Himpunan penyelesaian dari pertidaksamaan \(|2x + 1| \leq 3\) adalah:

a. \(-1 \leq x \leq 3\)

b. \(-2 \leq x \leq 2\)

c. \(-2 \leq x \leq 1\)

d. \(-1 \leq x \leq 1\)

e. \(0 \leq x \leq 1\)

Tolong jelaskan caranya.

∣2 x + 1∣ ≤ 3 2 x + 1 ≤ 3 ∧ 2 x + 1 ≥ − 3 2 x ≤ 2 ∧ 2 x ≥ − 4 x ≤ 1 ∧ x ≥ − 2 − 2 ≤ x ≤ 1 ⇒ C

Answered by konrad509 | 2024-06-10

The solution to the given inequality , |2x + 1| ≤ 3, is the set { -2 ≤ x ≤ 1 }, which corresponds to answer choice (c). This is achieved by solving two separate inequalities, 2x + 1 ≤ 3 and - (2x + 1) ≤ 3. ;

Answered by qwmoon | 2024-06-18

Himpunan penyelesaian dari pertidaksamaan ∣2 x + 1∣ ≤ 3 adalah − 2 ≤ x ≤ 1 , yang sesuai dengan pilihan c.
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Answered by konrad509 | 2024-10-07