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3x_5y=09x=2y+7

Answer» \tBy Elimination method,\tThe given system of equations is :\t3 x - 5 y - 4 = 0............(1)\t9 x = 2 y + 7\t9 x - 2 y - 7 = 0.............(2)\tMultiplying equation (1) by 3, we get\t9 x - 15 y - 12 = 0.............(3)\tSubtracting equation (3) from equation (2) , we get\t13 y + 5 = 0\t{tex}\\Rightarrow \\quad 13 y = - 5 \\Rightarrow y = \\frac { - 5 } { 13 }{/tex}\tSubstituting this value of y in equation (1), we get{tex}3 x - 5 \\left( \\frac { - 5 } { 13 } \\right) - 4 = 0{/tex}\t{tex}\\Rightarrow \\quad 3 x + \\frac { 25 } { 13 } - 4 = 0 \\Rightarrow 3 x - \\frac { 27 } { 13 } = 0{/tex}\t{tex}\\Rightarrow \\quad 3 x = \\frac { 27 } { 13 } \\Rightarrow x = \\frac { 9 } { 13 }{/tex}\tSo, the solution of the given system of equation is{tex}x = \\frac { 9 } { 13 } , y = \\frac { - 5 } { 13 }{/tex}\tBy Substitution method:\tThe given system of equation is:\t3 x - 5 y - 4 = 0.............(1)\t9 x = 2 y + 7...................(2)\tFrom equation (2),\t{tex}x = \\frac { 2 y + 7 } { 9 }{/tex}..................(3)\tSubstituting this value of x in equation(1), we get\t{tex}3 \\left( \\frac { 2 y + 7 } { 9 } \\right) - 5 y - 4 = 0{/tex}\t{tex}\\Rightarrow \\quad \\frac { 2 y + 7 } { 3 } - 5 y - 4 = 0{/tex}\t{tex}\\Rightarrow \\quad 2 y + 7 - 15 y - 12 = 0{/tex}\t{tex}\\Rightarrow{/tex}\xa0-13y - 5 = 0{tex}\\Rightarrow{/tex}\xa013y = -5\t{tex}\\Rightarrow \\quad y = \\frac { - 5 } { 13 }{/tex}\tSubstituting this value of y in equation(3), we get\t{tex}x = \\frac { 2 \\left( \\frac { - 5 } { 13 } \\right) + 7 } { 9 } = \\frac { - \\frac { 10 } { 13 } + 7 } { 9 } = \\frac { - 10 + 91 } { 117 } = \\frac { 81 } { 117 } = \\frac { 9 } { 13 }{/tex}\u200b\u200b\u200b\u200b


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