Question

in exercises 1–6, tell whether the ordered pair is a solution of the inequality.

1. ( x + y > 5; (3,2) )
2. ( x - y ge 2; (5,3) )
3. ( x + 2y le 4; (-1,2) )
4. ( 5x + y < 7; (2,-2) )
5. ( 3x - 4y > 6; (-1,-1) )
6. ( -x - 2y ge 5; (-2,-3) )

in exercises 7–18, graph the inequality in a coordinate plane.

1. ( y < 4 )
2. ( y > -1 )
3. ( x > 3 )
4. ( x le -1 )
5. ( y < -2 )
6. ( x > -2 )

Explanation:

Problem 1: \( x + y > 5; (3,2) \)

Step 1: Substitute \( x = 3 \), \( y = 2 \)

Substitute the values of \( x \) and \( y \) from the ordered pair into the inequality. So we get \( 3 + 2>5 \).

Step 2: Simplify the left - hand side

Calculate \( 3 + 2=5 \). Now the inequality becomes \( 5>5 \).

Step 1: Substitute \( x = 5 \), \( y = 3 \)

Substitute the values into the inequality: \( 5-3\geq2 \).

Step 2: Simplify the left - hand side

Calculate \( 5 - 3 = 2 \). The inequality is \( 2\geq2 \), which is true.

Step 1: Substitute \( x=-1 \), \( y = 2 \)

Substitute into the inequality: \( - 1+2\times2\leq4 \).

Step 2: Simplify the left - hand side

First, calculate \( 2\times2 = 4 \), then \( -1 + 4=3 \). The inequality is \( 3\leq4 \), which is true.

Answer:

The ordered pair \( (3,2) \) is not a solution of the inequality \( x + y>5 \) since \( 5\) is not greater than \( 5 \).

Problem 2: \( x - y\geq2; (5,3) \)