First, what I previously told you is from the “carry and equation way.” It is just much faster.
[MATH]S + L = R + 10w, \ w + Y + L = E + 10x, \ x + A + E = T + 10y,\\
y + L + W = E + 10z, \ z + P + 0 = E + 10u, \text { and } u + 0 + 0 = B.[/MATH]Plus we know that u, w, x, y, and z are each either zero or one, that each distinct letter represents a distinct decimal digit, and that
[MATH]B > 0, \ P > 0, \text { and } P > 0.[/MATH]
Now as I explained before
[MATH]u + 0 + 0 = B > 0 \implies u = 1 \implies B = 1.[/MATH]
[MATH]z + P + 0 = E + 10u \implies P = E + 10 \text { if } z = 0, \text { which is impossible because } P \text { and } E \text { are digits.}[/MATH]
[MATH]\therefore z = 1 \implies 1 + P = E + 10 \implies P - 9 = E \implies P = 9 \text { and } E = 0 \text { because both are digits.}[/MATH]
You do not have to go through all this work if you simply recognize that it will obviously be true when you add a n digit number and a (n +1) digit number and get a sum that is a (n + 2) digit number.
Given that, you now have
[MATH]S + L = R + 10w,\ w + Y + L = 10x, \ x + A = T + 10y, \text { and } y + L + W = 10.[/MATH]
Now solve it.