system of 3 equation with 3 unknown, with two 2nd order equation
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Hi,
 I am trying to solve a system of 3 equation with 3 unknown, with two 2nd order equation using this code, but unfortunately I don't understand why I do not obtain the result. Thank you in advance
syms QV3 QV2 QVl
eqn1 = 17.79  == QV3 + QV2+QVl;
eqn2 = 0.25*QV3.^2 == 0.14 * QV2^2;
eqn3 = 0.25*QV3.^2 == 3.89* QVl^2;
sol = solve([eqn1, eqn2, eqn3], [QV3 , QV2, QVl])
QV3Sol = sol.QV3
QV2Sol = sol.QV2
QVLsol = sol.QVl
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  Star Strider
      
      
 2020년 2월 23일
        The resuillts are strictly numerical (not containing symbolic variables), so use vpasolve instead of solve: 
syms QV3 QV2 QVl
eqn1 = 17.79  == QV3 + QV2+QVl;
eqn2 = 0.25*QV3.^2 == 0.14 * QV2^2;
eqn3 = 0.25*QV3.^2 == 3.89* QVl^2;
sol = vpasolve([eqn1, eqn2, eqn3], [QV3 , QV2, QVl])
QV3Sol = sol.QV3
QV2Sol = sol.QV2
QVLsol = sol.QVl
producing: 
QV3Sol =
 -214.86507282291207239983881906658
  6.8692131509642582629301134641865
 -30.161934858502498504240076448371
  8.5414023987755051080751389638705
QV2Sol =
  287.1255310312749278205504662485
 9.1793721884393279626540627283888
 40.305580843825110030042867255874
 11.413929063852511018159550534661
QVLsol =
 -54.470458208362855420711647181924
  1.7414146605964137744158238074247
  7.6463540146773884741972091924965
 -2.1653314626280161262346894985318
 These are however still symbolic values.  To use them in other calculations, convert them to double-precision values with the double function.  
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  Star Strider
      
      
 2020년 2월 23일
				My pleasure.  
It produces strictly numerical results because they are not functions of any other symbolic variables.  Any variables would appear in the vpasolve solutions if they existed (and if so, it wouild not be possible to use the double function to convert them to double-precision values).  
There are 4 solutions for each variable because each variable is a  -order polynomial.
-order polynomial.  
 -order polynomial.
-order polynomial.  추가 답변 (0개)
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