Step-by-step explanation:
[tex]\sf \leadsto \dfrac{5y - 2}{4(y + 2)} = \dfrac{1}{3}[/tex]
[tex]\sf \leadsto \dfrac{5y - 2}{4y + 8} = \dfrac{1}{3}[/tex]
[tex]\sf \leadsto 3(5y - 2) = 1(4y + 8)[/tex]
[tex]\sf \leadsto 15y - 6 = 4y + 8[/tex]
[tex]\sf \leadsto 15y - 4y = 8 + 6[/tex]
[tex]\sf \leadsto 11y = 14[/tex]
[tex]\sf \leadsto y = \dfrac{14}{11}[/tex]
