x+y=a+b
(x-a)/b=(y-b)/a
(x-a)/b=(y-b)/a
-
Note: the second equation disallows a = 0 or b = 0 (division by zero is not defined).
Multiply both sides of the second equation by a*b to get rid of the fractions:
a(x-a) = b(y-b)
Expand:
ax - a² = by - b²
ax - by = a² - b²
Multiply both sides of the first equation by b and add it to the above result:
bx + by = ab + b²
ax - by = a² - b²
--------------------------------------
(b+a)x = ab + a² = a(b+a)
Assuming b+a ≠ 0, divide both sides by (b+a); you get
x = a
Then from the first equation, y = b.
Suppose that a+b = 0 (so b = -a). Put that into the original system:
x + y = 0
(x-a)/(-a) = (y-(-a))/a
Multiply both sides of the second equation by a:
a - x = y + a
0 = x + y
which just repeats the first equation. Therefore, if a+b = 0, then there are infinitely many solutions for x and y: x = k, y = -k for all values of k.
Multiply both sides of the second equation by a*b to get rid of the fractions:
a(x-a) = b(y-b)
Expand:
ax - a² = by - b²
ax - by = a² - b²
Multiply both sides of the first equation by b and add it to the above result:
bx + by = ab + b²
ax - by = a² - b²
--------------------------------------
(b+a)x = ab + a² = a(b+a)
Assuming b+a ≠ 0, divide both sides by (b+a); you get
x = a
Then from the first equation, y = b.
Suppose that a+b = 0 (so b = -a). Put that into the original system:
x + y = 0
(x-a)/(-a) = (y-(-a))/a
Multiply both sides of the second equation by a:
a - x = y + a
0 = x + y
which just repeats the first equation. Therefore, if a+b = 0, then there are infinitely many solutions for x and y: x = k, y = -k for all values of k.