Talk:Intermediate Algebra/Solving Inequalities

In the "Special cases" section where the technique of switching signs is mentioned, there is an error in the given example. It says x => -6 and then switched to x <= -6 After the switch, the -6 should become positive. Isn't that right?


 * Yes, but it appears that was intentional after reading further. 67.111.137.227 (talk) 22:35, 27 January 2010 (UTC)


 * I think the point is that the first answer of x => -6 was wrong. If you multiply or divide both sides of an inequality by a negative number, then you need to turn round the inequality. e.g. 4 => 3 becomes -4 <= -3 etc. Perhaps the expression "switching the sign" is confusing. Recent Runes (talk) 00:14, 28 January 2010 (UTC)

Inequalities-special cases.
The presenter recommends solving 2/(x-1)<1 in a somewhat lengthy way. I suggest it be replaced by the following. x-1=0 when x=1, when x>1 denominator >0 and multiplying both sides by (x-1) gives 2<2x-2 or x>2 which requires x>2. When x<1 denominator <0 and multiplying by (x-1) gives 2>2x-2 or x<2 which requires x<1. Overall this requires x<1 and (or) x>2. I think an expression such as 1>x>2 should be avoided.