Answer
$$\left\{ {\left( {{{b - ab} \over {1 - ab}},{{a - ab} \over {1 - ab}}} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& x + by = b \cr
& ax + y = a \cr
& {\rm{Given \,the\, system}} \cr
& \left\{ {\matrix{
{{a_1}x + {b_1}y = {c_1}} \cr
{{a_2}x + {b_2}y = {c_2}} \cr
} } \right. \cr
& D = \left| {\matrix{
{{a_1}} & {{b_1}} \cr
{{a_2}} & {{b_2}} \cr
} } \right|,\,\,\,{D_x} = \left| {\matrix{
{{c_1}} & {{b_1}} \cr
{{c_2}} & {{b_2}} \cr
} } \right|,\,\,{D_y} = \left| {\matrix{
{{a_1}} & {{c_1}} \cr
{{a_2}} & {{c_2}} \cr
} } \right| \cr
& {\rm{First\, find\, }}D,{\rm{ If }}D \ne 0,\,{\rm{ then\, find\, }}{D_x}{\rm{ and }}{\,D_y} \cr
& D = \left| {\matrix{
1 & b \cr
a & 1 \cr
} } \right| = 1 - ab \cr
& {D_x} = \left| {\matrix{
b & b \cr
a & 1 \cr
} } \right| = b - ab \cr
& {D_y} = \left| {\matrix{
1 & b \cr
a & a \cr
} } \right| = a - ab \cr
& {\rm{Using \,the\, Cramer's\, rule}} \cr
& x = {{{D_x}} \over D} = {{b - ab} \over {1 - ab}} \cr
& \,\,y = {{{D_y}} \over D} = {{a - ab} \over {1 - ab}} \cr
& {\rm{The\, solution\, set\, is}} \cr
& \left\{ {\left( {{{b - ab} \over {1 - ab}},{{a - ab} \over {1 - ab}}} \right)} \right\} \cr} $$
