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Solve for x and y given x≠0,y≠0, 2/x+2/3=1/6, 3/x+2/y=0 hence find p for which y=px-4 |
| Answer» Taking\xa0{tex} \\frac { 1 } { x } = u{/tex}\xa0and\xa0{tex} \\frac { 1 } { y } = v.{/tex}The given system of equations become{tex} 2 u + \\frac { 2 } { 3 } v = \\frac { 1 } { 6 }{/tex}Therefore,\xa0{tex} 12u+4v=1{/tex}............(i)and, {tex}3u+2v=0{/tex}..........(ii)Multiplying (ii) by 2 and subtracting from (i), we get{tex} 6 u = 1 \\Rightarrow u = \\frac { 1 } { 6 }{/tex}Putting\xa0{tex} u = \\frac { 1 } { 6 }{/tex}in (i), we get{tex} 2 + 4 v = 1 \\Rightarrow v = - \\frac { 1 } { 4 }{/tex}Hence,\xa0{tex} x = \\frac { 1 } { u } = 6{/tex}\xa0and\xa0{tex} y = \\frac { 1 } { v } = - 4{/tex}So. the solution of the given system of equations is {tex}x=6,y=-4{/tex}\xa0Putting x = 6, y = -4 in {tex}y=ax-4{/tex}, we get{tex}-4=6a-4{/tex}{tex} \\Rightarrow a=0{/tex}\xa0 | |