(3x-y)^2+(x-5)^2=0 solve for x and y

Question

Need to solve for x and y

Answer 1

(3x-y)²+(x-5)²=0 is only valid when x=5 and y=3x=15, because the sum of two squares is always positive and is only zero when each square itself is zero.

SOLUTION: x=5, y=15

This is the only real solution but there are complex solutions:

3x-y=(x-5)i,

3x-y=ix-5i, 

y=3x-ix+5i,

y=x(3-i)+5i. This linear complex equation represents many solutions for x and y. Graphically, it's a line in 3-dimensional space: x=y/3=5-z which intersects the x-y plane (the "real plane") at x=5, y=15.