Math Systems of linear Equations

Math Systems of linear Equations
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  • For y we have the problem y=3(3)-5 and we get 9-5 which is 4 so y=4
  • What about y
  • Ok so you multiply the 2 to the parenthesis and the problem becomes 2x+4y=4 over 2x+4y=4. Everything crosses out and the answer is 0=0. 
  • The problem is  2(x=2y=2) over 2x+4y=4.
  • Ya, I can do it.
  • Are you ready for the last one.
  • Ok, thats easy.
  • So the last is pretty easy and it is a one solution.
  • So first i can bring the 3x-5 down to the y and the problem is 3x-5=(-x)+7. We can add the 5 to both sides and we get 3x=(-x)+12 and we can add the x to 3x to get 4x, so we have 4x=12. if we divide on both sides we get x=3.
  • The problem is y=3x-5 over y=(-x)+7
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