Please help me solve step by step: (1 + i)/(2 - 3i) minus 3 / (4-i)

(1 + i) / (2 - 3i) - 3 / (4 - i)
((1 + i) * (4 - i) - 3 * (2 - 3i)) / ((2 - 3i) * (4 - i)) =>
(4 - i + 4i - i^2 - 6 + 9i) / (8 - 2i - 12i + 3i^2) =>
(4 - 6 - (-1) - i + 4i + 9i) / (8 - 14i - 3) =>
(-2 + 1 + 12i) / (5 - 14i) =>
(-1 + 12i) / (5 - 14i) =>
(-1 + 12i) * (5 + 14i) / ((5 - 14i) * (5 + 14i)) =>
(-5 - 14i + 60i - 168) / (25 + 196) =>
(46i - 173) / 221

It is basically like fractions.
First you find a common denominator or cross multiply the opposite numerator and denominator
(1+i)/(2-3i) - 3/(4-i) = [(1+i)(4-i)-3(2-3i)]/[(2-3i)(4-i)]
Then you have to multiply and combine and remember that i^2=-1
[(1+i)(4-i)-3(2-3i)]/[(2-3i)(4-i)] = [(4+3i-i^2)-(6-9i)]/(8-14i+3i^2) = (-2+12i+1)/(8-14i-3) = (-1+12i)/(5-14i)
So now we combined them and we would have to multiply times the conjugate of the denominator to eliminate the i's (notice that I am not changing anything since the second fraction is technically 1)
(-1+12i)/(5-14i) * (5+14i)/(5+14i) = (-5+46i+168i^2)/(25-196i^2) = (-173+36i)/221 or -173/221 + 36/221 i
Hope it helps!