How do you solve this complex equation???/(Imaginary numbers)

(3+i)z+4-i=2i

The answer is z=(-9+13i)/10, but I dont get how do to it. Any help??

(3 + i)z + 4 - i = 2i

(3 + i)z = 2i - (4 - i)

(3 + i)z = 3i - 4

z = (3i - 4)/(3 + i)

Multiply and divide by the complex conjugate of the denominator, (3 - i)
z = (3i - 4)(3 - i) / (9 + 1)

z = (9i - 12 + 4i + 3) / 10

z = (13i - 9) / 10

(3 + i)z + 4 - i = 2i
(3 + i)z = 2i + i - 4
(3 + i)z = 3i - 4
z
= (3i - 4)/(3 + i)
= (3i - 4)(3 - i)/(3 + i)(3 - i)
= (9i - 12 - 3i^2 + 4i)/3^2 - i^2)
= [13i - 12 -3(-1)]/[9 - (-1)]
= (-9 + 13i)/10

(3+i)z =2i+i-4
z =(3i-4)/(3+i)
z =(-4+3i)*(3-i)/(3+i)(3-i)
z =(-12+4i+9i+3)/(9+1)
=(-9+13i)/10........Ans