SOLUTION: What is wrong with this arguement? Suppose x and y represent two real numbers, where x > y
2 > 1
2(y-x)> 1(y-x)
2y-2x>y-x
y-2x>-x
y>x
The final inequality, y>x, is imposs
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-> SOLUTION: What is wrong with this arguement? Suppose x and y represent two real numbers, where x > y
2 > 1
2(y-x)> 1(y-x)
2y-2x>y-x
y-2x>-x
y>x
The final inequality, y>x, is imposs
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Question 66699: What is wrong with this arguement? Suppose x and y represent two real numbers, where x > y
2 > 1
2(y-x)> 1(y-x)
2y-2x>y-x
y-2x>-x
y>x
The final inequality, y>x, is impossible because we were initally given x>y.
Extra info in problem possibly the "arguement"...This is a true statement . Multiply both sides by y-x. Use the distribtive property . Subtract from both sides. Add 2x to both sides. Answer by Nate(3500) (Show Source):
You can put this solution on YOUR website! A smaller number (y) minus a larger number (x) is a negative number ....
so: y - x is a negative number
2(y - x) > 1(y - x)
2(a negative number) > 1(a negative number)
2y - 2x < y - x
y - 2x < -x
y < x
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