If this is (a+bc)x=(d+ef)x, then a+bc=d+ef and x can have any value.
If this is a+b(cx)=d+e(fx), then there can be no definitive answer because constraints will be imposed on the constants.
b(cx)-e(fx)=d-a.
CASE 1: d=a
b(cx)=e(fx), (c/f)x=e/b, xlog(c/f)=log(e/b), x=log(e/b)/log(c/f).
It would appear that x=(log(e)-log(b))/(log(c)-log(f)), but if any of b, c, e or f are negative, x cannot be evaluated.
Furthermore, if either c or f is negative then there's a problem for certain values of x. For example, if x=0.5 and c or f is negative, then we would have √c or √f, which is an imaginary number, implying that the domain (x) is discontinuous. If, however, x=⅓, c and f can be negative (-c and -f) be cause their cube roots exist as -∛c and -∛f. The question doesn't qualify the constants, but here is a clear constraint: c,f≥0.
CASE 2: c=f=p>0
b(px)-e(px)=d-a, px(b-e)=d-a, px=(d-a)/(b-e), xlog(p)=log(d-a)-log(b-e), x=(log(d-a)-log(b-e))/log(p).
But this can only be evaluated if d>a and b>e; or d<a and b<e, so x=(log(a-d)-log(e-b))/log(p).
CASE 3: c=f=0
b and e are arbitrary constants and a=d. x can be any value.
CASE 4 a≠d, c,f>0
The way to check this out is to let y=d-a and plot y=b(cx)-e(fx). On the same graph, plot y=d-a (a horizontal line). If the line intersects the curve, there is a unique solution for x; otherwise there is no solution for x.
The four cases show that there can be no general formula for calculating x.