Question

solv for x, y ; 3/(x+y)+4/(x-y)=5 and 5/(x+y)+1/3(x-y)=2​

Answer 1

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

Given pair of equations are

$\rm \: \dfrac{3}{x + y} + \dfrac{4}{x - y} = 5$

and

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

Let assume that,

$\rm \: \dfrac{1}{x + y} = u - - - (1)$

and

$\rm \: \dfrac{1}{x - y} = v - - - (2)$

So, given equations can be rewritten as

$\rm \: \: 3u + 4v = 5 - - - (3)$

and

$\rm \: \: 5u + \dfrac{v}{3} = 2$

$\rm \: \: \dfrac{15u + v}{3} = 2$

$\rm \: 15u + v = 6 - - - (4)$

So, on multiply equation (4) by 4, we get

$\rm \: 60u + 4v = 24- - - (5)$

On Subtracting equation (3) from equation (5), we get

$\rm \: 57u = 19$

$\rm\implies \:u \: = \: \dfrac{1}{3}$

On substituting the value of u from equation (1), we get

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

$\rm\implies \:x + y = 3 - - - (6)$

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

$\rm \: 1 + 4v = 5$

$\rm \: 4v = 5 - 1$

$\rm \: 4v = 4$

$\rm\implies \:v = 1$

On substituting the value of v, from equation (2), we get

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

$\rm\implies \:x - y = 1 - - - (7)$

On adding equation (6) and (7), we get

$\rm \: 2x = 4$

$\rm\implies \:x = 2$

On substituting the value of x in equation (6), we get

$\rm \: 2 + y = 3$

$\rm\implies \:y = 1$

Hence,

The solution of equations

$\rm \: \dfrac{3}{x + y} + \dfrac{4}{x - y} = 5$

and

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

is

$\: \boxed{\bf{  \:x \: = \: 2 \: \: }} \: \: \: and \: \: \: \boxed{\bf{  \: \: y \: = \: 1 \: \: }}$

Verification :-

Consider

$\rm \: \dfrac{3}{x + y} + \dfrac{4}{x - y} = 5$

On substituting the value of x and y in above equation, we get

$\rm \: \dfrac{3}{2 + 1} + \dfrac{4}{2 - 1} = 5$

$\rm \: \dfrac{3}{3} + \dfrac{4}{1} = 5$

$\rm \: 1 + 4 = 5$

$\rm \: 5 = 5$

Hence, Verified