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

Algebra/Trig Review4x+ 534x−2x −2So the solution to this inequality will be x’s that are less than -2 or greater than1− .2Now, as I mentioned earlier you CAN NOT write the solution as the following doubleinequality.1− 2 > x >−2When you write a double inequality (as we have here) you are saying that x will be anumber that will simultaneously satisfy both parts of the inequality. In other words,in writing this I’m saying that x is some number that is less than -2 and AT THE1SAME TIME is greater than − . I know of no number for which this is true. So,2this is simply incorrect. Don’t do it. This is however, a VERY common mistake thatstudents make when solving this kinds of inequality.8. 4 −11x≥ 9SolutionNot much to this solution. Just be careful when you divide by the -11.4 −11x≤−9 4 −11x≥9−11x≤−13 or −11x≥513 5x≥x≤−11 119. 10x + 1 >− 4SolutionThis is another problem along the lines of Problems 3 and 6. However, the answerthis time is VERY different. In this case we are looking for x’s that when plugged inthe absolute value we will get back an answer that is greater than -4, but sinceabsolute value only return positive numbers or zero the result will ALWAYS begreater than any negative number. So, we can plug any x we would like into thisabsolute value and get a number greater than -4. So, the solution to this inequality isall real numbers.Trigonometry© 2006 Paul Dawkins 45http://tutorial.math.lamar.edu/terms.aspx