Algebra/Trig Review - Pauls Online Math Notes - Lamar University

Algebra/Trig ReviewAbsolute Value1. Evaluate 5 and − 123SolutionTo do these evaluations we need to remember the definition of absolute value.⎧ p if p≥0p = ⎨⎩ − p if p < 0With this definition the evaluations are easy.5 = 5 because 5 ≥ 0( )− 123 =− − 123 = 123 because -123 < 0Remember that absolute value takes any nonzero number and makes sure that it’spositive.2. Eliminate the absolute value bars from 3−8xSolutionThis one is a little different from the first example. We first need to address a verycommon mistake with these.3−8x≠ 3+8xAbsolute value doesn’t just change all minus signs to plus signs. Remember thatabsolute value takes a number and makes sure that it’s positive or zero. To convinceyourself of this try plugging in a number, say x = − 1083 = 83 = 3+ 80 = 3−8( −10) ≠ 3 + 8( − 10)= 3 − 80 =− 77There are two things wrong with this. First, is the fact that the two numbers aren’teven close to being the same so clearly it can’t be correct. Also note that if absolutevalue is supposed to make nonzero numbers positive how can it be that we got a -77of out of it? Either one of these should show you that this isn’t correct, but togetherthey show real problems with doing this, so don’t do it!That doesn’t mean that we can’t eliminate the absolute value bars however. We justneed to figure out what values of x will give positive numbers and what values of xwill give negative numbers. Once we know this we can eliminate the absolute valuebars. First notice the following (you do remember how to Solve Inequalities right?)33−8x≥0 ⇒ 3≥8x ⇒ x≤833− 8x< 0 ⇒ 3< 8x ⇒ x>833So, if x ≤ then 3−8x≥ 0 and if x > then 3− 8x< 0. With this information we88can now eliminate the absolute value bars.© 2006 Paul Dawkins 6http://tutorial.math.lamar.edu/terms.aspx