My expression is as shown: \({8y^2 - 2y - 3\over y^2 - 1}\)÷\({2y^2 - 3y - 2 \over 2y - 2}\)÷\({3 - 4y \over y + 1}\)

I know that when dividing rational expressions, you follow procedures similar to dividing fractions: **convert the division to multiplication by mulitiplying by the reciprocal of the divisor**.

However, I'm given three rational expressions... do I apply the same procedures to **both **of the second and third rational expressions? Or should I just "divide" the first two, then with my answer, divide the third expression?

I never came across a question like this so I wanted to make sure what I was doing was correct...

Guest Apr 3, 2018

#1**+3 **

a ÷ b ÷ c = (a ÷ b) ÷ c

So...

\(\begin{array}\ \frac{8y^2-2y-3}{y^2-1}\div\frac{2y^2-3y-2}{2y-2}\div\frac{3-4y}{y+1}\,&=&\,\Big(\frac{8y^2-2y-3}{y^2-1}\div\frac{2y^2-3y-2}{2y-2}\Big)\div\frac{3-4y}{y+1}\\~\\ &=&\,\Big(\frac{8y^2-2y-3}{y^2-1}\cdot\frac{2y-2}{2y^2-3y-2}\Big)\div\frac{3-4y}{y+1}\\~\\ &=&\,\Big(\frac{8y^2-2y-3}{y^2-1}\cdot\frac{2y-2}{2y^2-3y-2}\Big)\cdot\frac{y+1}{3-4y}\\~\\ &=&\,\frac{8y^2-2y-3}{y^2-1}\cdot\frac{2y-2}{2y^2-3y-2}\cdot\frac{y+1}{3-4y}\\~\\ &=&\,\frac{(8y^2-2y-3)(2y-2)(y+1)}{(y^2-1)(2y^2-3y-2)(3-4y)} \end{array}\)

Then we can factor the numerator and denominators and cancel the common factors.

\(\begin{array}\ \frac{8y^2-2y-3}{y^2-1}\div\frac{2y^2-3y-2}{2y-2}\div\frac{3-4y}{y+1}\,&=&\,\frac{(8y^2-2y-3)(2y-2)(y+1)}{(y^2-1)(2y^2-3y-2)(3-4y)}\\~\\ &=&\,\frac{(2y+1)(4y-3)(2)(y-1)(y+1)}{(y+1)(y-1)(y-2)(2y+1)(-1)(4y-3)}\\~\\ &=&\,\frac{(2)}{(y-2)(-1)} \\~\\ &=&\,-\frac{2}{y-2}\qquad\text{and }\,y\neq-1,1,-\frac12,\frac34 \end{array}\)

So in this case, flipping the latter two fractions and changing the signs to multiplication will work. But note that if problem were given as a ÷ (b ÷ c) , only the middle fraction would end up being flipped. So it is easiest to do them one at a time. Just remember that if there are no parenthesees given, it means the same as (a ÷ b) ÷ c .

hectictar Apr 3, 2018