Make x the subject of the formula in b(x-a)=x(c+d_e)
Assuming you meant:
b(x-a)=x(c+d-e)
bx - ab = cx + dx - ex
bx -cx - dx + ex = ab
x(b - c - d + e) = ab
x = ab/(b - c - d + e)
Expanding the brackets on the right side gives:
bx - ba = cx + dex
Collecting the x terms on one side and the constant terms on the other:
bx - cx = dex + ba
Factorizing out x:
x(b - c) = ba + de
Dividing both sides by (b - c):
x = (ba + de)/(b - c)