I'll use Gabriel Cramer's method to solve the system of equations.

First of all, let #x+y=p# and #x-y=q#

Then the first equation is equal to

#5/p-2/q+1=0#

or, #5/p-2/q=-1#

LET #a_1=5#, #b_1=-2#, #c_1=-1#

Also the second equation is equal to

#15/p+7/q-10=0#

or, #15/p+7/q=10#

LET #a_2=15#, #b_2=7#, #c_2=10#

Now we have,

#Delta=|(a_1,b_1),(a_2,b_2)|=|(5,-2),(15,7)|=7*5-(-2)**15#

#=35+30=65#

Also,

#Delta_1=|(c_1,b_1),(c_2,b_2)|=|(-1,-2),(10,7)|=-1*7-(-2)*10#

#=20-7=13#

Similarly, #Delta_2=|(a_1,c_1),(a_2,c_2)|=|(5,-1),(15,10)|=10*50-(-1)*15#

#=50+15=65#

According to Cramer's rule

#1/p=Delta_1/Delta#

or, #p=Delta/Delta_1=65/13=5#

Similarly , #1/q=Delta_2/Delta#

or #q=Delta/Delta_2=65/65=1#

Putting #p= x+y# and #q= x-y# we have

#x+y=5#....equation(1)

and #x-y=1#.....equation(2)

From equation(1) we have,

#x=5-y#....equation...(3)

Putting the value of #x# from e#qn(3)# to #eqn(2)# we get

#5-y-y=1#

#2y=4#

#y=2#

Putting the value of #y# in #eqn(3)# we get

#x=5-y=5-2=3#

So the value #x# is #3# and #y# is #2#.