How can I find a unique solution within tolerance using solve?

Hello,
So I am using solve to find values for 3 variables while having 3 equations. The problem is that my equations dont seem to have unique solution. How can I setup tolerance in a way that the solution only exits within 20k to 2M? For example for values in 700k,1.5M and 2M the right hand side is approximately equal to the left hand side.

syms Rx Risop Rison
V1=252.79315;
V2=287.20685;
V3=496.51688;
V4=43.483124;
V5=38.272892;
V6=501.72711;
R=100000000;
R1=68220;
R2=68220;
Vx1=17.206846;
Vx2=226.51688;
Vx3=231.72711;
equ1=(V1/Risop)+(V1/R)==(Vx1/Rx)+(V2/Rison)+(V2/R);
equ2= (V3/Risop)+(V3/R)==(V4/Rison)+(V4/R1)+(V4/R)-(Vx2/Rx);
equ3= (V5/Risop)+(V5/R)+(V5/R2)==(V6/Rison)+(Vx3/Rx)+(V6/R);
[Risop,Rison,Rx]=solve([equ1,equ2,equ3],[Risop,Rison,Rx]);
vpa(Risop)
vpa(Rison)
vpa(Rx)
Risop=1500000;
Rison=2000000;
Rx=700000;
LHS=(V1/Risop)+(V1/R);
RHS=(Vx1/Rx)+(V2/Rison)+(V2/R);
vpa(LHS)
vpa(RHS)
LHS=(V3/Risop)+(V3/R);
RHS=(V4/Rison)+(V4/R1)+(V4/R)-(Vx2/Rx);
vpa(LHS)
vpa(RHS)
LHS=(V5/Risop)+(V5/R)+(V5/R2);
RHS=(V6/Rison)+(Vx3/Rx)+(V6/R);
vpa(LHS)
vpa(RHS)Hello,
So I am using solve to find values for 3 variables while having 3 equations. The problem is that my equations dont seem to have unique solution. How can I setup tolerance in a way that the solution only exits within 20k to 2M? For example for values in 700k,1.5M and 2M the right hand side is approximately equal to the left hand side.

syms Rx Risop Rison
V1=252.79315;
V2=287.20685;
V3=496.51688;
V4=43.483124;
V5=38.272892;
V6=501.72711;
R=100000000;
R1=68220;
R2=68220;
Vx1=17.206846;
Vx2=226.51688;
Vx3=231.72711;
equ1=(V1/Risop)+(V1/R)==(Vx1/Rx)+(V2/Rison)+(V2/R);
equ2= (V3/Risop)+(V3/R)==(V4/Rison)+(V4/R1)+(V4/R)-(Vx2/Rx);
equ3= (V5/Risop)+(V5/R)+(V5/R2)==(V6/Rison)+(Vx3/Rx)+(V6/R);
[Risop,Rison,Rx]=solve([equ1,equ2,equ3],[Risop,Rison,Rx]);
vpa(Risop)
vpa(Rison)
vpa(Rx)
Risop=1500000;
Rison=2000000;
Rx=700000;
LHS=(V1/Risop)+(V1/R);
RHS=(Vx1/Rx)+(V2/Rison)+(V2/R);
vpa(LHS)
vpa(RHS)
LHS=(V3/Risop)+(V3/R);
RHS=(V4/Rison)+(V4/R1)+(V4/R)-(Vx2/Rx);
vpa(LHS)
vpa(RHS)
LHS=(V5/Risop)+(V5/R)+(V5/R2);
RHS=(V6/Rison)+(Vx3/Rx)+(V6/R);
vpa(LHS)
vpa(RHS) Hello,
So I am using solve to find values for 3 variables while having 3 equations. The problem is that my equations dont seem to have unique solution. How can I setup tolerance in a way that the solution only exits within 20k to 2M? For example for values in 700k,1.5M and 2M the right hand side is approximately equal to the left hand side.

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