1/3b+1/3c=4;1/6b-1/3c=4

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Solution for 1/3b+1/3c=4;1/6b-1/3c=4 system of equations: 


  • /| (1/3)*b+(1/3)*c-4 = 0| (1/6)*b-((1/3)*c)-4 = 0
  • We try to solve the equation: (1/3)*b+(1/3)*c-4 = 0
  • 1/3*b+1/3*c-4 = 0 // - 1/3*c-4
  • 1/3*b = -(1/3*c-4) // : 1/3
  • b = (-(1/3*c-4))/1/3
  • b = 3*(4-(1/3*c))
  • We insert the solution into one of the initial equations of our system of equations
  • We get a system of equations:
  • /| 3*(1/6)*(4-(1/3*c))-((1/3)*c)-4 = 0| b = 3*(4-(1/3*c))
  • 1/2*(4-1/3*c)-c/3-4 = 0
  • 2-1/2*c-4 = 0
  • -1/2*c-2 = 0
  • -1/2*c-2 = 0 // + 2
  • -1/2*c = 2 // : -1/2
  • c = 2/(-1/2)
  • c = -4
  • We insert the solution into one of the initial equations of our system of equations
  • For b = 3*(4-(1/3*c)): 
  • b = 3*(4-(-4*1/3))
  • b = 3*(4/3+4)
  • b = 3*16/3
  • b = 16
  • We get a system of equations:
  • /| b = 16| c = -4

 

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