2x + 1 = 5 or 2x + 1 = -5
2x = 4 or 2x = -6
x = 2 or x = -3
The solution set is {2 , -3}.
|2x - 3| = -1 There is no solution since the absolute value of a number is never negative.
2x - 3 = x + 2 or 2x - 3 = -x - 2
2x = x + 5 or 2x = -x + 1
x = 5 or 3x = 1
x = 1/3
The solution set is {1/3, 5}
-1 < x - 3 < 1 -1 + 3 < x - 3 + 3 < 1 + 3 2 < x < 4 The inequality is true in the open interval (2, 4)
2x - 1 < -3 or 3 < 2x - 1 2x < -2 or 4 < 2x x < -1 or 2 < x The inequality is true in the open intervals (-∞, -1) or (2, +∞)
x - 2 < 0 or 0 < x - 2
x < 2 or 2 < x
The inequality is true in the open intervals (-∞, 2) or (2, +∞).
That is, x may take any value except 2; represented on
the real line as follows:
=========================o==================
-1 0 1 2 3 4
-5 ≤ 3 - 2x ≤ 5 -8 ≤ -2x ≤ 2 -2 ≤ 2x ≤ 8 -1 ≤ x ≤ 4 The inequality is true in the closed interval [-1, 4]
-5 ≤ 5 - x ≤ 5 -10 ≤ -x ≤ 0 0 ≤ x ≤ 10 The inequality is true in the closed interval [0, 10]
-4 < 2x - 10 < 4 6 < 2x < 14 3 < x < 7 The inequality is true in the open interval (3, 7)
4|x - 4| ≥ 8 |x - 4| ≥ 2 x - 4 ≤ -2 or 2 ≤ x - 4 x ≤ 2 or 6 ≤ x The inequality is true in the half open intervals (-∞, 2], [6, +∞)
4|2x - 7| ≤ 12 |2x - 7| ≤ 3 -3 ≤ 2x - 7 ≤ 3 4 ≤ 2x ≤ 10 2 ≤ x ≤ 5 The inequality is true in the closed interval [2, 5]
(a) |x - 2| < 4
-4 < x - 2 < 4
-2 < x < 6
(b) |x - 2| > 1
x - 2 < -1 or 1 < x - 2
x < 1 or 3 < x
(a) (-----------------------------------------------)
(b) -------------------) (----------------------
-|-----|-----|-----|-----|-----|-----|-----|-----|------
-2 -1 0 1 2 3 4 5 6
The inequality is true in the open intervals (-2, 1), (3, 6)
(a) |x - 1| ≤ d
-d ≤ x - 1 ≤ d
1 - d ≤ x ≤ 1 + d
(a) |x - 1| > e
x - 1 < -e or e < x - 1
x < 1 - e or 1 + e < x
(a) [-----------------------------------]
(b) ---------------) (----------------------
--|-------------|---|---|-------------|----------------
1-d 1-e 1 1+e 1+d
The inequality is true in the half open intervals [1 - d, 1 - e), (1 + e, 1 + d]
Find all the critical points
-3 < x2 - 3x - 1 or x2 - 3x - 1 < 3
0 < x2 - 3x + 2 or x2 - 3x - 4 < 0
0 < (x - 1)(x - 2) or (x + 1)(x - 4) < 0
(x - 1) - | + | +
(x - 2) - | - | +
is > 0 ==================) (==================
(x + 1) - | + | +
(x - 4) - | - | +
is < 0 (=============================)
both true (===========) (===========)
-------|-----------|-----|-----------|------
-1 0 1 2 3 4
The inequality is true in the open intervals (-1, 1), (2, 4)
_____________
|z + i| = √ x2 + (y + 1)2 = 2
The solution is the circle with radius 2 and center at: 0 - i
_____________
0 < √(x + 2)2 + y2 < 1
The solution is all points excluding the point -2 + 0i and within the circle with
radius 1 and center at: -2 + 0i.