`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

The given equations
`(1)/(2(2x + 3y))+ (12)/(7(3x - 2y)) = (1)/(2)` ….(1)
and `(7)/(2x + 3y) + (4)/(3x - 2y) = 2` …. (2)
are not the linear equations. So, first of all we shall make these as a pair of linear equations, by putting `(1)/(2x + 3y) = a` and `(1)/(3x - 2y) = b`.
Therefore, equations (1) and (2) become,
`(1)/(2) a + (12)/(7) b = (1)/(2) ....(3)`
and 7a + 4b = 2 ....(4)
Multiplying equation (3) by 14, we get
7a + 24b = 7 .....(5)
Subtracting equation (5) from (4), we get
7a + 24b = 7
and `{:(7a+24b=7),(7a+4b=2),("- - -"),(bar(" "20b=5)):}`
`implies b = (1)/(4)`
putting `b = (1)/(4)` in equation (4), we get
`7a + 4 ((1)/(4)) = 2 implies 7a = 2 -1 implies a = (1)/(7)`
Since, `a=(1)/(7) implies (1)/(2x + 3y) = (1)/(7) implies 2x + 3y = 7 " "....(6)`
and `b = (1)/(4) implies (1)/(3x - 2y) = (1)/(4) implies 3x - 2y = 4 " "....(7)`
Multiplying equation (6) by 2 equation (7) by 3, we get
`{:(4x+6y=14" ....(8)",),(ul(9x-6y=12)" ....(9)",):}`
On adding, we get 13 x = 26 implies x = 2
Putting x = 2 in equation (8), we get
4 (2) + 6y = 14 implies 6y = 14 - 8 implies y = 1
Hence, the solution is `{:(x = 2),(y = 1):}}`.