given that (1-5i)z1 - 2z2 =3-7i . find z1 and z2 as conjugate complex numbers?

Let z = x + iy, then it's conjugate must be: z̅ = x - iy.
Substitute "z" and it's conjugate "z̅"
into the equation:

(1 - 5i)(x + iy) - 2 (x - iy) = 3 - 7i
x + iy - 5ix + 5y - 2x + 2iy = 3 - 7i
(- x + 5y) + (- 5x + 3y)i = 3 - 7i

Now equate real and imaginary parts to find x and y:
- x + 5y = 3 and - 5x + 3y = - 7

Solving these simultaneously gives:
- 5 (5y - 3) + 3y = - 7
- 25y + 15 + 3y = - 7
- 22y = - 22
y = 1

x = 5(1) - 3 = 2

Thus z = 2 + i and it's conjugate z̅ = 2 - i
Hope this helps !!!!!!!!!!!!!!!!

Let z₁ = a + bi
Then, z₂ = a - bi

(1 - 5i) z₁ - 2 z₂ = 3 - 7i
(1 - 5i) (a + bi) - 2 (a - bi) = 3 - 7i
(a + bi) - 5i (a + bi) - 2a + 2bi = 3 - 7i
a + bi - 5ai + 5b - 2a + 2bi = 3 - 7i
(5b - a) + (3b - 5a)i = 3 - 7i

Compare the real parts and the imaginary parts on the both sides :
5b - a = 3 ...... [1]
3b - 5a = -7 ...... [2]

From [1] :
a = 5b - 3 ...... [3]

Substitute [3] into [2] :
3b - 5(5b - 3) = -7
3b - 25b + 15 = -7
-22b = -22
b = 1

Put b = 1 into [3] :
a = 5(1) - 3
a = 2

z₁ = 2 + i
z₂ = 2 - i