Question

solve the following pair of linear equations by reducing the them to a pair of linear equations:
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Answer 1

Answer :

The required values are

x = 1 and y = 1

Given :

The equations are :

$\sf \longrightarrow \dfrac{1}{3x + y}+ \dfrac{1}{3x - y} = \dfrac{3}{4} \\\\ \sf \longrightarrow \dfrac{1}{2(3x+y)} - \dfrac{1}{2(3x - y)} = \dfrac{-1}{8}$

Task :

• To solve the given pair of equations

Solution :

$\sf Let \: us \: assume \: the \: following$

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

Thus the pair of equations becomes:

$\sf \implies u + v = \dfrac{3}{4} \longrightarrow (1) \\\\ \sf \implies \dfrac{u}{2} - \dfrac{v }{2} = \dfrac{-1}{8} \\\\ \sf \implies \dfrac{1}{2}(u - v) = \dfrac{-1}{8} \\\\ \sf \implies u - v = \dfrac{-1}{4} \longrightarrow (2)$

Adding (1) and (2) we have :

$\sf \implies u + v + u - v = \dfrac{3}{4} + \dfrac{-1}{4} \\\\ \sf \implies 2u = \dfrac{3 - 1}{4} \\\\ \sf \implies u = \dfrac{2}{4\times2} \\\\ \sf \implies u = \dfrac{1}{4}$

Putting the value of u in (1) we have

$\sf \implies \dfrac{1}{4} + v = \dfrac{3}{4} \\\\ \sf \implies v = \dfrac{3}{4} - \dfrac{1}{4} \\\\ \sf \implies v = \dfrac{3 - 1}{4} \\\\ \sf\implies v = \dfrac{2}{4} \\\\ \sf \implies v = \dfrac{1}{2}$

Now from our assumption we have :

$\sf \implies \dfrac{1}{3x+y} = u \\\\ \implies \dfrac{1}{3x + y }=\dfrac{1}{4} \\\\ \sf \implies 3x + y = 4 \longrightarrow (3) \\\\ \sf and, \\ \sf \implies \dfrac{1}{3x-y}= v \\\\ \sf \implies \dfrac{1}{3x-y}=\dfrac{1}{2} \\\\ \sf \implies 3x - y = 2 \longrightarrow (4)$

Adding (3) and (4) we have :

$\sf \implies 3x + y + 3x - y = 4+2 \\\\ \sf \implies 6x = 6 \\\\ \sf \implies x = 1$

Putting the value of x in (3)

$\sf \implies 3\times 1 + y = 4 \\\\ \sf \implies y = 4-3 \\\\ \sf \implies y = 1$

Thus , the value of x is 1 and y is 1

Answer 2

Gd evening.........................