1/3x+4/5y=2;1/2x-9/5y=7

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Solution for 1/3x+4/5y=2;1/2x-9/5y=7 system of equations: 


  • /| (1/3)*x+(4/5)*y-2 = 0| (1/2)*x-((9/5)*y)-7 = 0
  • We try to solve the equation: (1/3)*x+(4/5)*y-2 = 0
  • 1/3*x+4/5*y-2 = 0 // - 1/3*x-2
  • 4/5*y = -(1/3*x-2) // : 4/5
  • y = (-(1/3*x-2))/4/5
  • y = 5/4*(2-(1/3*x))
  • We insert the solution into one of the initial equations of our system of equations
  • We get a system of equations:
  • /| (1/2)*x-(5/4*(9/5)*(2-(1/3*x)))-7 = 0| y = 5/4*(2-(1/3*x))
  • x/2-1*9/4*(2-1/3*x)-7 = 0
  • 5/4*x-7-9/2 = 0
  • 5/4*x-23/2 = 0
  • 5/4*x-23/2 = 0 // + 23/2
  • 5/4*x = 23/2 // : 5/4
  • x = 23/2/5/4
  • x = 46/5
  • We insert the solution into one of the initial equations of our system of equations
  • For y = 5/4*(2-(1/3*x)): 
  • y = 5/4*(2-(1/3*46/5))
  • y = 5/4*(2-46/15)
  • y = -16/15*5/4
  • y = -4/3
  • We get a system of equations:
  • /| y = -4/3| x = 46/5

 

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