Let i=sqrt-1 and r and w represent real numbers. If (r-4i)(5+3i) = 2+wi then find the value of (r+w)

Question

Let i=sqrt-1 and r and w represent real numbers. If (r-4i)(5+3i) = 2+wi then find the value of (r+w)

Answer 1

(r - 4i)(5 + 3i) = 2 + wi

5r - 20i + 3ri +12 = 2 + wi

5r + 12 + (-20 + 3r)i = 2 + wi

5r + 12 = 2

5r = -10

r = -2

-20 + 3r = w

-20 + 3(-2) = w

-20 - 6 = w

-20 = w

Answer 2

(r - 4i)(5 + 3i) = 2 + wi find the value of find the value of (r+w)

5r - 20i + 3ir -12i² = 2 + wi i²=-1

5r - 20i + 3ir -12(-1) = 2 + wi

5r - 20i + 3ir +12 = 2 + wi group real and imaginary

5r +12 +3ir-20i= 2+ wi factor i

5r+12 + i(3r-20) = 2+wi set up two equations:

equation 1: equation 2: i(3r-20)=wi drop the i's

5r + 12 = 2 subtract 12 on both sides -20 + 3r = w remember r= -2

-12 -12 -20 +3(-2)= w

5r = -10 divide by 5 -20 - 6 = w

r = -2 -20= w

find the value of find the value of (r+w)= -22

-2+ -20