[tex]\displaystyle\iint_R(9x-3y)(2x-y)\,\mathrm dA[/tex]
[tex]\begin{cases}u=x-3y\\v=2x-y\end{cases}\implies\begin{cases}x=\frac{3v-u}5\\y=\frac{v-2u}5\end{cases}\implies\mathrm dA=|\det J|\,\mathrm du\,\mathrm dv[/tex]
where [tex]J[/tex] is the Jacobian matrix for the transformation,
[tex]J=\begin{bmatrix}u_x&u_y\\v_x&v_y\end{bmatrix}=\begin{bmatrix}-\dfrac15&\dfrac35\\\\-\dfrac25&\dfrac15\end{bmatrix}\implies|\det J|=\dfrac15[/tex]
We have
[tex]9x-3y=8x+x-3y=\dfrac85(3v-u)+u=\dfrac{24v-3u}5[/tex]
so that the integral is
[tex]\displaystyle\frac1{25}\int_9^{10}\int_0^{10}(24v-3u)v\,\mathrm du\,\mathrm dv=\boxed{\frac{4051}5}[/tex]
