Is Algebraic Equality True? $\frac{x+y}{(x^2+y^2)} = \frac{1}{x+y}$

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The equation \(\frac{x+y}{(x^2+y^2)} = \frac{1}{x+y}\) is not algebraically correct. A specific example shows that \(\frac{2 + 3}{2^2 + 3^2} = \frac{5}{13}\) does not equal \(\frac{1}{5}\). The confusion arises from the misconception that \((x+y)^2\) equals \(x^2+y^2\), which is false. This misunderstanding is commonly referred to as the "freshman's dream." Therefore, the original equation is incorrect.
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Is \frac{x+y}{(x^2+y^2)} = \frac{1}{x+y}

would that be algebraically correct?
 
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whatlifeforme said:
Is \frac{x+y}{(x^2+y^2)} = \frac{1}{x+y}

would that be algebraically correct?

No, it is not.
 
whatlifeforme,
Would this be correct?

$$ \frac{2 + 3}{2^2 + 3^2} = \frac{1}{2 + 3}$$

More simply, this is asking whether 5/13 is equal to 1/5.
 
And I want to add that your mistake is the so-called "freshman's dream", that is, the incorrect idea that ##(x+y)^2=(x^2+y^2)##, which it does not.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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