Re: Help with this!!!! please

*To*: mathgroup at smc.vnet.net*Subject*: [mg4190] Re: [mg4139] Help with this!!!! please*From*: Allan Hayes <hay at haystack.demon.co.uk>*Date*: Thu, 13 Jun 1996 23:08:53 -0400*Sender*: owner-wri-mathgroup at wolfram.com

"Erick Houli Katz" <ehk at netpoint.net> [mg4139] Help with this!!!! please writes >How can I solve this exercise??? > >|1-2X|+|3X+1|=2 Erick: |a| = a when a >= 0 and -a when a <=0 (or -a>=0) So there are any solutin of your equation will satisfy one of the following linear equations -1 (1-2x) + -1(3x+1) == 2 , -1(1-2x) + 1 (3x+1) == 2 , 1 (1-2x) + -1(3x+1) == 2 , -1(1-2x) + -1 (3x+1) == 2 One method is to solve all of these and then keep only those solutions that satisfy the original equation. eqn = Abs[1-2x]+Abs[3x+1]==2; lst = {1-2x, 3x+1}; mult = Flatten[Outer[List,{-1,1},{-1,1}],1] (*pairs of -1's and 1's*) {{-1, -1}, {-1, 1}, {1, -1}, {1, 1}} redeqs = (lst.# == 2)&/@mult (*all the linear equations*) {-2 - x == 2, 5 x == 2, -5 x == 2, 2 + x == 2} pos = Solve/@red (*solutions of equations redeqs*) 2 2 {{{x -> -4}}, {{x -> -}}, {{x -> -(-)}}, {{x -> 0}}} 5 5 Cases[pos, s_/;(eqn/.s} == {True}] (*keep those that satisfy eqn*) 2 {{{x -> -(-)}}, {{x -> 0}}} 5 We can put the last few steps together: Cases[ Solve[lst.# == 2]&/@Flatten[Outer[List,{-1,1},{-1,1}],1], s_/;(eqn/.s) == {True} ] 2 {{{x -> -(-)}}, {{x -> 0}}} 5 A second way is based on the following idea Let R[-1,-1] be the region of the real line where -1(1-2x) >=0 and -1 (3x+1) >=0 Any solution to eqn in this region will be a solution to -1(1-2x) + -1 (3x+1) = 2 The only solution to the last equation is -4. But -1(1-2 (-4)) = -9, which is <0; so there is no solution to eqn in this region. Let R[1,-1] the region of the real line where 1(1-2x) >=0 and -1 (3x+1) >=0 Any solution to eqn in this part will be a solution to 1(1-2x) + -1 (3x+1) = 2 The only solution to the last equation is -2/5. Now 1(1-2 (-2/5)) = 9/5, and -1 (3(-2/5)+1) = 1/5; so -2/5 is a solution of eqn in this region Similarly for the anologously defined R[-1,1] and R[1,1]. Since these four regions cover the line we will in this way find all the solutions of eqn. . The work can be done by Mathematica: If[ Min[(lst #)/.(sl = Solve[lst.# == 2,x])]>=0, sl, Unevaluated[Sequence[]] (*empty sequence*) ]&/@Flatten[Outer[List,{-1,1},{-1,1}],1] 2 {{{x -> -(-)}}, {{x -> 0}}} 5 Allan Hayes hay at haystack.demon.co.uk 08-06-96 ==== [MESSAGE SEPARATOR] ====