For this question, I've borrowed the system of equations presented in Solving for a single variable across multiple equations. I've converted all values to exact numbers, to provide both Solve and Reduce with exact input.
I've found that, when Simplify is applied to the solution provided by Reduce, the solution disappears; this appears to be because Simplify equates the solution to zero. Simplify does not do this for the solution produceSolutions returned by Solve, even though its solution is equivalent.
The following (undocumented) syntax means: "Using reduce, solve the given system of equations for $\sigma \mathrm M$, while elminating VF and Vs are provided as Rules:
redsolnRule1 = Reduce[
Log10[fy -> 1/(3 (1 -+ fx)] == pH) - pKa(-1 + F2 Vsx)/(23/106 R(1 T)- &&x \[Sigma]M+ ==x^2))
+ 2/(
cF VF && Ee3 -(1 Epzc+ ==1/3 VF(-1 + Vs, \[Sigma]M, {Vs,2 VF}]x)^2));
If I apply Simplify to the above, the solution for $\sigma \mathrm M$ is simplified, but it's no longer displayedSolutions returned by Reduce are provided as being the solution to $\sigma \mathrm M$; instead, it's equated to zeroequations:
simpRedsolnEqn1 = Simplify@red
This can be seen if we pull out the solution for $\sigma \mathrm M$ and apply Simplify specifically to it:
redSigmaM = red[[3, 2]]
Simplify@redSigmaM
I don't know why Simplify is doing this, because when I apply Simplify just to the RHS of Reduce's solution for $\sigma \mathrm M$ (RHSredSigmaM), it doesn't equate it to zero. While RHSredSigmaM does equal zero under the specific conditions that cF = 0, or Ee = Epzc and either R = 0 or T = 0, it does not appear that it is generically equal to zero:
RHSredSigmaMy === redSigmaM[[2]]
simpRHSredSigmaM1/(3 =(1 Simplify@RHSredSigmaM
RHSredSigmaM+ /.x)) cF- (->1 0
RHSredSigmaM+ 2 x)/.(6 {Ee(1 -> Epzc,x R+ ->x^2)) 0}
RHSredSigmaM+ 2/.(
{Ee -> Epzc, T3 (1 + 1/3 (->1 0}+ 2 x)^2));
Note It is my understanding that thisSimplify's standard behavior is not due to my usetreat both of undocumented syntax with Reduce, sincethese the same thing happens with standard syntax. I chose to use the undocumented syntax because it allowed me to more clearly focus just on the solution to $\sigma \mathrm M$. With the following standard syntax, I get other solutions as well:
Simplify@solnRule1
Simplify@solnEqn1
However, makingthat is not the case for a messier output (suppressed here) and thus a less readable postthese identical solutions:
red2 solnRule2 = Reduce[
Log10[f/(1y -> f)]-((-10 ==cF pHEe -F pKaLog[10] + 10 cF Epzc F Vs/(23/10Log[10] R- T)
&& \[Sigma]M == 23 cF VFpH &&R
T Log[10] + 23 EecF -pKa EpzcR ==T VFLog[10] + Vs,
{\[Sigma]M, Vs, VF}];
Simplify@red2
I don't see this issue with the output of Solve. Solve's solution for $\sigma \mathrm M$ is initially in a different form from Reduce's, which could explain why Simplify doesn't show the same behavior with Solve.
sol = Solve[
23 cF Log10[fR T Log[f/(1 - f)])/(10 ==F pHLog[10]));
solnEqn2 =
y == -((-10 pKacF Ee F Log[10] + 10 cF Epzc F Vs/(Log[10] -
23/10 cF pH R T) &&Log[10] \[Sigma]M+ ==23 cF VFpKa &&R
T Log[10] + Ee
- Epzc == VF +23 Vs,cF \[Sigma]M,R {Vs,T VF}Log[f/(1 - f)]
simpSol)/(10 =F Simplify@solLog[10]));
When Simplify is applied to the rule-based version of the above solution, it acts normally:
Simplify@solnRule2
Nevertheless
But when Simplify is applied to the equation-based version, it equates the solution to zero:
Simplify@solnEqn2
I don't know why Simplify is doing this. When I convert solnEqn2 to a comparison of simpRHSredSigmaM and simpRHSsolSigmaM using SameQ shows they are identicalrule-based format, it behaves normally:
simpRHSsolSigmaM = simpSol[[1, 1, 2]]ToRules@solnEqn2
simpRHSredSigmaM === simpRHSsolSigmaMSimplify@%
[simpRHSsolSigmaM $\equiv$ applyThis can cause problems when applying Simplify to the RHSoutput of Solve'sReduce: Reduce can provide solutions to several different variables, and when one solution for $\sigma \mathrm M$; simpRHSredSigmaM $\equiv$ apply Simplifyis equated to zero it can be difficult to determine the RHS of Reduces's solution for $\sigma \mathrm M$]variable it represents.
