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Question

Solve the following pairs of equations by reducing them to a pair of linear equations:
$$\dfrac{1}{3x+ y } + \dfrac{1}{3x-y} = \dfrac{3}{4}$$
$$\dfrac{1}{2(3x+y)} - \dfrac{1}{2(3x-y)} = \dfrac{-1}{8}$$

Solution
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$$\dfrac{1}{3x+ y } + \dfrac{1}{3x-y} = \dfrac{3}{4}$$ $$...(i)$$
$$\dfrac{1}{2(3x+y)} - \dfrac{1}{2(3x-y)} = \dfrac{-1}{8}$$$$........(ii)$$
$$Let$$ $$p = \dfrac{1}{3x + y}$$ $$and$$ $$q = \dfrac{1}{3x - y}$$
$$\therefore p + q = \dfrac{3}{4}...(iii)$$
$$\dfrac{p}{2} - \dfrac{q}{2} = -\dfrac{1}{8}$$
$$p - q = - \dfrac{1}{8} \times 2$$
$$\therefore p - q = -\dfrac{1}{4}$$$$....(iv)$$
$$Adding \space eqn. (iii)\space and \space eqn. (iv)\space$$
$$2p=\dfrac{2}{4}$$ $$\implies p=\dfrac{1}{4}$$
$$Subtracting \space (iii) \space from \space (iv)$$
$$(p+q)-(p-q)=\dfrac{3}{4}-\dfrac{-1}{4}$$
$$2q=1$$ $$\implies q=\dfrac{1}{2}$$
$$\dfrac{1}{3x+y}=\dfrac{1}{4}$$ $$and$$ $$\dfrac{1}{3x-y}=\dfrac{1}{2}$$
$$\implies 3x+y=4 \space and \space 3x-y=2$$
$$Adding \space we \space get $$
$$6x=6 \implies x=1$$
$$Substituting \space in \space 3x+y=4 \space we \space get$$
$$3+y=4 \implies y=4-3=1$$
$$Hence \space solution (1,1)$$

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