Takumi and Kaliska manipulated the rational expression \dfrac{-9y^2 - 3y}{-6y^2 - 6y}

−6y

2

−6y

−9y

2

−3y

start fraction, minus, 9, y, squared, minus, 3, y, divided by, minus, 6, y, squared, minus, 6, y, end fraction. Their responses are shown below.

Takumi Kaliska

\dfrac{-3y(3y+1)}{-6y(y+1)}

−6y(y+1)

−3y(3y+1)

start fraction, minus, 3, y, left parenthesis, 3, y, plus, 1, right parenthesis, divided by, minus, 6, y, left parenthesis, y, plus, 1, right parenthesis, end fraction \dfrac{3y+1}{2y+2}

2y+2

3y+1

start fraction, 3, y, plus, 1, divided by, 2, y, plus, 2, end fraction

Question

contestada

Respuesta :

pinquancaro pinquancaro · Answer 1 · 2020-04-23T08:48:51+02:00

Answer:

Both Takumi and Kaliska expression is equivalent to the original expression.

Step-by-step explanation:

Given : Takumi and Kaliska manipulated the rational expression [tex]\frac{-9y^2-3y}{-6y^2-6y}[/tex]

Takumi Kaliska

[tex]\frac{-3y(3y+1)}{-6y(y+1)}[/tex] [tex]\frac{(3y+1)}{(2y+2)}[/tex]

To find : Which student write an expression that is equivalent to the original expression ?

Solution :

Simplifying the rational expression,

[tex]\frac{-9y^2-3y}{-6y^2-6y}[/tex]

Taking common term out,

[tex]=\frac{-3y(3y+1)}{-6y(y+1)}[/tex]

Cancel the like terms in numerator and denominator,

[tex]=\frac{(3y+1)}{2(y+1)}[/tex]

[tex]=\frac{(3y+1)}{(2y+2)}[/tex]

On simplifying we see that both Takumi and Kaliska expression is equivalent to given expression.

Therefore, Both Takumi and Kaliska expression is equivalent to the original expression.