Re: Help with this!!!! please

*To*: mathgroup at smc.vnet.net*Subject*: [mg4202] Re: [mg4139] Help with this!!!! please*From*: Robert Pratt <rpratt at math.unc.edu>*Date*: Thu, 13 Jun 1996 23:11:01 -0400*Sender*: owner-wri-mathgroup at wolfram.com

In my previous post, I neglected to explain why two answers were extraneous in Method 1. METHOD 1: The given equation is really shorthand for four equations. (1) 1 - 2x + 3x + 1 = 2 when 1 - 2x >= 0 and 3x + 1 >= 0 -1/3 <= x <= 1/2 (2) 1 - 2x - 3x - 1 = 2 when 1 - 2x >= 0 and 3x + 1 <= 0 x <= -1/3 (3) -1 + 2x + 3x + 1 = 2 when 1 - 2x <= 0 and 3x + 1 >= 0 x >= 1/2 and x <= -1/3 (no x works) (4) -1 + 2x - 3x - 1 = 2 when 1 - 2x <= 0 and 3x + 1 <= 0 x >= 1/2 The solutions are (1) x = 0 (2) x = -2/5 (3) x = 2/5, but 2/5 < 1/2 and 2/5 > -1/3, so not a valid solution (4) x = -4, but -4 < 1/2, so not a valid solution However, only the first two actually satisfy the original equation since the solutions for (3) and (4) do not lie in the proper intervals. METHOD 2: | 1 - 2x | = 2 - | 3x + 1 | Square both sides. (1 - 2x)^2 = 4 - 4 | 3x + 1 | + (3x + 1)^2 Expand, transpose, and collect like terms to get 4 | 3x + 1 | = 5x^2 + 10x + 4 | 12x + 4 | = 5x^2 + 10x + 4 Now we really have two equations: (1) 5x^2 + 10x + 4 = 12x + 4 (2) 5x^2 + 10x + 4 = -12x - 4 Solve however you like (factoring or quadratic formula) (1) x = 0, x = 2/5 (2) x = -2/5, x= -4 Again, two answers are extraneous. We introduced them when we squared both sides of the equation above. Rob Pratt Department of Mathematics The University of North Carolina at Chapel Hill CB# 3250, 331 Phillips Hall Chapel Hill, NC 27599-3250 rpratt at math.unc.edu On Fri, 7 Jun 1996, Erick Houli Katz wrote: > How can I solve this exercise??? > > |1-2X|+|3X+1|=2 > > Thanks for any help. > > Erick. > > P.S.: Please don't forget to give me all the theory you use to solve it. > Thanks again. > > > ==== [MESSAGE SEPARATOR] ====