Question

2/x+2/3y=1/6;3/x+2/y=0​

Answer 1

I hope ,it is right

Answer 2

Answer:

x=6 and y =-4

Step-by-step explanation:

$\dfrac{2}{x}+\dfrac{2}{3y}=\dfrac{1}{6}.....equ(1)\\let\:\dfrac{1}{x}\:=p\:and\dfrac{1}{y}=q\\\\\implies\:2p+\dfrac{2q}{3}=\dfrac{1}{6}\\\\\implies\:\dfrac{6p+2q}{3}=\dfrac{1}{6}\\\\\implies6p+2q=\dfrac{1}{2}....equ(2)$$\dfrac{3}{x}+\dfrac{2}{y}=0\\\\\implies3p+2q=0......equ(3)\\\\subtract\:equ(3)from\:equ(2)\\\\\implies6p+2q=\dfrac{1}{2}\\-(3p+2q=0)\\..........................\\\\after\:it\:we\:get\:,\:3p=\dfrac{1}{2}\implies\:p=\dfrac{1}{6}\\\\substitute\:the\:values\:in\:equ(3)\\\\3p+2q=0\\\\\implies3*\dfrac{1}{6}+2q=0\\\\\implies\dfrac{1}{2}+2q=0\:\implies\:2q=-\dfrac{1}{2}\:\implies\:q=-\dfrac{1}{4}\\\\substitute\:values\:for\:x\:and\:y\\\\\implies\:\dfrac{1}{x}=p\:\implies\:\dfrac{1}{x}=\dfrac{1}{6}\:\implies\:x=6$$similarly\\\\\dfrac{1}{y}=q\:\implies\:\dfrac{1}{y}=-\dfrac{1}{4}\\\\\implies\:y=-4$