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

STEP-BY-STEP EXPLANATION :

.

$\sf \frac{1}{3x + y} \: + \frac{1}{3x - y} = \frac{3}{4} \: \: \: \: \: \And \: \: \: \: \: \frac{1}{2(3x + y)} - \frac{1}{2(3x - y)} = \frac{1}{8}$

$\sf Let's,$

• $\sf \frac{1}{3x + y} = p$
• $\sf \frac{1}{3x - y} = q$

$\sf p + q = \frac{3}{4} \: \: \: ...(1)$

$\implies \sf \frac{p}{2} - \frac{q}{2} = \frac{1}{8}$

$\implies \sf \frac{p - q}{2} = \frac{1}{8}$

$\implies \sf {p - q} = \frac{1}{4} \: \: \: ...(2)$

$\sf Add \: Both \: Eq [1] \: and \: Eq [2],$

$\implies \sf p + q + (p - q)= \frac{3}{4} + \frac{1}{4}$

$\implies \sf p \cancel{ + q} + p \cancel{- q}= \frac{3 + 1}{4}$

$\implies \sf 2p = 1$

$\implies \sf p = \frac{1}{2}$

$\sf Substitute \: this \: value \: of \: p \: i n \: Eq [1],$

$\implies \sf p + q = \frac{3}{4} \: \: \: ...(1)$

$\implies \sf q = \frac{3}{4} - \frac{1}{2}$

$\implies \sf q = \frac{1}{4}$

$\sf Here,$

$\implies \sf \frac{1}{3x + y} = p$

$\implies \sf \frac{1} {3x + y} = \frac{1} {2}$

$\implies \sf {3x + y} = {2} \: \: \: ...(3)$

$\implies \sf \frac{1}{3x - y} = q$

$\implies \sf \frac{1}{3x - y} = \frac{1}{4}$

$\implies \sf {3x - y} = {4} \: \: \: ...(4)$

$\sf We \: have,$

• $\sf {3x + y} = {2} \: \: \: ...(3)$
• $\sf {3x - y} = {4} \: \: \: ...(4)$

$\sf Add \: Both \: Eq [3] \: and \: Eq [4],$

$\implies \sf {3x + y} + (3x - y) = {2} + 4$

$\implies \sf {3x + y} + 3x - y= {2} + 4$

$\implies \sf 6x = 6$

$\sf Substitute \: this \: value \: of \: x \: in \: Eq [3],$

$\implies \sf {3x + y} = {2} \: \: \: ...(3)$

$\implies \sf {3 + y} = {2}$

$\implies \sf y = - 1$

$\sf Therefore,$

$\sf x = 1 \: \And \: y = -1$

REQUIRED ANSWER,

.

• $\sf x = 1 \: \And \: y = -1$

Answer 2

To FiNd,

• $\color{red} \boxed{ \sf \: The \: value \: of \: x \: and \: y \: }$

SoLuTiOn,

$\sf \frac{1}{3x + y} \: + \frac{1}{3x - y} = \frac{3}{4} \: \: \: \: \: \And \: \: \: \: \: \frac{1}{2(3x + y)} - \frac{1}{2(3x - y)} = \frac{1}{8}$

$\sf Let's,$

$\sf \frac{1}{3x + y} = p$

$\sf \frac{1}{3x - y} = q$

$\sf p + q = \frac{3}{4} \: \: \: ...(1)$

$\implies \sf \frac{p}{2} - \frac{q}{2} = \frac{1}{8}$

$\implies \sf \frac{p - q}{2} = \frac{1}{8}$

$\implies \sf {p - q} = \frac{1}{4} \: \: \: ...(2)$

$\sf Add \: Both \: Eq [1] \: and \: Eq [2],$

$\implies \sf p + q + (p - q)= \frac{3}{4} + \frac{1}{4}$

$\implies \sf p \cancel{ + q} + p \cancel{- q}= \frac{3 + 1}{4}$

$\implies \sf 2p = 1$

$\implies \sf p = \frac{1}{2}$

$\sf Substitute \: this \: value \: of \: p \: i n \: Eq [1],$

$\implies \sf p + q = \frac{3}{4} \: \: \: ...(1)$

$\implies \sf q = \frac{3}{4} - \frac{1}{2}$

$\implies \sf q = \frac{1}{4}$

$\sf Here,$

$\implies \sf \frac{1}{3x + y} = p$

$\implies \sf \frac{1} {3x + y} = \frac{1} {2}$

$\implies \sf {3x + y} = {2} \: \: \: ...(3)$

$\implies \sf \frac{1}{3x - y} = q$

$\implies \sf \frac{1}{3x - y} = \frac{1}{4}$

$\implies \sf {3x - y} = {4} \: \: \: ...(4)$

$\sf We \: have,$

$\sf {3x + y} = {2} \: \: \: ...(3)$

$\sf {3x - y} = {4} \: \: \: ...(4)$

$\sf Add \: Both \: Eq [3] \: and \: Eq [4],$

$\implies \sf {3x + y} + (3x - y) = {2} + 4$

$\implies \sf {3x + y} + 3x - y= {2} + 4$

$\implies \sf 6x = 6$

$\sf Substitute \: this \: value \: of \: x \: in \: Eq [3],$

$\implies \sf {3x + y} = {2} \: \: \: ...(3)$

$\implies \sf {3 + y} = {2}$

$\implies \sf y = - 1$

$\sf Therefore,$

$\sf x = 1 \: \And \: y = -1$

FiNaL AnSwEr,

$\color{red}\boxed{\sf x = 1 \: \And \: y = -1}$

Hope you get your AnSwEr