Question

Drag the tiles to the correct boxes to complete the pairs.

Answer 1

Answer:

Part 1)  -7 →  $\dfrac{y^2-y+6}{-2(y+7)}$  

Part 2) 3/2 → $\dfrac{y^2-2y+1}{2y-3}$ 

Part 3) 1  → $\dfrac{5y^2-6y+1}{-5(y-1)}$

Part 4) -1/4 → $\dfrac{y(y+5)}{4y+1}$

Step-by-step explanation:

The complete question in the attached figure

we know that

To find out the non permissible replacements for y, equate the denominator of each expression equal to 0.

step 1    

we have

$\dfrac{y^2-2y+1}{2y-3}$

Equate (2y-3) equal to 0.

$2y-3=0$

solve for y

Adds 3 both sides.

$2y=3$

Divide both sides by 2.

$y=\dfrac{3}{2}$

therefore

3/2 is the non permissible replacement for y.

step 2

we have

$\dfrac{y(y+5)}{4y+1}$

Equate (4y+1) equal to 0.

$4y+1=0$

Subtract 1 both sides

$4y=-1$

Divide by 4 both sides

$y=-\dfrac{1}{4}$

therefore

-1/4 is the non permissible replacement for y.

step 3

we have

$\dfrac{5y^2-6y+1}{-5(y-1)}$

Equate -5(y-1) equal to 0.

$-5(y-1)=0$

Divide by -5 both sides

$y-1=0$

Adds 1 both sides

$y=1$

therefore

1 is the non permissible replacement for y.

step 4

we have

$\dfrac{y^2-y+6}{-2(y+7)}$

Equate -2(y+7) equal to 0.

$-2(y+7)=0$

Divide by -2 both sides

$y+7=0$

Subtract 7 both sides

$y=-7$

therefore

-7 is the non permissible replacement for y