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

Algebra/Trig Review−4 ≤7x−10 ≤46 ≤7x≤146≤ x ≤27In solving these make sure that you remember to add the 10 to BOTH sides of theinequality and divide BOTH sides by the 7. One of the more common mistakes hereis to just add or divide one side.5. 1− 2x< 7SolutionThis one is identical to the previous problem with one small difference.− 7< 1− 2x< 7− 8 x >−3Don’t forget that when multiplying or dividing an inequality by a negative number (-2in this case) you’ve got to flip the direction of the inequality.6. x −9 ≤− 1SolutionThis problem is designed to show you how to deal with negative numbers on theother side of the inequality. So, we are looking for x’s which will give us a number(after taking the absolute value of course) that will be less than -1, but as withProblem 3 this just isn’t possible since absolute value will always return a positivenumber or zero neither of which will ever be less than a negative number. So, thereare no solutions to this inequality.7. 4x + 5 > 3SolutionAbsolute value inequalities involving > and ≥ are solved as follows.p > d ⇒ pdp ≥d ⇒ p≤−d or p≥dNote that you get two separate inequalities in the solution. That is the way that itmust be. You can NOT put these together into a single inequality. Once I get thesolution to this problem I’ll show you why that is.Here is the solution© 2006 Paul Dawkins 44http://tutorial.math.lamar.edu/terms.aspx