Answer
$$\left\{ {\left( {1,0} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& {b^2}x + {a^2}y = {b^2} \cr
& ax + by = a \cr
& {\rm{Given\, the\, system}} \cr
& \left\{ {\matrix{
{{a_1}x + {b_1}y = {c_1}} \cr
{{a_2}x + {b_2}y = {c_2}} \cr
} } \right. \cr
& D = \left| {\matrix{
{{a_1}} & {{b_1}} \cr
{{a_2}} & {{b_2}} \cr
} } \right|,\,\,\,{D_x} = \left| {\matrix{
{{c_1}} & {{b_1}} \cr
{{c_2}} & {{b_2}} \cr
} } \right|,\,\,{D_y} = \left| {\matrix{
{{a_1}} & {{c_1}} \cr
{{a_2}} & {{c_2}} \cr
} } \right| \cr
& {\rm{First\, find\, }}D,{\rm{ If }}D \ne 0,\,{\rm{ \,then\, find \,}}{D_x}{\rm{ \,and\, }}{D_y} \cr
& D = \left| {\matrix{
{{b^2}} & {{a^2}} \cr
a & b \cr
} } \right| = {b^3} - {a^3} \cr
& {D_x} = \left| {\matrix{
{{b^2}} & {{a^2}} \cr
a & b \cr
} } \right| = {b^3} - {a^3} \cr
& {D_y} = \left| {\matrix{
{{b^2}} & {{b^2}} \cr
a & a \cr
} } \right| = a{b^2} - a{b^2} = 0 \cr
& {\rm{Using \,the \,Cramer's\, rule}} \cr
& x = {{{D_x}} \over D} = {{{b^3} - {a^3}} \over {{b^3} - {a^3}}} = 1 \cr
& \,\,y = {{{D_y}} \over D} = {0 \over {{b^3} - {a^3}}} = 0 \cr
& {\rm{The\, solution\, set\, is}} \cr
& \left\{ {\left( {1,0} \right)} \right\} \cr} $$
