
MODERATE EXERCISES - Absolute Value and Distance
(1) Solve for x.
(a)(b)
(c)
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(a) |x2 - 1| = 3 (b) |x2 + 3x| = 0
(a)(b) |7r - 7s| - |r - s| - 2|3(r - s)| (c) a - b + |a-b| - |b - a|

(1) Solve for x.
(a)(b)
(c)
![]()
(a) |x2 - 1| = 3 (b) |x2 + 3x| = 0
(a)(b) |7r - 7s| - |r - s| - 2|3(r - s)| (c) a - b + |a-b| - |b - a|