(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
