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