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(b) x  9 £ 0 [Solution]
(c) 2 x  4 ³ 0 [Solution]
(d) 3 x  9 > 0 [Solution]
Solution
These four examples seem to cover all our bases.
(a) Now we know that p ³ 0 and so can’t ever be less than zero. Therefore, in this case there is
no solution since it is impossible for an absolute value to be strictly less than zero (i.e. negative).
(b) This is almost the same as the previous part. We still can’t have absolute value be less than
zero, however it can be equal to zero. So, this will have a solution only if x9 = 0
and we know how to solve this from the previous section. x9 = 0 Þ x=9 (c) In this case let’s again recall that no matter what p is we are guaranteed to have p ³ 0 . This
means that no matter what x is we can be assured that 2 x  4 ³ 0 will be true since absolute
values will always be positive or zero.
The solution in this case is all real numbers, or all possible values of x. In inequality notation this
would be ¥ < x < ¥ .
(d) This one is nearly identical to the previous part except this time note that we don’t want the
absolute...
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 Spring '12
 MrVinh

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