$\text{a)}2x-\left|x+1\right|=5$  

Z definicji wartości bezwzględnej:

$\left|x+1\right|=\{\begin{array}{ll}x+1& \text{dla}x\ge -1\\ -x-1& \text{dla}x<-1\end{array}$   

Stąd:

$\{\begin{array}{l}x\ge -1\\ 2x-\left(x+1\right)=5\end{array}\vee \{\begin{array}{l}x<-1\\ 2x-\left(-x-1\right)=5\end{array}$  

$\{\begin{array}{l}x\ge -1\\ 2x-x-1=5\end{array}\vee \{\begin{array}{l}x<-1\\ 2x+x+1=5\end{array}$  

$\{\begin{array}{l}x\ge -1\\ x=6\end{array}\vee \{\begin{array}{l}x<-1\\ 3x=4\text{/}:3\end{array}$  

$\{\begin{array}{l}x\ge -1\\ x=6\end{array}\vee \{\begin{array}{l}x<-1\\ x=\frac{4}{3}>-1\end{array}$  

$\text{Odp.}x=6.$