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# Explaining how both sides of this equation are equal

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1. Say I have:

(e^x - y / x - e^y) = (y-e^x / e^y - x)

By plugging and chugging numbers in it works, just not sure how.

Just something I came across when studying implicity dy/dx which I never gave more thought originally.

thnx
2. (Original post by girlwonder17)
Say I have:

(e^x - y / x - e^y) = (y-e^x / e^y - x)

By plugging and chugging numbers in it works, just not sure how.

Just something I came across when studying implicity dy/dx which I never gave more thought originally.

thnx
Think about a common factor in the numerator and the denominator of one side.

3. Multiplying by is essentially just multiplying by 1.

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