Putting {tex}\\frac 1x{/tex}= u and {tex}\\frac 1y{/tex}\xa0= v, the given equations become{tex}2u+ 3v\xa0= 13{/tex} …. (i){tex}5u – 4v = -2{/tex} …….(ii){tex}\\text{Multiplying (i) by 4 and (ii) by 3 and adding the results, we get}{/tex}{tex}8u\xa0+ 15u\xa0= 52 – 6{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}23u = 46{/tex}{tex}\\Rightarrow \\quad u = \\frac { 46 } { 23 } = 2{/tex}{tex}\\text{Putting u\xa0= 2 in (i), we get}{/tex}(2 {tex}\\times {/tex}\xa02) + 3v\xa0= 13 {tex}\\Rightarrow{/tex}\xa03v\xa0= 13 – 4 = 9 {tex}\\Rightarrow{/tex}\xa0v\xa0= 3.Now,u\xa0= 2{tex}\\Rightarrow \\frac { 1 } { x } = 2 \\Rightarrow 2 x = 1 \\Rightarrow x = \\frac { 1 } { 2 }{/tex}And, v\xa0= 3 {tex}\\Rightarrow \\frac { 1 } { y } = 3 \\Rightarrow 3 y = 1 \\Rightarrow y = \\frac { 1 } { 3 }{/tex}Hence,\xa0{tex}x = \\frac { 1 } { 2 } \\text { and } y = \\frac { 1 } { 3 }{/tex}