Question

consider u = {x|x is a real number}.
a = {x|x ∈ u and x + 2 > 10}
b = {x|x ∈ u and 2x > 10}
which pair of statements is correct?
○ 5 ∉ a; 5 ∈ b
○ 6 ∈ a; 6 ∉ b
○ 8 ∉ a; 8 ∈ b
○ 9 ∈ a; 9 ∉ b

Explanation:

Step1: Analyze set A's condition

For set \( A=\{x|x\in U \text{ and } x + 2>10\} \), solve the inequality \( x + 2>10 \). Subtract 2 from both sides: \( x>10 - 2=8 \). So \( A=\{x|x\in\mathbb{R},x > 8\} \).

Step2: Analyze set B's condition

For set \( B=\{x|x\in U \text{ and } 2x>10\} \), solve the inequality \( 2x>10 \). Divide both sides by 2: \( x > \frac{10}{2}=5 \). So \( B=\{x|x\in\mathbb{R},x > 5\} \).

Step3: Check each option

• Option 1: Check \( 5\in A \)? \( 5>8 \)? No, so \( 5

otin A \). Check \( 5\in B \)? \( 5 > 5 \)? No, \( 5 = 5 \), so \( 5
otin B \). This option is wrong.

• Option 2: Check \( 6\in A \)? \( 6>8 \)? No, \( 6

otin A \). Check \( 6\in B \)? \( 6>5 \)? Yes, but the option says \( 6
otin B \), so this option is wrong.

• Option 3: Check \( 8\in A \)? \( 8>8 \)? No, \( 8

otin A \). Check \( 8\in B \)? \( 8>5 \)? Yes, so \( 8\in B \). This matches the option.

• Option 4: Check \( 9\in A \)? \( 9>8 \)? Yes, \( 9\in A \). Check \( 9\in B \)? \( 9>5 \)? Yes, but the option says \( 9

otin B \), so this option is wrong.

Answer:

8 ∉ A; 8 ∈ B (the third option)