Question

What is the quotient [3y+2/3y] / [6y^2+4y/3y+2]

Answer 1

[3y+2/3y] / [6y^2+4y/3y+2]
Steps
First, divide fractions
= (3y + 2)(3y + 2)/3y(6y^2 + 4y)
Then... factor 3y(6y^2 + 4y): 6y^2(3y + 2)
= (3y + 2)(3y + 2)/6y^2(3y + 2)
Then, cancel the common factor: 3y + 2
= 3y + 2/6y^2

Answer 2

Answer:$\frac{3y+2}{6y^2}$

Step-by-step explanation: Given rational expression : $\frac{\frac{3y+2}{3y} }{\frac{6y^2+4y}{3y+2} }$.

Rewriting it in proper form.

$\frac{3y+2}{3y}$ ÷$\frac{6y^2+4y}{3y+2}$

Converting division sign into multiplication and flipping second fraction, we get

$\frac{3y+2}{3y}$×$\frac{3y+2}{6y^2+4y}$

Factoring out 2y from$6y^2+4y$

$6y^2+4y = 2y(3y+2)$

$\frac{3y+2}{3y}$×$\frac{3y+2}{2y(3y+2)}$

Crossing out common factors (3y+2) in second fraction, we get

=$\frac{3y+2}{3y}$×$\frac{1}{2y}$

=$\frac{3y+2}{6y^2}$.