1A+4C=85;3A+2C=105

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Solution for 1A+4C=85;3A+2C=105 system of equations:


  • /| 1*A+4*C-85 = 0| 3*A+2*C-105 = 0
  • We try to solve the equation: 1*A+4*C-85 = 0
  • A+4*C-85 = 0 / - 4*C-85
  • A = -(4*C-85)
  • A = 85-(4*C)
  • We insert the solution into one of the initial equations of our system of equations
  • We get a system of equations:
  • /| 3*(85-(4*C))+2*C-105 = 0| A = 85-(4*C)
  • 3*(85-4*C)+2*C-105 = 0
  • 255-10*C-105 = 0
  • 150-10*C = 0
  • 150-10*C = 0 / - 150
  • -10*C = -150 / : -10
  • C = -150/(-10)
  • C = 15
  • We insert the solution into one of the initial equations of our system of equations
  • For A = 85-(4*C):
  • A = 85-(4*15)
  • A = 85-60
  • A = 25
  • We get a system of equations:
  • /| A = 25| C = 15

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