Lets take a look at a couple of examples example 1

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Unformatted text preview: b) 4 x + 3 = 3 - x This one will work in pretty much the same way so we won’t put in quite as much explanation. 4 x + 3 = - ( 3 - x ) = -3 + x 4x + 3 = 3 - x or or 3 x = -6 x = -2 or 5x = 0 x=0 Now, before we check each of these we should give a quick warning. Do not make the assumption that because the first potential solution is negative it won’t be a solution. We only © 2007 Paul Dawkins 144 http://tutorial.math.lamar.edu/terms.aspx College Algebra exclude a potential solution if it makes the portion without absolute value bars negative. In this case both potential solutions will make the portion without absolute value bars positive and so both are in fact solutions. So in this case, unlike the first example, we get two solutions : x = -2 and x = 0 . [Return to Problems] (c) 2 x - 1 = 4 x + 9 This case looks very different from any of the previous problems we’ve worked to this point and in this case the formula we’ve been using doesn’t really work at all. However, if we think about this a little we can see that we’ll still do something similar here to get a solution. Both sides of the equation have contain absolu...
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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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