Respuesta :

1) (y^2-2y+1)/(2y-3)

2y-3=0
Solving for y:
2y-3+3=0+3
2y=3
2y/2=3/2
y=3/2

2) y(y+5)/(4y+1)

4y+1=0
Solving for y:
4y+1-1=0-1
4y=-1
4y/4=-1/4
y=-1/4

3) (5y^2-6y+1)/[-5(y-1)]

-5(y-1)=0
Solving for y:
-5(y-1)/(-5)=0/(-5)
y-1=0
y-1+1=0+1
y=1

4) (y^2-y-6)/[-2(y+7)]

-2(y+7)=0
-2(y+7)/(-2)=0/(-2)
y+7=0
y+7-7=0-7
y=-7

Pairs

-7     →   (y^2-y+6)/[-2(y+7)]

3/2    →   (y^2-2y+1)/(2y-3)

1       →   (5y^2-6y+1)/[(-5(y-1)]

-1/4   →   y(y+5)/(4y+1)

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