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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.
>
>
>
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