- d.Note that |a|cif and only if (acor
a–c). Using this fact, we see that the values
ofx that satisfy the inequality |x|3 are pre-
cisely those values ofxthat satisfy either x 3
or x–3. So, the solution set is (–∞,–3)∪(3,∞).
- a.First, note that |–2x| = |–1||2x| = |2x|.
Also, |a|cif and only if (acor a–c).
The values ofx that satisfy the inequality
|2x|0 are those that satisfy either 2x0 or
2 x0. Dividing both of these inequalities by
2 yields x0 or x0. So, the solution set is
(–∞,0)∪(0,∞).
- c.First, note that the inequality –|–x– 1| 0
is equivalent to |–x– 1| 0. Moreover, since
|–x – 1| = |–(x+ 1)| = |–1||x+ 1| = |x+ 1|,
this inequality is also equivalent to |x+ 1| 0.
The left side must be nonnegative since it is
the absolute value of a quantity. The only way
that it can be less than or equal to zero is if it
actually equalszero. This happens only when x
- 1 = 0, which occurs when x = –1.
- c.Note that |a| c if and only if (a cor a
- c). Using this fact, we see that the values ofx
that satisfy the inequality |8x+ 3| 3 are pre-
cisely those values ofxthat satisfy either 8x+
3 3 or 8x+ 3 –3. We solve these two
inequalities separately:
8 x+ 3 38 x+ 3 –3
8 x 08 x –6
x 0 x – ^68 = –^34
Thus, the solution set is [0,∞)∪(–∞,–^34 ]
- d.Note that |a|cif and only if –cac.
Using this fact, we see that the values ofx that
satisfy the inequality |2x–3|5 are precisely
those values ofxthat satisfy –5 2 x– 35.
We solve this compound inequality as follows:
–5 2 x–3 5
–2 2 x 8
–1x 4
Thus, the solution set is (–1,4).
- a.First, we rewrite the given inequality in an
equivalent form:
2 – (1 – (2 – |1 – 2x|))–6
2 – (1 – 2 + |1 – 2x|)–6
2 – (–1 + |1 – 2x|)–6
2 + 1 – |1 – 2x|–6
3 – |1 –2x|–6
–|1 – 2x|–9
|1 – 2x| 9
Now, note that |a|cif and only if if –ca
c.Using this fact, we see that the values ofx
that satisfy the inequality |1 –2x|9 are pre-
cisely those values ofxthat satisfy –91 – 2x
9. We solve this compound inequality:
–91 – 2x 9
–10–2x 8
5 x–4
So, the solution set is (–4,5).
- c.First, we rewrite the given inequality in an
equivalent form:
–7|1 – 4x| + 20 –2|1 – 4x| – 15
–7|1 – 4x| + 35 –2|1 – 4x|
35
5|1–4x|
7
|1 – 4x|
The last inequality is equivalent to |1 – 4x| 7.
Now, |a| cif and only if (a cor a – c).
Using this fact, we see that the values ofx that
satisfy the inequality |1 – 4x| 7 are precisely
those values ofxthat satisfy either 1 – 4x 7
ANSWERS & EXPLANATIONS–
