Let $$A = \left[\begin{array}{cc} 5 &-10\cr 7 &-13\cr 1 &-4 \end{array}\right] \ \mbox{ and } \ \vec{b} = \left[\begin{array}{c} -35\cr -47\cr -11 \end{array}\right].$$ A linear transformation $T: {\mathbb R}^2 \rightarrow {\mathbb R}^3$ is defined by $T(x) = Ax$. Find an $\vec{x}$ in ${\mathbb R}^2$ whose image under $T$ is $\vec{b}$.

$\left[\begin{array}{c} x_{1}\cr x_{2}\cr \end{array}\right] =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$