\(4x +\frac{y}{3} =\frac {8}{3}\)
⇒ 12 x + y = 8 — (1) ( on multiplying both sides by 3)
and \(\frac{x}{2}+\frac{y}{4}=-\frac{5}{2}\)
⇒ 2x + y = -10 — (2) ( on multiplying both sides by 4)
subtract equation (2) from (1) , we get
(12x+y) - (2x+y) = 8 - (-10)
⇒ 10x = 18
⇒ x = \(\frac{18}{10}=\frac{9}{5}
\)
∴ from (2), we obtain
\(\frac{18}{5}+y=-10\)
⇒ y = -10 - \(\frac{18}{5}\) = \(\frac{−50−18}{5}=\frac{−68}{5}\)