Note as well that we also have 0 0 mathematical

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Unformatted text preview: ator : x = 1 The number line for this problem is, So, as with the polynomial inequalities we can just assume that the regions will always alternate in sign. Also, note that while the middle two regions do give negative values in the rational expression we need to avoid x = 1 to make sure we don’t get division by zero. This means that we will have to write the answer as two inequalities and/or intervals. -4 < x < 1 and 1< x < 4 ( -4,1) and (1, 4 ) Once again, it’s important to note that we really do need to test each region and not just assume that the regions will alternate in sign. Next we need to take a look at some examples that don’t already have a zero on one side of the inequality. Example 4 Solve 3x + 1 ³ 1. x+4 Solution The first thing that we need to do here is subtract 1 from both sides and then get everything into a single rational expression. © 2007 Paul Dawkins 137 College Algebra 3x + 1 -1 ³ 0 x+4 3x + 1 x + 4 ³0 x+4 x+4 3x + 1 - ( x + 4 ) ³0 x+4 2x - 3 ³0 x+4 In this case...
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