$4x - 3y = 3,3x - 5y = 7.$

The given system of equations can be written in the form $AX = B$,

where
$A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&{ - 3}\\3&{ - 5}\end{array}} \right],X = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x\\y\end{array}} \right]$

and $B = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3\\7\end{array}} \right]$

Now, $|A| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&{ - 3}\\3&{ - 5}\end{array}} \right| = - 20 + 9 = - 11 \ne 0$

$\Rightarrow$ A is a non-singular matrix and so the given system has a unique solution.

Cofactors of elements of A are given by

${A_{11}} = {( - 1)^{1 + 1}}( - 5) = - 5,{A_{12}} = {( - 1)^{1 + 2}}(3) = - 3,$

${A_{21}} = {( - 1)^{2 + 1}}( - 3) = 3,{A_{22}} = {( - 1)^{2 + 2}}(4) = 4$

% $\therefore$ $adj\;A = {\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 5}&{ - 3}\\3&4\end{array}} \right]^{\prime}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 5}&3\\{ - 3}&4\end{array}} \right]$

and ${A^{ - 1}} = \cfrac{1}{{|A|}}(adj\;A) = \cfrac{1}{{ - 11}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 5}&3\\{ - 3}&4\end{array}} \right]$

Solution of given system is given by $X = {A^{ - 1}}B.$

$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x\\y\end{array}} \right] = - \cfrac{1}{{11}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 5}&3\\{ - 3}&4\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3\\7\end{array}} \right]$

$= - \cfrac{1}{{11}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 15 + 21}\\{ - 9 + 28}\end{array}} \right] = \cfrac{1}{{ - 11}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}6\\{19}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\cfrac{{ - 6}}{{11}}}\\{\cfrac{{ - 19}}{{11}}}\end{array}} \right]$

Hence, $x = \cfrac{{ - 6}}{{11}},y = \cfrac{{ - 19}}{{11}}$