#{(5x+2y=12" S1"),(-6x-2y=-14" S2"):}#

Looking at the equation we can see that in #S1# we have #ay# and in #S2# we have #-ay#, thus we could use the Linear Systems with Addition or Subtraction to transform the system as :

#{(S1),(S1+S2):}#

#{(5x+2y=12),(-x+0y=-2):}=>{(5x+2y=12),(x=2):}#

Now you can put the #x# value in #S1# to find the #y# value

#{(5*2+2y=12),(x=2):}=>{(10+2y=12),(x=2):}=>{(5x+2y=12),(x=2):}#

#=>{(2y=12-10),(x=2):}=>{(y=cancel(2)^1/cancel(2)^1),(x=2):}#

#=>{(x=2),(y=1):}#

graph{(5x+2y-12)(-6x-2y+14)=0 [-1.413, 5.518, -1.016, 2.45]}