Answer
$$\left\{ {\left( { - 1,2} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& ax + by = c \cr
& dx + ey = f \cr
& {\rm{Given \,the\, system}} \cr
& \left\{ {\matrix{
{{a_1}x + {b_1}y = {c_1}} \cr
{{a_2}x + {b_2}y = {c_2}} \cr
} } \right. \cr
& D = \left| {\matrix{
{{a_1}} & {{b_1}} \cr
{{a_2}} & {{b_2}} \cr
} } \right|,\,\,\,{D_x} = \left| {\matrix{
{{c_1}} & {{b_1}} \cr
{{c_2}} & {{b_2}} \cr
} } \right|,\,\,{D_y} = \left| {\matrix{
{{a_1}} & {{c_1}} \cr
{{a_2}} & {{c_2}} \cr
} } \right| \cr
& {\rm{First\, find \,}}D,{\rm{ If }}D \ne 0,\,{\rm{ then\, find\, }}{D_x}{\rm{ and }}{D_y} \cr
& D = \left| {\matrix{
a & b \cr
d & e \cr
} } \right| = ae - bd \cr
& {D_x} = \left| {\matrix{
c & b \cr
f & e \cr
} } \right| = ce - bf \cr
& {D_y} = \left| {\matrix{
a & c \cr
d & f \cr
} } \right| = af - cd \cr
& {\rm{Using \,the \,Cramer's\, rule}} \cr
& x = {{{D_x}} \over D} = {{ce - bf} \over {ae - bd}} \cr
& \,\,y = {{{D_y}} \over D} = {{af - cd} \over {ae - bd}} \cr
& {\rm{Let\, }}a = 1,\,\,b = 2,\;c = 3,\,\,d = 4,\,\,\,e = 5,\,\,f = 6 \cr
& x = {{15 - 12} \over {5 - 8}} = - 1 \cr
& y = {{6 - 12} \over {5 - 8}} = 2 \cr
& {\rm{The \,solution\, set\, is}} \cr
& \left\{ {\left( { - 1,2} \right)} \right\} \cr} $$
