We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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