abs(2x-1 -abs(x+5=3
thank you
Re-write
$\displaystyle |2x-1|=\left\{\begin{array}{lcr}2x-1&if&x\geq\dfrac12 \\ -(2x-1)&if& x<\dfrac12\end{array}\right.$
$\displaystyle |x+5|=\left\{\begin{array}{lcr}x+5&if&x\geq-5\\ -(x+5)&if& x<-5\end{array}\right.$
Combine the domains (x<-5; -5<x<1/2; x>1/2). You'll get 3 different linear equations.
The final solution is $\displaystyle x=-\dfrac73~\vee~x=9$
As I've told you there are three distinct intervals. To each interval belongs one equation without any absolute values:
A)$\displaystyle x<-5:~~~~~~~~-(2x-1)-(-(x+5))=3$
B) $\displaystyle -5\leq x < \frac12:~-(2x-1)-(x+5)=3$
C) $\displaystyle x\geq \frac12:~~~~~~~~~~~(2x-1)-(x+5)=3$
Expand the brackets and solve for x. Afterwards you have to check if the solution is possible:
A) $\displaystyle -(2x-1)-(-(x+5))=3~\iff~-x+6=3~\iff~x=3$.......BUT: 3 is not smaller than -5
B) $\displaystyle -(2x-1)-(x+5)=3~\iff~-3x-4=3~\iff~-3x=7~\iff~x=-\frac73$
I'll leave the last one for you.
Btw: If you need some additional informations ask in the original thread, please.