Hi Alia,
I believe there must be a picture of angle(s) that accompanies the question. From your question, there can be several possibility on where point X can be. Based on the location of point X and the relationship between (< for angle) <ABC and <CBX, we can write an equation to solve for X.
Case 1: Point X is somewhere on ray BA, then m<ABC = m<CBX. By substitution, 7x-4 = 7-x. Add x to both sides, we get 8x-4 = 7. Add 4 to both sides, we get 8x = 11. Divide both sides by 8, we get x= 11/8 or 1 3/8.
Case 2: Point X is on another ray with starting point at point B, forming an angle <CBX sharing one side, BC, with <ABC. If <ABX is an right angle (90 degrees), then m<ABC + m<CBX = 90. By substitution, (7x-4) + (7-x) = 90. Combining like terms, we get 6x+3 = 90. Subtract 3 from both sides, we get 6x = 87. Divide both sides by 6, we get x=87/6 or 14 1/2 or 14.5 degrees. If <ABX is a straight angle (180 degrees), then m<ABC + m<CBX = 180. By substitution, (7x-4) + (7-x) = 180. Combining like terms, 6x + 3 = 180. Subtract 3 from both sides, 6x = 177. Divide both sides by 6, we get x = 177/6 or 29 3/6 or 29 1/2 or 29.5 degrees.
Because I don't see the picture, these are the only cases I can easily imagine. If they don't match the picture you have, please let me know. With a better description of the drawn picture/figure you have, we can then go from there. Let me know also if one of these cases matches your picture, then let me know if the solution I provided is of help to you.
Faye Huey-Ming H.