Let x = a+bi and y = a-bi.
So, (1+4i)x - 4y = -1+2i becomes
(1+4i)(a+bi) - 4(a-bi) = -1+2i, or
a + 4ai + bi - 4b - 4a + 4bi = -1+2i, or
-3a - 4b + (4a+5b)i = -1+2i.
Equate real & imaginary parts:
-3a - 4b = -1, and
(4a+5b)i = 2i, or 4a + 5b = 2.
Multiply the first equation by 4 and the second equation by 3, then add them so as to find a simultaneous solution.
-12a - 16b = -4
+(12a + 15b = 6)
------------------------
-b = 2, or
b = -2.
Substitute this value:
4a + 5(-2) = 2, or
4a = 12, or
a = 3.
x = (3-2i), y = (3+2i).
So, (1+4i)x - 4y = -1+2i becomes
(1+4i)(a+bi) - 4(a-bi) = -1+2i, or
a + 4ai + bi - 4b - 4a + 4bi = -1+2i, or
-3a - 4b + (4a+5b)i = -1+2i.
Equate real & imaginary parts:
-3a - 4b = -1, and
(4a+5b)i = 2i, or 4a + 5b = 2.
Multiply the first equation by 4 and the second equation by 3, then add them so as to find a simultaneous solution.
-12a - 16b = -4
+(12a + 15b = 6)
------------------------
-b = 2, or
b = -2.
Substitute this value:
4a + 5(-2) = 2, or
4a = 12, or
a = 3.
x = (3-2i), y = (3+2i).