#(3x+y)/8 = (x-y)/5 = (x^2-y^2)/5#

Start with #(x-y)/5 = (x^2-y^2)/5#. Multiply by #5# and factor the right side:

#(x-y) = (x - y)(x+y)#.

Collect on one side:

#(x - y)(x+y) - (x-y) = 0 #.

Factor #(x-y)#

#(x - y)(x+y - 1) = 0 #.

So #x-y=0# or #x+y-1 = 0#

This gives us: #y=x# or #y = 1-x#

Now use the first two expressions together with these solutions for #y#.

#(3x+y)/8 = (x-y)/5#

Leads to: #15x+5y=8x-8y#.

So #7x+13y =0#

**Solution 1**

Now, when #y=x#, we get #20x = 0#, so #x=0# and thus #y=0#

**Solution 2**

When #y=1-x#, we get

#7x+13(1-x)=0#

#7x + 13 -13x =0#

#-6x = -13#

#x=13/6# and

#y = 1-x = 1- 13/6 = -7/6#

**Checking these solutions**

#(3x+y)/8 = (x-y)/5 = (x^2-y^2)/5#

For #(0,0)#, we get

#0/8 = 0/5 =0/5#

For #(13/6, -7/6)#, we get:

#(3(13/6)+(-7/6))/8 = (39-7)/48 = 32/48 = 2/3#

#((13/6)-(-7/6))/5 = 20/30 = 2/3#

#((13/6)^2-(-7/6)^2)/5 = (169 - 49)/(36*5) = 120/(36*5) = 20/(6*5) = 2/3#