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