[tex] {\qquad\qquad\huge\underline{{\sf Answer}}} [/tex]
let's solve ~
[tex]\qquad \sf \dashrightarrow \: \cfrac{4}{y - 6} + \cfrac{5}{y + 3} = \cfrac{7y - 4}{ {y}^{2} - 3y - 18} [/tex]
[tex]\qquad \sf \dashrightarrow \: \cfrac{4(y + 3) + 5(y - 6)}{(y - 6)(y + 3)} = \cfrac{7y - 4}{ {y}^{2} - 3y - 18} [/tex]
[tex]\qquad \sf \dashrightarrow \: \cfrac{4y + 12 + 5y - 30}{ {y}^{2} + 3y - 6y - 18 } = \cfrac{7y - 4}{ {y}^{2} - 3y - 18} [/tex]
[tex]\qquad \sf \dashrightarrow \: \cfrac{9y - 18}{ {y}^{2} - 3y - 18 } = \cfrac{7y - 4}{ {y}^{2} - 3y - 18} [/tex]
[tex]\qquad \sf \dashrightarrow \: 9y - 18 = 7y - 4[/tex]
[ denominator is same, so numerator must have same value to be equal ]
[tex]\qquad \sf \dashrightarrow \: 9y - 7y = - 4 + 18[/tex]
[tex]\qquad \sf \dashrightarrow \: 2y = 14[/tex]
[tex]\qquad \sf \dashrightarrow \: y = 7[/tex]
