Solve the system
$$xy=-64 \\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{4} $$

Question

Solve the system 
$$xy=-64 \\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{4}   $$

Uniteasy Admin · Accepted Answer

Let
$$a=xy$$
$$b=x+y$$
Then,
$$a = -64$$n
$$\dfrac{1}{x}-\dfrac{1}{y} =\dfrac{y-x}{xy} =\dfrac{1}{4} $$
Square both sides of the equation
$$\dfrac{(x-y)^2}{x^2y^2} = \dfrac{1}{16} $$
$$\dfrac{(x+y)^2-4xy}{x^2y^2} = \dfrac{1}{16} $$
Plugging a,b gives
$$\dfrac{b^2-4a}{a^2}=\dfrac{1}{16} $$n
Substitute the value of $$a$$ in (1) and solve for $$b$$
$$b = 0$$
Then we get the system
$$\begin{cases} xy=-64 \ x+y = 0 \end{cases}$$
Solving for $$x,y$$ yields
$$\begin{cases} x=8 \ y = -8 \end{cases} \quad	ext{or}\quad \begin{cases} x=-8 \ y = 8 \end{cases}$$

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