Statement-I: If 11x - 9 le 68, then x∈ (-∞, 7). Statement-II: If an inequality consist of signs le or ge, then the point on the line are also included in the solution region.

Question

Statement-I: If 11x - 9 le 68, then x∈ (-∞, 7). Statement-II: If an inequality consist of signs le or ge, then the point on the line are also included in the solution region. (A) If both Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement-I (B) If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement-I (C) If Statement-I is true but Statement-II is false (D) If Statement-I is false and Statement-II is true

Tardigrade Admin · Accepted Answer

We have, 11x - 9 le 68 ⇒ 11x le 77 ⇒ x le ∴ 7x∈ (-∞, 7) So, statement-I is false but statement-II is true.

Q. Statement-I: If $11 x - 9 \leq 68$ , then $x \in (- \infty, 7)$ .
Statement-II: If an inequality consist of signs $\leq$ or $\geq$ , then the point on the line are also included in the solution region.

If both Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement-I

If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement-I

If Statement-I is true but Statement-II is false

If Statement-I is false and Statement-II is true

We have, $11 x - 9 \leq 68$
$\Rightarrow 11 x \leq 77 \Rightarrow x \leq ∴ 7 x \in (- \infty, 7)$
So, statement-I is false but statement-II is true.

Q. Statement-I: If 11x−9≤68, then x∈(−∞,7). Statement-II: If an inequality consist of signs ≤ or ≥, then the point on the line are also included in the solution region.