There is a problem with the second one however if we

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Unformatted text preview: olute value does not just make all minus signs into plus signs! To solve these we’ve got to use the formula above since in all cases the number on the right side of the equal sign is positive. (a) 2 x - 5 = 9 There really isn’t much to do here other than using the formula from above as noted above. All we need to note is that in the formula above p represents whatever is on the inside of the absolute value bars and so in this case we have, 2 x - 5 = -9 2x - 5 = 9 or At this point we’ve got two linear equations that are easy to solve. 2 x = -4 x = -2 2 x = 14 x=7 or or So, we’ve got two solutions to the equation x = -2 and x = 7 . [Return to Problems] (b) 1 - 3t = 20 This one is pretty much the same as the previous part so we won’t put as much detail into this one. 1 - 3t = -20 - 3t = -21 or or t=7 or The two solutions to this equation are t = - 1 - 3t = 20 - 3t = 19 19 t=3 19 and t = 7 . 3 [Return to Problems] © 2007 Paul Dawkins 142 http://tutorial.math.lamar.edu/terms.aspx College Algebra (c) 5 y - 8 = 1 Again, not much more to this one. 5 y...
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