• 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.
>
>
>

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