Please explain the algebra done between the highlighted steps in both part A and part B. I understand the distance needs to be isolated but how is the second distance variable cancelled out?

25. Point charges of 5.00 uC and - 3.00 uC are placed 0.250 m apart. (a) Where can a third charge be placed so that the net force on it is zero? (b) What if both charges are positive? Solution (a) We know that since the negative charge is smaller, the third charge should be placed to the right of the negative charge if the net force on it is to be zero. So if we want Fret = F, + F2 = 0, we can use F = k 4172 to write the forces in terms of distances: kq,92 k929 = kq 91 - 92 = 0, or since r, = 0.250 m + d andr2 = d, 5x10-C 3x10-6 (0.250 m + d ) d 2 , or d = 3(0.250m)+ 3d so that OpenStax College Physics Instructor Solutions Manual Chapter 18 (0.250m) and finally, d =- = 0.859m 1 - V3 5 The charge must be placed at a distance of 0.859 m to the far side of the negative charge. (b) This time we know that the charge must be placed between the two positive charges and closer to the 3 uC charge for the net force to be zero. So if we want Fnet = F, + F2 = 0, we can again use F = k 4192 to write the forces in terms of distances: 2 kq29 = kq *9 192 + - + 92 = 0 5x 10- C 3x10-6 Or since r, = 0.250m - 12 (0.250-72 )2 or 12 2 = 2 (0.250 m - 12) , or (0.250m) uilw (0.250 m - r2), and finally 72 = = 0.109 m 1+