In each quadrant we have the following signs for x

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Unformatted text preview: value to ever be zero. So, we don’t care what value the absolute value takes as long as it isn’t zero. This means that we just need to avoid value(s) of x for which we get, 3x - 9 = 0 Þ 3x - 9 = 0 Þ x=3 The solution in this case is all real numbers except x = 3 . Now, let’s do a quick set of examples with negative numbers. Example 4 Solve each of the following. (a) 4 x + 15 < -2 and 4 x + 15 £ -2 (b) 2 x - 9 ³ -8 and 2 x - 9 > -8 Solution Notice that we’re working these in pairs, because this time, unlike the previous set of examples the solutions will be the same for each. © 2007 Paul Dawkins 150 College Algebra Both (all four?) of these will make use of the fact that no matter what p is we are guaranteed to have p ³ 0 . In other words, absolute values are always positive or zero. (a) Okay, if absolute values are always positive or zero there is no way they can be less than or equal to a negative number. Therefore, there is no so...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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