If [tex]x[/tex] and [tex]y-1[/tex] have the same sign, then either
[tex]x>0,y>1 \implies 2|x| + 3|y-1| = 2x + 3(y-1)=6 \implies 2x + 3y = 9[/tex]
or
[tex]x<0,y<1 \implies 2|x| + 3|y-1| = -2x - 3(y-1) = 6 \implies 2x + 3y = -3[/tex]
If [tex]x[/tex] and [tex]y-1[/tex] have opposite sign, then
[tex]x>0,y<1 \implies 2|x| + 3|y-1| = 2x - 3(y-1) = 6 \implies 2x -3y = 3[/tex]
or
[tex]x<0,y>1 \implies 2|x| + 3|y-1| = -2x + 3(y-1) = 6 \implies 2x-3y = -9[/tex]
This is to say that the region has boundaries given by these two sets of parallel lines, so we can equivalently describe the region with the set
[tex]R = \left\{(x,y) \mid -3\le2x+3y\le9 \text{ and } -9\le2x-3y\le3\right\}[/tex]
The area of [tex]R[/tex] is given by the double integral
[tex]\displaystyle \iint_R dx\,dy[/tex]
To compute the area, change the variables to
[tex]\begin{cases}u = 2x + 3y \\ v = 2x - 3y\end{cases} \implies \begin{cases}x = \frac14(u+v) \\ y = \frac16(u-v)\end{cases}[/tex]
The Jacobian for this transformation is
[tex]J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix}1/4 & 1/4 \\ 1/6 & -1/6\end{bmatrix}[/tex]
with determinant [tex]\det(J) = -\frac1{12}[/tex]. Then the integral transforms to
[tex]\displaystyle \iint_R dx\,dy = \iint_R |J| \, du \, dv = \frac1{12} \int_{-3}^9 \int_{-9}^3 dv\, du[/tex]
which is 1/12 the area of a square with side length 12. Hence the integral evaluates to
[tex]\displaystyle \iint_R dx\,dy = \frac1{12}\times12^2 = \boxed{12}[/tex].
