Question

Match the expressions with the nonpermissible replacements for y

Answer 1

-7 _D_

$\frac{3}{2}$ _A_

1 _C_

$\frac{-1}{4}$ _B_

_A_ $\frac{y^{2} -2y+1}{2y-3}$

_B_ $\frac{y(y+5)}{4y+1}$

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

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

Answer 2

Answer:

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

3/2  →      $\dfrac{y^2-2y+1}{2y-3}$

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

-1/4  →      $\dfrac{y(y+5)}{4y+1}$

Step-by-step explanation:

We need to find the nonpermissible replacements for y.

Equate the denominator of each expression equal to 0, to find the nonpermissible replacements for y.

First expression is

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

Equate (2y-3) equal to 0.

$2y-3=0$

Add 2 on both sides.

$2y=3$

Divide both sides by 3.

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

Therefore, 3/2 is the nonpermissible replacements for y.

Second expression is

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

Equate (4y+1) equal to 0.

$4y+1=0$

$4y=-1$

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

Therefore, -1/4 is the nonpermissible replacements for y.

Third expression is

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

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

$-5(y-1)=0$

$y-1=0$

$y=1$

Therefore, 1 is the nonpermissible replacements for y.

Fourth expression is

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

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

$-2(y+7)=0$

$y+7=0$

$y=-7$

Therefore, -7 is the nonpermissible replacements for y.