Question

If $x - \frac{1}{x} = y$, $y - \frac{1}{y} = z$, $z - \frac{1}{z} = x$, then which of the following equations are true? A. $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$ B. $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 8$ C. $\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} = -3$ Choose the correct answer from the options given below:

Accepted Answer

B. Adding the three equations: $x + y + z - (\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = y + z + x \Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$. So A is correct. Squaring both sides of $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$ gives $\sum \frac{1}{x^2} + 2\sum \frac{1}{xy} = 0$. Also squaring original equations $x^2 + \frac{1}{x^2} - 2 = y^2$, summing gives $\sum \frac{1}{x^2} = 6$. So B is incorrect. Then $6 + 2\sum \frac{1}{xy} = 0 \Rightarrow \sum \frac{1}{xy} = -3$. So C is correct. Thus, A and C only.

Algebraic System of Equations

Explanation

Practise this under exam conditions