Center of mass coordinate question

A 3.39 kg particle has the xy coordinates (-1.02 m, 0.0699 m), and a 4.31 kg particle has the xy coordinates (0.523 m, -0.834 m). Both lie on a horizontal plane. At what (a)x and (b)y coordinates must you place a 4.56 kg particle such that the center of mass of the three-particle system has the coordinates (-0.424 m, -0.121 m)?
Have had several goes at this and can't grasp it. Just need to see the calculations being worked through i think. Thanks in advance!

Treat the x coordinates and the y coordinates separately; measuring from the origin, the x coordinate of center of mass is given by

xc = (x1*m1 + x2+m2 + x3*m3)/(m1 + m2 + m3)

-0.424 = (-1.02*3.39 + 0.523*4.31 + x?*4.56)/(3.39 + 4.31 + 4.56) solve for x?

yc = (y1*m1 + y2+m2 + y3*m3)/(m1 + m2 + m3)

-0.121 = (0.0699*3.39 - 0.834*4.31 + y?*4.56)/(3.39 + 4.31 + 4.56) solve for y?

(x?, y?) =-0.876, 0.411

I think you generally have to sum up the x and y values individually and multiply them by their mass then divide by their total mass. It's similar to to finding the centroid but without given area, which is replaced by mass.

For x, where A is the variable of the unknown's x-> (-1.02*3.39+0.523*4.31+4.56A)/total mass=-0.424
For y, where B is the variable of the unknown's y-> (0.0699*3.39-0.834*4.31+4.56B)/total mass=-0.121

Answer=final position (A,B)