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Solve : 2/x+3/3y =1/6 and 3/x+2/y =0

Answer» 3(2/x +3/3y = 1/6)2(3/x+2/y = 0)6/x +3/y = 1/26/x +4/y = 0(-) (-)0 + (-y) = 1/2y = -22/x - 3/6 = 1/62/x = 1/6 + 3/62/x = 4/64x = 12x = 3
This is a very simple question of linear equations in two variables chapter, I think you have to practice more on fractions arithmetic.\xa0{tex}\\large Question:\\frac 2x +\\frac 3{3y}=\\frac 16\\space and\\space \\frac 3x +\\frac 2y=0,\\space find\\space x,y?{/tex}Solution: 		\t\t\t{tex}\\large\\implies\\boxed {\\frac {2\\times 3y +3\\times x}{x\\times 3y}=\\frac 16 \\space \\dots eq.(i)}{/tex}{tex}\\large{\\implies {\\frac {6y +3x}{3xy}=\\frac 16 \\space }}{/tex}{tex}\\large {\\underline {Step\\space 2:\\space Cross-multiplication:} }{/tex}{tex}\\large{\\implies {6\\times{(6y +3x)}=1\\times {3xy} \\space }}{/tex}{tex}\\large{\\implies {36y +18x}={3xy} \\space }{/tex}{tex}\\large{\\implies {36y +18x}-{3xy}=0 \\space }{/tex}{tex}\\large {\\implies\\boxed {\\frac {3\\times y +2\\times x}{x\\times y}=0 \\space\\dots eq.(ii) }}{/tex}{tex}\\large {\\implies\\frac {3y +2x}{xy}=0 \\space }{/tex}{tex}\\large {\\underline {Step\\space 2:\\space Cross-multiplication:} }{/tex}{tex}\\large {\\implies{3y +2x}=0\\times {xy}=0 \\space }{/tex}{tex}\\large {\\implies\\boxed {{3y}=-2x \\space \\dots eq.(iii)}}{/tex}\xa0\xa0\t\t{tex}\\large {\\underline {Step\\space 3:\\space Putting\\space relation\\space of\\space eq.(iii):} }{/tex}{tex}\\large{\\implies {{\\underline {3y}}(12-x) +18x}=0 \\space }{/tex}{tex}\\large{\\implies {\\underline{(-2x)}(12-x) +18x}=0 \\space }{/tex}{tex}\\large{\\implies {-24x+2x^2 +18x}=0 \\space }{/tex}{tex}\\large{\\implies {2x^2-6x}=0 \\space }{/tex}{tex}\\large{\\implies {2x^2=6x \\space} }{/tex}{tex}\\large{\\implies\\boxed {x=2 \\space} }{/tex}{tex}\\large {\\underline {Step\\space 4:\\space Putting\\space value\\space of\\space \'x\'\\space in\\space eq.(iii):} }{/tex}{tex}\\large {\\implies{3y}=-2x \\space }\\\\\\large {\\implies{3y}=-2\\times 2 \\space }\\\\\\large\\implies\\boxed {{y}={-4\\over3} \\space }\\\\{/tex}\t\xa0


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