Let x and y be real numbers. If (x+yi)-(2-3i)=-6+4i, what are the values of x and y?

So far I got

x+yi-2+3i=-6+4i
x+yi+3i=-6+2+4i
x+yi+3i=-4+4i
x+yi=-4+i

the real parts of the left side must match the real parts of the right side, and the imaginary parts ....

your steps are correct, so
x = -4, y = 1

Thanks

To find the values of x and y, we need to simplify the expression (x + yi) - (2 - 3i) and equate it to the given value -6 + 4i.

Given:
(x + yi) - (2 - 3i) = -6 + 4i

We can simplify this expression by distributing the negative sign to (2 - 3i):
x + yi - 2 + 3i = -6 + 4i

Next, we group the real terms (x - 2) and imaginary terms (yi + 3i) together:
(x - 2) + (y + 3)i = -6 + 4i

Now, we have two equations:
1) x - 2 = -6 -----> equation (1)
2) y + 3 = 4 -----> equation (2)

From equation (1), we can solve for x by adding 2 to both sides:
x - 2 + 2 = -6 + 2
x = -4

From equation (2), we can solve for y by subtracting 3 from both sides:
y + 3 - 3 = 4 - 3
y = 1

Therefore, the values of x and y are x = -4 and y = 1, respectively.