Drag the tiles to the correct boxes to complete the pairs.
Match the expressions with the nonpermissible replacements for y.
947 97
y(y + 5)
41 +1
5y2 – 6y +
1
-5(y – 1)
32 – 2y + 1
2y - 3

Respuesta :

Answer:

Part 1)  -7 →  [tex]\dfrac{y^2-y+6}{-2(y+7)}[/tex]  

Part 2) 3/2 → [tex]\dfrac{y^2-2y+1}{2y-3}[/tex] 

Part 3) 1  → [tex]\dfrac{5y^2-6y+1}{-5(y-1)}[/tex]

Part 4) -1/4 → [tex]\dfrac{y(y+5)}{4y+1}[/tex]

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

[tex]\dfrac{y^2-2y+1}{2y-3}[/tex]

Equate (2y-3) equal to 0.

[tex]2y-3=0[/tex]

solve for y

Adds 3 both sides.

[tex]2y=3[/tex]

Divide both sides by 2.

[tex]y=\dfrac{3}{2}[/tex]

therefore

3/2 is the non permissible replacement for y.

step 2

we have

[tex]\dfrac{y(y+5)}{4y+1}[/tex]

Equate (4y+1) equal to 0.

[tex]4y+1=0[/tex]

Subtract 1 both sides

[tex]4y=-1[/tex]

Divide by 4 both sides

[tex]y=-\dfrac{1}{4}[/tex]

therefore

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

step 3

we have

[tex]\dfrac{5y^2-6y+1}{-5(y-1)}[/tex]

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

[tex]-5(y-1)=0[/tex]

Divide by -5 both sides

[tex]y-1=0[/tex]

Adds 1 both sides

[tex]y=1[/tex]

therefore

1 is the non permissible replacement for y.

step 4

we have

[tex]\dfrac{y^2-y+6}{-2(y+7)}[/tex]

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

[tex]-2(y+7)=0[/tex]

Divide by -2 both sides

[tex]y+7=0[/tex]

Subtract 7 both sides

[tex]y=-7[/tex]

therefore

-7 is the non permissible replacement for y

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