SOLUTION: For what real number k is (8 + ki) + (7 - 3i) equal to a real number?

Instead of doing your problem, I'll do one exactly like it:

For what real number k is (9 + ki) + (6 - 4i) equal to a real number?

To have a real number, all the imaginary parts (terms with i) must equal 0.

ki - 4i = 0

Divide through by i

k - 4 = 0

    k = 4

Checking:

(9 + ki) + (6 - 4i) 
(9 + 4i) + (6 - 4i)
9 + 4i + 6 - 4i
15

And 15 is a real number.

Now do your problem the exact same way.

Edwin