Area enclosed by the curve $\left|x+y-1\right|+\left|2x+y-1\right|=1$  is

If both positive, then$x+y-1+2x+y-1=1$ 

$\Rightarrow 3x+2y=3\Rightarrow \frac{x}{1}+\frac{y}{3/2}=1$

First positive and second negative, then$\left(x+y-1\right)-\left(2x+y-1\right)=1$

                                                $\Rightarrow -x-1=0$

                                                $\Rightarrow x=1$

If first negative and second positive, then$-\left(x+y-1\right)+\left(2x+y-1\right)=1$

                                               $\Rightarrow x-1=0$

                                             $\Rightarrow x=1$

If both negative, then $-\left(x+y-1\right)+\left(2x+y-1\right)=1$

                                          $\Rightarrow -3x-2y+1=0$

                                          $\Rightarrow \frac{x}{1/3}+\frac{y}{1/2}=1$

$\therefore$  Required Area $=\left(\frac{1}{2}\times 2\times 3\right)-\frac{1}{2}\left(\frac{4}{3}\times 2\right)+\frac{1}{2}\times 1\times \frac{2}{3}$

                                $=3-\frac{4}{3}+\frac{1}{3}=3-1$

                                   = 2 sq unit.

Alternative method:

Required Area $={\displaystyle {\int }_{-1}^1\left[\left(\frac{3}{2}-x\right)-\left(\frac{1}{2}-\frac{3x}{2}\right)\right]dx}$

                            $={\displaystyle {\int }_{-1}^{1}1dx+\frac{1}{2}{\displaystyle {\int }_{-1}^{1}xdx}}$

                              = 2+0 = 2 sq unit