Use the information in the right-hand frame to solve the following:
Prob 2.4.9': Find the solution set for |x| ³ 7.
Which case is this?
There will be two intervals.
Express the solution set.
The solution set is x
£ -7 or x ³ 7Show the solution on the number line.
-7 7 <—————•.....•—————>
Prob 2.4.23':
Solve |a - 7| < 4Will the (a - 7) be between -4 and 4 or will it be outside that interval?
(a - 7) will be between -4 and 4.
Set this up.
-4 < a - 7 < 4
What next?
Isolate a by adding 7 everywhere.
Do it.
-4 + 7 < a < 4 + 7, or
3 < a < 11
Check with values near the ends of this interval in |a - 7| < 4.
Try a = 4: |4 - 7| = | -3 | = 3 < 4 OK
Try a = 10: |10 - 7| = | 3 | = 3 < 4 OK
Show the solution set on the number line.
3 11 <.....•———————•.....>
Prob 2.4.39':
Solve |6 -5a| < 10Will (6 - 5a) be between -10 and 10, or outside this interval?
(6 - 5a) lies between -10 and 10.
Set this up.
-10 < 6 - 5a < 10
Subtract 6 everywhere.
-10 - 6 < -5a < 10 - 6, or
-16 < -5a < 4
What do we do next?
Divide by -5.
Is that all?
No, we have to reverse the direction of the inequalities when we divide by a negative quantity.
Do it
, or
Check an easy point near the left end of the interval.
is near the left end:
OK
Check an easy value near the right end of the interval.
3 is near the right end:
|6 - 5*3| = |6 - 15| = | -9 | = 9 < 10 OK
Show the solution set on the number line.
-0.8 3.2 <......•———————•.......>