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1/2(x+y) + 5/3(3x-2y)=-3/25/4(x+2y) - 3/5(3x-2y)=61/60 |
| Answer» The given equations are{tex}\\frac { 1 } { 2 ( x + 2 y ) } + \\frac { 5 } { 3 ( 3 x - 2 y ) } = - \\frac { 3 } { 2 }{/tex}.....(1)and\xa0{tex}\\frac { 5 } { 4 ( x + 2 y ) } - \\frac { 3 } { 5 ( 3 x - 2 y ) } = \\frac { 61 } { 60 }{/tex}....(2)Putting\xa0{tex}\\frac 1{x+2y}{/tex}=u and\xa0{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:-\xa0{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}\xa0or\xa0{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\xa0{tex}\\mathrm { x } = \\frac { 1 } { 2 } , \\mathrm { y } = \\frac { 5 } { 4 }{/tex} | |