`Sol v e :1/(2(2x+3y))+(12)/(7(3x-2y))=1/2` `7/(2x+3y)+4/(3x-2y)=2` where `2x+3y!=0a n d3x-2y!=0.`

Explanation

Putting ` (1)/(( 2x + 3y )) = u and (1)/(3 x - 2y ) = v `, the given equations become
` " " (u)/(2) + (12 v) /(7) = (1)/(2) rArr 7u + 24 v = 7 " " `… (i)
and ` 7u + 4v = 2" " `… (ii)
Substracting (ii) from (i), we get
` 20 v = 5 rArr v = ( 5)/(20) = (1)/(4)`
Putting ` v = (1)/(4) ` in (i), we get
` 7u+ ( 24 xx (1)/(4)) = 7`
` rArr 7u + 6 = 7 rArr 7u = 1 rArr u = (1)/(7)`
Now, `u = (1)/(7) rArr (1)/((2x + 3y)) = (1)/(7) rArr 2x +3y = 7 " " `... (iii)
and ` v = (1)/(4) rArr (1)/((3 x - 2y )) = (1)/(4) rArr 3x - 2y = 4 " " `... (iv)
Multiplying (iii) by 2 and (iv) by 3 and adding the results, we get
` 4x + 9x = 14 + 12 `
` rArr 13 x = 26 rArr x = 2 `.
Putting ` x = 2 ` in (iii), we get
` ( 2xx 2 ) + 3y = 7 rArr 3y = ( 7 - 4) = 3 rArr y = 1 `
Hence, ` x= 2 and y = 1 `