Question

1) 5y+1 / y + 1 = y+2 / 2
2) y(в квадрате )/ y + 3= y / y+3

Answer 1

1) 5y+1/y+1=y+2/2
5y+1/y+1=y+2+2, y≠-1
2(5y+1)=(y+2)(y+1)
2(5y+1)-(y+2)(y+1)=0
10y+2-(y²+y+2y+2)=0
10y+2-(y²+3y+2)=0
10y+2-y²-3y-2=0
7y-y=0
y(7-y)=0
y=0
7-y=0
y=0
y=7,  y≠-1
y₁=0
y₂=7