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