Question

$\green{\underbrace{\huge\underline{ \red{\sf Question:-}}}}$

Solve the following simultaneous Equation
$\\ \sf \: \frac{1}{3x + y} \: + \frac{1}{3x - y} = \frac{3}{4} \: \: \: ; \: \: \frac{1}{2(3x + y)} - \frac{1}{2(3x - y)} = \frac{1}{8}$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given pair of simultaneous equations

$\rm :\longmapsto\:\dfrac{1}{3x + y} \: + \dfrac{1}{3x - y} = \dfrac{3}{4} - - - (1)$

and

$\rm :\longmapsto\:\dfrac{1}{2(3x + y)} - \dfrac{1}{2(3x - y)} = \dfrac{1}{8}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{1}{3x + y} - \dfrac{1}{3x - y} = \dfrac{1}{4} - - - (2)$

Let assume that

$\rm :\longmapsto\:\dfrac{1}{3x + y} \: = \: u \: \: and \: \: \dfrac{1}{3x - y} = v$

So, equation (1) and (2) can be rewritten as

$\rm :\longmapsto\:u + v = \dfrac{3}{4} - - - (3)$

and

$\rm :\longmapsto\:u - v = \dfrac{1}{4} - - - (4)$

On adding equation (3) and (4), we get

$\rm :\longmapsto\:2u = \dfrac{3}{4} + \dfrac{1}{4}$

$\rm :\longmapsto\:2u = \dfrac{3 + 1}{4}$

$\rm :\longmapsto\:2u = \dfrac{4}{4}$

$\rm :\longmapsto\:2u =1$

$\rm \implies\:u = \dfrac{1}{2} - - - (5)$

On substituting the value of u in equation (3), we get

$\rm :\longmapsto\:\dfrac{1}{2} + v = \dfrac{3}{4}$

$\rm :\longmapsto\:v = \dfrac{3}{4} - \dfrac{1}{2}$

$\rm :\longmapsto\:v = \dfrac{3 - 2}{4}$

$\rm \implies\:v = \dfrac{1}{4} - - - - (6)$

Now, from equation (5),

$\rm \implies\:u = \dfrac{1}{2}$

$\rm \implies\:\dfrac{1}{3x + y} = \dfrac{1}{2}$

$\rm \implies\:3x + y = 2 - - - (7)$

Now, from equation (6), we have

$\rm \implies\:v = \dfrac{1}{4}$

$\rm \implies\:\dfrac{1}{3x - y} = \dfrac{1}{4}$

$\rm \implies\:3x - y = 4 - - - - (8)$

On adding equation (7) and (8), we get

$\rm :\longmapsto\:6x = 6$

$\bf\implies \:x = 1$

On substituting the value of x in equation (7), we get

$\rm :\longmapsto\:3(1) + y = 2$

$\rm :\longmapsto\:3 + y = 2$

$\rm :\longmapsto\:y = 2 - 3$

$\bf\implies \:y = - 1$

Hence, solution of simultaneous equations are

$\red{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{x = 1} \\ \\ &\sf{y = - 1} \end{cases}\end{gathered}\end{gathered}}$

Answer 2

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Step-by-step explanation:

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