X/2+2y/3 =-1 and x-y/3=3

{tex}\frac{x}{2} + \frac{{2y}}{3} = - 1{/tex} ....(1)
{tex}x - \frac{y}{3} = 3{/tex} ...(2)

1. Elimination method: Multiplying equation (2) by 2, we get (3)
   {tex}2x - \frac{2}{3}y = 6{/tex} ....(3)
   {tex}\frac{x}{2} + \frac{{2y}}{3} = - 1{/tex} ....(1)
   Adding (3) and (1), we get
   {tex}\frac{5}{2}x = 5{/tex}
    ⇒ x =2
   Putting value of x in (2), we get
   2− {tex}\frac{y}{3}{/tex}= 3
   ⇒ y =−3
   Therefore, x =2 and y =−3
2. Substitution method:{tex}\frac{x}{2} + \frac{{2y}}{3} = - 1{/tex} ....(1)
   {tex}x - \frac{y}{3} = 3{/tex} ....(2)
   From equation (2), we can say that {tex}x = 3 + \frac{y}{3} = \frac{{9 + y}}{3}{/tex}
   Putting this in equation (1), we get
   {tex}\frac{{9 + y}}{6} + \frac{2}{3}y = - 1{/tex}
   _{{tex} \Rightarrow \;\frac{{9 + y + 4y}}{6} = - 1{/tex}}
   ⇒ 5y +9=−6
   ⇒ 5y =−15
   ⇒ y =−3
   Putting value of y in (1), we get
   {tex}\frac{x}{2} + \frac{2}{3}( - 3) = - 1{/tex}
   ⇒ x = 2
   Therefore, x =2 and y =−3.