(-2 + 2i)z1 + (3 + 3i)z2 = 6 + 2i

--> z1 = [(6+2i)-(3+3i)z2]/(-2+2i)

(3 + 4i)z1 + (8 - 6i)z2 = -6 + 17i

(3+4i)[(6+2i)-(3+3i)z2]/(-2+2i) + (8-6i)z2 = (-6+17i)

[ (3+4i)(6+2i)-(3+4i)(3+3i)z2 ] / (-2+2i) + (8-6i)z2 = (-6+17i)

[ (10+30i)-(-3+21i)z2 ] / (-2+2i) + (8-6i)z2 = (-6+17i)

(10+30i)/(-2+2i) - (-3+21i)z2/(-2+2i) + [(-2+2i)(8-6i)z2]/(-2+2i) = (-6+17i)

z2 = [ (-6+17i)-[(10+30i)/(-2+2i)] ] * (-2+2i)/[(-3+21i)+(-2+2i)(8-6i)

z2 = [ (-6+17i)(-2+2i)-(10+30i) ]/(-2+2i) ] * (-2+2i)/(-7+49i)

z2 = [ (-6+17i)(-2+2i)-(10+30i) ]/(-7+49i)

z2 = [ (-22-46i)-(10+30i) ]/(-7+49i)

z2 = (-12-76i)/(-7+49i)

z2 = (12+76i)/(7-49i)

z2 = (12+76i)/(7-49i) * (7+49i)/(7+49i)

z2 = (-3640+1120i)/(-2352)

z2 = (-3640+1120i)/(-2352)

z2 = (-3640/-2352) + i(1120/2352)

Therefore Re(z2)=(3640/2352) and Ir(z2)=(1120/2352)

(You can simplify fractions).

To find Re(z1) and Ir(z1) plug z2 = (3640/2352) + i(1120/2352) back into either of the two original equations and solve.

!

Note: I would check all my work, since the probability of a mistake is extremely high. Hand solving is much easier.