SOLUTION: x+y=z then x+y-z=0
2(x+y)=2(z) 3(x+y)=3(z)
2x+2y=2z 3x+3y=3z
2x+2y-2z=0 3x+3y-3z=0
2(x+y-z)=0 3(x+y-z)=0
then 2(x+y-z)=3(x+y-z)
then 2=3
Surely this can
Algebra ->
Linear-equations
-> SOLUTION: x+y=z then x+y-z=0
2(x+y)=2(z) 3(x+y)=3(z)
2x+2y=2z 3x+3y=3z
2x+2y-2z=0 3x+3y-3z=0
2(x+y-z)=0 3(x+y-z)=0
then 2(x+y-z)=3(x+y-z)
then 2=3
Surely this can
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Question 8051: x+y=z then x+y-z=0
2(x+y)=2(z) 3(x+y)=3(z)
2x+2y=2z 3x+3y=3z
2x+2y-2z=0 3x+3y-3z=0
2(x+y-z)=0 3(x+y-z)=0
then 2(x+y-z)=3(x+y-z)
then 2=3
Surely this can't be? Answer by ichudov(507) (Show Source):
You can put this solution on YOUR website! Yep, that can't be. when you change 2(x+y-z)=3(x+y-z) to 2=3, you divide both paths of the equation by x+y-z, which, as you know is zero. Division by zero is an illegal operation. You can get all sorts of false results by dividing by zero.
