On solving $\mathrm{x}+2 \mathrm{y}=2, \mathrm{y}-\mathrm{x}=1$ and $2 \mathrm{x}+\mathrm{y}=7$,
$\therefore$ Pts of intersection are $(0,1),(2,3)$ and $(4,-1)$


$$
2(y-1)+y=7
$$
$\therefore$ Required area
$=\int_{-1}^1-(2-2 y) d y+\int_{-1}^3 \frac{(7-y)}{2} d y-\int_1^3(y-1) d y$
$=\left[-2 y+\frac{2 y^2}{2}\right]_{-1}^1+\left[\frac{7 y}{2}-\frac{y^2}{2.2}\right]_{-1}^3-\left[\frac{y^2}{2}-y\right]_1^3$
$=\left[-2+\frac{2}{2}-2-\frac{2}{2}\right]+\left[\frac{21}{2}-\frac{9}{4}+\frac{7}{2}+\frac{1}{4}\right]-\left[\frac{9}{2}-3-\frac{1}{2}+1\right]$
$=-4+12-2=6$ sq units