \) (since the test point below the line did not satisfy the inequality, the solution is above the line).

Part b: Identifying Solution Ordered Pairs

To determine if an ordered pair \( (x, y) \) is a solution to \( y > -2x + 5 \), we substitute the \( x \) and \( y \) values into the inequality and check if it holds true.

i. \( (6, 0) \)

Substitute \( x = 6 \), \( y = 0 \):
\( 0 > -2(6) + 5 \)
\( 0 > -12 + 5 \)
\( 0 > -7 \)
This is true. So, \( (6, 0) \) is a solution.

ii. \( (-3, -1) \)

Substitute \( x = -3 \), \( y = -1 \):
\( -1 > -2(-3) + 5 \)
\( -1 > 6 + 5 \)
\( -1 > 11 \)
This is false. So, \( (-3, -1) \) is not a solution.

iii. \( (4, -3) \)

Substitute \( x = 4 \), \( y = -3 \):
\( -3 > -2(4) + 5 \)
\( -3 > -8 + 5 \)
\( -3 > -3 \)
This is false (since it's "greater than," not "greater than or equal to"). So, \( (4, -3) \) is not a solution.

iv. \( (7, 5) \)

Substitute \( x = 7 \), \( y = 5 \):
\( 5 > -2(7) + 5 \)
\( 5 > -14 + 5 \)
\( 5 > -9 \)
This is true. So, \( (7, 5) \) is a solution.

v. \( (0, 4) \)

Substitute \( x = 0 \), \( y = 4 \):
\( 4 > -2(0) + 5 \)
\( 4 > 5 \)
This is false. So, \( (0, 4) \) is not a solution.

vi. \( (0, 5) \)

Substitute \( x = 0 \), \( y = 5 \):
\( 5 > -2(0) + 5 \)
\( 5 > 5 \)
This is false (since it's "greater than," not "greater than or equal to"). So, \( (0, 5) \) is not a solution.

Answer:

(Part b):
The ordered pairs that are solutions are:
i. \( (6, 0) \)
iv. \( (7, 5) \)

(For Part a, the graph should have a dashed line through \( (0, 5) \) and \( (2, 1) \) (or other points on \( y = -2x + 5 \)) with the region above the line shaded.)