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Absolute Value Inequalities: Example 3
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Solve |6 –5a| < 10
Use the button to refer to the "Help" page.
Will (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 in the usual order of smallest to largest we get
, or
–0.8 < a < 3.2
Are the points at x = – 0.8 and x = 3.2 included in the intervals?
No.
How do we indicate that in a diagram?
We show open circles at those points not included in the interval.
Show the solution set on the number line.
–0.8 3.2 <......°———————°.......>
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
The end. If you found this helpful and would recommend that I create more pages like this one, please let me know:
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General Contents
Detailed Contents
Index