For Algebra Fans
You lost ONE pound last week?
Or was it TWO?
No matter; they are the same.
Here's proof:
I can assume that x = y.
As long as I perform the same mathematical operation on both sides of the equals sign, I've not changed the equality.
So, multiply both sides by y to get
xy = y^2
(where y^2 is read as y squared)
Now, subtract x^2 from both sides to get
xy - x^2 = y^2 - x^2
There is a common x term on the left side, so I can factor it out to get
x (y - x)
The right side of the equation can be factored into the product (y - x) (y + x)
Making these substitutions, the equality becomes
x (y - x) = (y - x) (y + x)
Note that there is a common term (y - x) on both sides, so I can divide it out (or cancel it) to get
x = (y + x)
The original assumption was that x and y were equal, so I'll substitute x for the y term
x = x + x
or, simplifying
x = 2x
Since there is an x on each side, I'll divide both sides by x, and recognizing that x / x = 1, this leads to
1 = 2.
Or was it TWO?
No matter; they are the same.
Here's proof:
I can assume that x = y.
As long as I perform the same mathematical operation on both sides of the equals sign, I've not changed the equality.
So, multiply both sides by y to get
xy = y^2
(where y^2 is read as y squared)
Now, subtract x^2 from both sides to get
xy - x^2 = y^2 - x^2
There is a common x term on the left side, so I can factor it out to get
x (y - x)
The right side of the equation can be factored into the product (y - x) (y + x)
Making these substitutions, the equality becomes
x (y - x) = (y - x) (y + x)
Note that there is a common term (y - x) on both sides, so I can divide it out (or cancel it) to get
x = (y + x)
The original assumption was that x and y were equal, so I'll substitute x for the y term
x = x + x
or, simplifying
x = 2x
Since there is an x on each side, I'll divide both sides by x, and recognizing that x / x = 1, this leads to
1 = 2.
0
Replies
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yeah...I failed Algebra....0
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hmmm...that's cool..i did it out and you didn't screw anything up.
x and y are both zero
:bigsmile: then it works
but then that means you lost NO weight! bwahahahaha!0 -
Ah, I remember this one...always a good one. Had to stare at it a while the first time before I got it. I won't ruin it for it everyone.0
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x and y are both zero
:bigsmile: then it works
Doesn't quite work if x and y are both zero...hint hint.0 -
There are issues with the algebraic expression; however, it was entertaining to try to solve for the heck of it.
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Fail!0
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The title alone made my head hurt...then I read it...and I am pretty sure I burned the same calories reading it as I do running for two minutes on the treadmill!!
Thanks for posting!0 -
x = 2x
Since there is an x on each side, I'll divide both sides by x, and recognizing that x / x = 1, this leads to
1 = 2.
Common mistake. If x = 0 then you just divided by zero and only Chuck Norris can do that.
(not sure if this works for the (y - x) term too further up the equation. That one seems plausible.)
Anyway, you would actually wanna subtract x from both sides which gets you 0 = x
Thanks for the math exercise! I miss math!0 -
The title alone made my head hurt...then I read it...and I am pretty sure I burned the same calories reading it as I do running for two minutes on the treadmill!!
Thanks for posting!
Bwahahahaha!!! Yup, soon as I saw Algebra and fan together. I ran away. Which had to burn something.0 -
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x (y - x) = (y - x) (y + x)
Note that there is a common term (y - x) on both sides, so I can divide it out (or cancel it) to get
x = (y + x)
The original assumption was that x and y were equal, so I'll substitute x for the y term
x = x + x
or, simplifying
x = 2x
Since there is an x on each side, I'll divide both sides by x, and recognizing that x / x = 1, this leads to
1 = 2.0 -
Not meaning to be a jerk, but do you get outside much? :bigsmile:
I almost went to High School on the five year plan...
Ted0
This discussion has been closed.
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