Please read the documentation of Simplify. I clearly states, that it only simplifies expression:
performs a sequence of algebraic and other transformations on expr, and returns the simplest form it finds.
When you want to solve your equation your first thought should be Solve or Reduce and in some other cases maybe NSolve or FindRoot, depending on what your problem is and what you try to achieve.
Reduce tries to solve your equation analytically and gives you additionally required conditions for a solution. Therefore, this should be your first try
Reduce[y == (1/(1/4 + y) - 1/(3/4 - y))/((1/4 + y) + 1/(3/4 - y)), y]
(*
y == Root[32 - 147 #1 - 84 #1^2 - 16 #1^3 + 64 #1^4 &, 1] ||
y == Root[32 - 147 #1 - 84 #1^2 - 16 #1^3 + 64 #1^4 &, 2] ||
y == Root[32 - 147 #1 - 84 #1^2 - 16 #1^3 + 64 #1^4 &, 3] ||
y == Root[32 - 147 #1 - 84 #1^2 - 16 #1^3 + 64 #1^4 &, 4]
*)
You see your equation has 4 solutions which are the roots of the polynomial $$32 - 147 y - 84 y^2 - 16 y^3 + 64 y^4.$$ You can verify that by subtracting y on both sides of your equation
0 == (1/(1/4 + y) - 1/(3/4 - y))/((1/4 + y) + 1/(3/4 - y)) - y
If you use Together on this equation you get $$0=\frac{-64 y^4+16 y^3+84 y^2+147 y-32}{(4 y+1) \left(16 y^2-8
y-19\right)}$$
and you see that the exact same polynomial appears in the numerator. To get numeric values of your solutions you can apply N to the Root expressions or you use Solve like you did in the first place
Reduce[y == (1/(1/4 + y) - 1/(3/4 - y))/((1/4 + y) + 1/(3/4 - y)), y] // N
(*
y == 0.195639 ||
y == 1.70683 ||
y == -0.826235 - 0.902601 I ||
y == -0.826235 + 0.902601 I
*)
Simplify. You're assumingy. Also check thatyhas not been defined. I do not get the same output as you. Further,Simplifywill not yield solutions. Perhaps you wantReduce? – Michael E2 Mar 15 at 17:15True == (32 - 128 True)/(19 + 84 True + 16 True^2 - 64 True^3). The reason for this is that the second argument of Simplify is an assumption. Assumingymeans assuming thaty === Truefor Mathematica. This is meaningful whenyis part of a logical expression such asx && (y || z). – Szabolcs Mar 15 at 21:34