Solve |3*-2|s|x-l|.
This is equivalent to -|x-1|<3*-2=s |*-1|. Case 1. Jt-l>0. Then -(x - I)s3x -2 -* + l<3*-2<*-l; the first inequality is equivalent to | impossible. Case 2. *-l<0. -x + l>3x-2s=*-l; the first inequality is equivalent to jc s f and
the second to jt > |. Hence, we have f •& x s |. Answer
Solve |* - 2| + |x - 5| = 9.
Case 1. x>5. Then jr-2 + jt-5 = 9, 2*-7 = 9, 2* = 16, x = 8. Case 2. 2 jt-2 + 5-x = 9, 3 = 9, which is impossible. Case 3. x<2. Then 2-x + 5-x = 9, l-2x = 9,
2x = —2, x=—\. So, the solutions are 8 and-1.
Solve 4-*s:|5x + l|.
Case 1. Sx + laO, that is, Jta-j. Then 4-*>5j: + l, 3>6^:, i>^:. Thus, we obtain the
solutions -^ <*:-5x-l, 4x>-5, xs-|.
Thus, we obtain the solutions -!<*<-$. Hence, the set of solutions is [- 5, I] U [- f, - j) = [-1, |].
Prove |a-&|<|«| + |&|.
By the triangle inequality, \a-b\ = |o + (-fc)|< |«| + \-b\ = \a\ + \b\.
Solve the inequality \x - 1| a |jc -3|.
We argue geometrically from Fig. 2-2. \x — 1| is the distance of x from 1, and \x — 3| is the distance of x
from 3. The point x = 2 is equidistant from 1 and 3. Hence, the solutions consist of all x a 2.
Fig. 2-2
8
CHAPTER 2
2.32
2.33
2.34
2.35
2.36