If frac{x+y}{x y}=2 and frac{x-y}{x y}=6, the | vedclass

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Question

(D) Given equations are:$(1) \frac{x+y}{x y} = 2 \implies \frac{1}{y} + \frac{1}{x} = 2$$(2) \frac{x-y}{x y} = 6 \implies \frac{1}{y} - \frac{1}{x} =

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(D) Given equations are:$(1) \frac{x+y}{x y} = 2 \implies \frac{1}{y} + \frac{1}{x} = 2$$(2) \frac{x-y}{x y} = 6 \implies \frac{1}{y} - \frac{1}{x} = 6$Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$.The equations become:$v + u = 2$$v - u = 6$Adding the two equations:$(v + u) + (v - u) = 2 + 6$$2v = 8 \implies v = 4$Substituting $v = 4$ into $v + u = 2$:$4 + u = 2 \implies u = -2$Since $u = \frac{1}{x} = -2$,we get $x = -\frac{1}{2}$.

If $\frac{x+y}{x y}=2$ and $\frac{x-y}{x y}=6,$ then $x=\ldots \ldots \ldots \ldots .$

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