Given
(a) 8 + |x -2| = 8
(b) |x + 1| +5 = 2
(c) 4|x - 3|= -16
(d) 3|x +10| = 0
Find
x
Solution
(a) |x -2| = 0 . . . . . subtract 8
... x -2 = 0 . . . . . . . only one solution, because -0 = 0
... x = 2
(b) |x +1| = -3 . . . . subtract 5. No solution. (an absolute value cannot be negative)
(c) No solution. (see (b))
(d) |x +10| = 0 . . . . divide by 3
... x +10 = 0 . . . . . . one solution. (see (a))
... x = -10