Question

5.7
in exercises 1 and 2, tell whether the ordered pair is a solution of the system of
linear inequalities.

1. (2, 0); ( y > x - 5 )

( y leq 2x + 1 )

1. (1, 4); ( y < 2x + 2 )

( y geq -3x + 4 )
in exercises 3–8, graph the system. identify a solution, if possible.

1. ( x + y leq 2 )

( y leq 1 )

1. ( 3x + y > 4 )

( y < -3x + 1 )

1. ( x - y < 3 )

( -x - y geq -1 )

1. ( y leq \frac{1}{3}x + 2 )

( y > -\frac{1}{2}x + 5 )

1. ( x > -2 )

( y < 3 )
( y geq 2x - 1 )

1. ( x + y > 4 )

( x - y < -1 )
( y > 7 )
in exercises 9 and 10, write a system

Answer:

To solve this, we'll take each exercise one by one. Let's start with Exercise 1:

Exercise 1: Check if (2, 0) is a solution to the system \( y > x - 5 \) and \( y \leq 2x + 1 \)

Step 1: Substitute into \( y > x - 5 \)

Substitute \( x = 2 \) and \( y = 0 \) into \( y > x - 5 \):
\( 0 > 2 - 5 \)
\( 0 > -3 \)
This is true.

Step 2: Substitute into \( y \leq 2x + 1 \)

Substitute \( x = 2 \) and \( y = 0 \) into \( y \leq 2x + 1 \):
\( 0 \leq 2(2) + 1 \)
\( 0 \leq 4 + 1 \)
\( 0 \leq 5 \)
This is also true.

Since both inequalities are satisfied, (2, 0) is a solution.

Exercise 2: Check if (1, 4) is a solution to the system \( y < 2x + 2 \) and \( y \geq -3x + 4 \)

Step 1: Substitute into \( y < 2x + 2 \)

Substitute \( x = 1 \) and \( y = 4 \) into \( y < 2x + 2 \):
\( 4 < 2(1) + 2 \)
\( 4 < 2 + 2 \)
\( 4 < 4 \)
This is false (4 is not less than 4).

Even though we could check the second inequality, since the first one fails, (1, 4) is not a solution.

Exercise 3: Graph \( x + y \leq 2 \) and \( y \leq 1 \), and identify a solution

Step 1: Graph \( x + y \leq 2 \)

Rewrite as \( y \leq -x + 2 \). This is a line with slope -1, y-intercept 2. Shade below the line (since \( \leq \)).

Step 2: Graph \( y \leq 1 \)

This is a horizontal line at \( y = 1 \). Shade below the line (since \( \leq \)).

Step 3: Find the intersection (solution region)

The solution region is where both shadings overlap. A solution can be (0, 0):

• For \( x + y \leq 2 \): \( 0 + 0 = 0 \leq 2 \) (true).
• For \( y \leq 1 \): \( 0 \leq 1 \) (true).

Exercise 4: Graph \( 3x + y > 4 \) and \( y < -3x + 1 \), check for solutions

Step 1: Graph \( 3x + y > 4 \)

Rewrite as \( y > -3x + 4 \). Dashed line (since \( > \)), shade above.

Step 2: Graph \( y < -3x + 1 \)

Dashed line (since \( < \)), shade below.

Step 3: Check for overlap

The lines \( y = -3x + 4 \) and \( y = -3x + 1 \) are parallel (same slope -3). The region above \( y = -3x + 4 \) and below \( y = -3x + 1 \) has no overlap (since \( -3x + 4 > -3x + 1 \) for all \( x \)). Thus, no solution.

Exercise 5: Graph \( x - y < 3 \) and \( -x - y \geq -1 \), identify a solution

Step 1: Graph \( x - y < 3 \)

Rewrite as \( y > x - 3 \). Dashed line, shade above.

Step 2: Graph \( -x - y \geq -1 \)

Rewrite as \( y \leq -x + 1 \). Solid line, shade below.

Step 3: Find a solution

A solution could be (0, 0):

• \( 0 - 0 = 0 < 3 \) (true).
• \( -0 - 0 = 0 \geq -1 \) (true).

Exercise 6: Graph \( y \leq \frac{1}{3}x + 2 \) and \( y > -\frac{1}{2}x + 5 \), identify a solution

Step 1: Graph \( y \leq \frac{1}{3}x + 2 \)

Solid line, slope \( \frac{1}{3} \), y-intercept 2. Shade below.

Step 2: Graph \( y > -\frac{1}{2}x + 5 \)

Dashed line, slope \( -\frac{1}{2} \), y-intercept 5. Shade above.

Step 3: Find a solution

Check (3, 3):

• \( 3 \leq \frac{1}{3}(3) + 2 = 1 + 2 = 3 \) (true, since \( \leq \)).
• \( 3 > -\frac{1}{2}(3) + 5 = -1.5 + 5 = 3.5 \)? No, 3 < 3.5.

Check (6, 4):

• \( 4 \leq \frac{1}{3}(6) + 2 = 2 + 2 = 4 \) (true).
• \( 4 > -\frac{1}{2}(6) + 5 = -3 + 5 = 2 \) (true).

Exercise 7: Graph \( x > -2 \), \( y < 3 \), \( y \geq 2x - 1 \), identify a solution

Step 1: Graph \( x > -2 \)

Dashed vertical line at \( x = -2 \), shade right.

Step 2: Graph \( y < 3 \)

Dashed horizontal line at \( y = 3 \), shade below.

Step 3: Graph \( y \geq 2x - 1 \)

Solid line, slope 2, y-intercept -1. Shade above.

Step 4: Find a solution

Check (0, 0):

• \( 0 > -2 \) (true).
• \( 0 < 3 \) (true).
• \( 0 \geq 2(0) - 1 = -1 \) (true).

Exercise 8: Graph \( x + y > 4 \), \( x - y < -1 \), \( y > 7 \), check for solutions

Step 1: Graph \( x + y > 4 \)

\( y > -x + 4 \), dashed, shade above.

Step 2: Graph \( x - y < -1 \)

\( y > x + 1 \), dashed, shade above.

Step 3: Graph \( y > 7 \)

Dashed horizontal line, shade above.

Step 4: Check for overlap

Find where all three shadings overlap. Let’s solve \( y > 7 \), \( y > x + 1 \), \( y > -x + 4 \).

From \( y > 7 \) and \( y > x + 1 \): \( 7 > x + 1 \implies x < 6 \).

From \( y > 7 \) and \( y > -x + 4 \): \( 7 > -x + 4 \implies x > -3 \).

A solution could be (0, 8):

• \( 0 + 8 = 8 > 4 \) (true).
• \( 0 - 8 = -8 < -1 \) (true, since \( -8 < -1 \)).
• \( 8 > 7 \) (true).

Exercises 9–10: Write a system (not shown, but process: define inequalities based on context, e.g., real-world constraints like "number of items \( x, y \) with \( x + y \leq 10 \), \( 2x + 3y \geq 15 \)").

Final Answers (Key Results):

1. (2, 0) is a solution.
2. (1, 4) is not a solution.
3. Solution: (0, 0) (example).
4. No solution.
5. Solution: (0, 0) (example).
6. Solution: (6, 4) (example).
7. Solution: (0, 0) (example).
8. Solution: (0, 8) (example).

(Note: Exercises 9–10 require specific context, not provided here.)