Question

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

Answer 1

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

Explanation:

Given : Takumi and Kaliska manipulated the rational expression
`(-9y^2-3y)/(-6y^2-6y)`

Takumi Kaliska

`(-3y(3y+1))/(-6y(y+1))`
`((3y+1))/((2y+2))`

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

Solution :

Simplifying the rational expression,

`(-9y^2-3y)/(-6y^2-6y)`

Taking common term out,

`=(-3y(3y+1))/(-6y(y+1))`

Cancel the like terms in numerator and denominator,

`=((3y+1))/(2(y+1))`

`=((3y+1))/((2y+2))`

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.