1/2(x+y) + 5/3(3x-2y)=-3/2 5/4(x+2y) - 3/5(3x-2y)=61/60

The given equations are
{tex}\frac { 1 } { 2 ( x + 2 y ) } + \frac { 5 } { 3 ( 3 x - 2 y ) } = - \frac { 3 } { 2 }{/tex}.....(1)
and {tex}\frac { 5 } { 4 ( x + 2 y ) } - \frac { 3 } { 5 ( 3 x - 2 y ) } = \frac { 61 } { 60 }{/tex}....(2)
Putting {tex}\frac 1{x+2y}{/tex}=u and {tex}\frac 1{3x-2y}{/tex}= v in equation (1) & equation (2) so that we may get two linear equations in the variables u & v as following:- 
{tex}\frac { 1 } { 2 } u + \frac { 5 } { 3 } v = - \frac { 3 } { 2 }{/tex}.....................(1)
{tex}\frac { 5 } { 4 } u - \frac { 3 } { 5 } v = \frac { 61 } { 60 }{/tex}.................(2)

Multiplying (1) by 36 and (2) by 100, we get
{tex}{/tex}
{tex}{/tex}
{tex}18u + 60v = -54{/tex}...............(3)
{tex}125 u - 60 v = \frac { 305 } { 3 }{/tex}.............(4)
Adding (3) and (4),we get
{tex}143 u = \frac { 305 } { 3 } - 54 = \frac { 305 - 162 } { 2 } = \frac { 143 } { 3 }{/tex}
{tex}\therefore \quad u = \frac { 1 } { 3 } = \frac { 1 } { x + 2 y }{/tex}
{tex}\therefore{/tex}{tex} x + 2y = 3{/tex}...........(5)
Putting value of u in (3), we get
{tex}1 + 10v = -9{/tex} (after dividing by 3)
{tex}\therefore 10 \mathrm { v } = - 10{/tex} or {tex}\mathrm { v } = - 1 {/tex}
{tex}\Rightarrow - 1 = \frac { 1 } { 3 x - 2 y }{/tex}
{tex}\Rightarrow 3 x - 2 y = - 1{/tex}......(6)
Adding (5) and (6), we get
{tex}4 x = 2 \quad{/tex}
{tex} \therefore x = \frac { 1 } { 2 }{/tex}
Putting value of x in (5),
{tex}\frac { 1 } { 2 } + 2 y = 3{/tex}
{tex} \text { or } 2 y = 3 - \frac { 1 } { 2 } = \frac { 5 } { 2 }{/tex}
{tex}\therefore \quad y = \frac { 5 } { 4 }{/tex}
The required solution is {tex}\mathrm { x } = \frac { 1 } { 2 } , \mathrm { y } = \frac { 5 } { 4 }{/tex}