Answer:
B. 2x + 3 ≥ 11 or 4x - 7 ≤ 1
Step-by-step explanation:
Given compound inequality,
In option A,
2x + 3 ≥ 11 and 4x - 7 ≤ 1
⇒ 2x ≥ 8 and 4x ≤ 8
⇒ x ≥ 4 and x ≤ 2
[tex]\implies [4,\infty)\cap (-\infty, 2][/tex]
[tex]=\phi[/tex]
2x + 3 ≥ 11 and 4x - 7 ≤ 1 can not be the correct inequality,
In option B,
2x + 3 ≥ 11 or 4x - 7 ≤ 1
⇒ 2x ≥ 8 or 4x ≤ 8
⇒ x ≥ 4 or x ≤ 2
[tex]\implies [4,\infty)\cup (-\infty, 2][/tex]
Which is shown in the given graph,
Thus, 2x + 3 ≥ 11 or 4x - 7 ≤ 1 is the correct compound inequality,
In option C,
2x + 3 > 11 or 4x - 7 < 1
⇒ 2x > 8 or 4x < 8
⇒ x > 4 or x < 2
[tex]\implies (4,\infty)\cup (-\infty, 2)[/tex]
So, which is not shown in the graph,
2x + 3 > 11 or 4x - 7 < 1 can not be the correct compound inequality,
In option D,
2x + 3 ≥ 11 or 4x - 7 ≥ 1
⇒ 2x ≥ 8 or 4x ≥ 8
⇒ x ≥ 4 or x ≥ 2
[tex]\implies [4,\infty)[/tex]
Which is not shown in the graph,
2x + 3 ≥ 11 or 4x - 7 ≥ 1 is not the correct compound inequality.
