Respuesta :

kphung76127
Cross multiply and solve
1(y^2-9)=6(y-3)

y^2-9= 6y-18
y^2-6y-9+18=0
y^2-6y+9=0

(y-3)(y-3)=0
(y-3)=0
y=3

When you plug y in however, it makes the denominator 0, making both fractions undefined.

Final answer: No real solution