Inequalities Part 2 Quiz

2(3x − 1) < 4x + 5 or, 6x - 2 < 4x + 5 or, 2x < 7 or, x < 7/2

|x − 4| ≤ 3.
or, -3 ≤ (x − 4) ≤ 3
or, 4 - 3 ≤ (x − 4 + 4) ≤ 3 + 4
or, 1 ≤ x ≤ 7

|2x + 1| > 5 means
=>2x + 1 > 5 or 2x + 1 < −5 => x > 2 or x < -3 x < −3 or x > 2

−2 ≤ 3x − 1 < 7 add +1 or, −2 + 1 ≤ 3x − 1 + 1 < 7 + 1 or, -1 ≤ 3x < 8 or, -1/3 ≤ x < 8/3

(x − 2)/(x + 1) > 0
The inequality is strict (> 0), so exclude where expression = 0 (x = 2) and where undefined (x = −1)
So the solution is x < −1 or x > 2

1/(x − 3) ≤ 0
The fraction is ≤ 0 when the numerator and denominator have opposite signs (or numerator = 0, but here numerator is 1 > 0, so never zero).
So the solution is x < 3

Solve |x + 2| ≥ 4
Case 1: x + 2 ≥ 4 → x ≥ 2
Case 2: x + 2 ≤ −4 → x ≤ −6
So the solution is: x ≤ −6 or x ≥ 2

5 − |x| > 2
Subtract 5 from both sides: −|x| > −3
or, |x| < 3 or, −3 < x < 3.

(2x − 1)(x + 3) ≤ 0
Critical points:
1) 2x − 1 = 0 → x = ½
2) x + 3 = 0 → x = −3
The product is ≤ 0 between the roots (including endpoints)
Solution: −3 ≤ x ≤ ½

3/(2x + 1) > 1
or, 3/(2x + 1) − 1 > 0
or, [3 − (2x + 1)] / (2x + 1) > 0
or, 2(1 − x) / (2x + 1) > 0
Critical points: x = 1 (numerator = 0), x = −1/2 (denominator = 0, undefined).
Solution: Positive when −1/2 < x < 1