solve the equatio

**hoger** · November 5th 2008, 03:52 PM

abs(2x-1 -abs(x+5=3

thank you

**earboth** · November 5th 2008, 10:32 PM

Originally Posted by hoger

abs(2x-1 -abs(x+5=3

thank you

Re-write

$|2x-1|=\left\{\begin{array}{lcr}2x-1&if&x\geq\dfrac12 \\ -(2x-1)&if& x<\dfrac12\end{array}\right.$

$|x+5|=\left\{\begin{array}{lcr}x+5&if&x\geq-5\\ -(x+5)&if& x<-5\end{array}\right.$

Combine the domains (x<-5; -5<x<1/2; x>1/2). You'll get 3 different linear equations.

The final solution is $x=-\dfrac73~\vee~x=9$

**earboth** · November 10th 2008, 04:45 AM

As I've told you there are three distinct intervals. To each interval belongs one equation without any absolute values:

A) $x<-5:~~~~~~~~-(2x-1)-(-(x+5))=3$

B) $-5\leq x < \frac12:~-(2x-1)-(x+5)=3$

C) $x\geq \frac12:~~~~~~~~~~~(2x-1)-(x+5)=3$

Expand the brackets and solve for x. Afterwards you have to check if the solution is possible:

A) $-(2x-1)-(-(x+5))=3~\iff~-x+6=3~\iff~x=3$ .......BUT: 3 is not smaller than -5

B) $-(2x-1)-(x+5)=3~\iff~-3x-4=3~\iff~-3x=7~\iff~x=-\frac73$

I'll leave the last one for you.

Btw: If you need some additional informations ask in the original thread, please.

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