If he is using WolframAlpha, where xa is the product of two variables, and, based on some experimentation, I suspect he may have meant to write x + (1 - x) e= .11 instead of just x + (1 - x) e and I translate this calculation into Mathematica just to see the result then it seems
In[1]:= eqns = {x a == 1/10, x b + 1 - x == 89/100, x c == 1/100, x a + (1 - x) d == 1/10, x b + (1 - x) e == 0,
x c + (1 - x) f == 9/10, (1 - x) d == 0, x + (1 - x) e == 11/100, (1 - x) f == 89/100};
sol = Reduce[eqns, {a, b, c, d, e, f, x}, Backsubstitution -> True]
Out[2]= b == 1/10 (10 - 11 a) && c == a/10 && d == 0 && -1 + 10 a != 0 && e == (-10 + 11 a)/(10 (-1 + 10 a)) &&
f == (89 a)/(10 (-1 + 10 a)) && a != 0 && x == 1/(10 a)
In[3]:= Simplify[eqns //. ToRules[sol]]
Out[3]= {True, True, True, True, True, True, True, True, True}
But this still doesn't address why WolframAlpha doesn't seem to provide this, even when you replace 'e' with 'g' or another name.