Question

enter = if the proposed identity holds, and n otherwise. (\frac{1}{x} + \frac{1}{y} square \frac{1}{x+y}). (\frac{1}{x} + \frac{1}{y} square \frac{x+y}{xy}). note: in order to get credit for this problem all answers must be correct. you have attempted this problem 0 times. you have unlimited attempts remaining.

Explanation:

Step1: Analyze the first identity

To add \(\frac{1}{x}+\frac{1}{y}\), we find a common denominator, which is \(xy\). So \(\frac{1}{x}+\frac{1}{y}=\frac{y}{xy}+\frac{x}{xy}=\frac{x + y}{xy}\). Now compare this with \(\frac{1}{x + y}\). Since \(\frac{x + y}{xy}\) is not equal to \(\frac{1}{x + y}\) (for example, take \(x = 1,y = 1\): \(\frac{1}{1}+\frac{1}{1}=2\), \(\frac{1}{1 + 1}=\frac{1}{2}\), \(2
eq\frac{1}{2}\)), the first identity does not hold.

Step2: Analyze the second identity

From Step 1, we have \(\frac{1}{x}+\frac{1}{y}=\frac{y}{xy}+\frac{x}{xy}=\frac{x + y}{xy}\), so this identity holds.

Answer:

First box: N, Second box: =