Simplify the complex fraction [(6y² + y − 1)/(4y² + 4y + 1)] ÷ [(3y² + 2y − 1)/(2y² − 7y − 4)].

Step-by-step solution

1. 1

   Read the complex fraction as a division. A fraction whose numerator and denominator are themselves fractions means

   $\frac{6y^{2}+y-1}{4y^{2}+4y+1}\div\frac{3y^{2}+2y-1}{2y^{2}-7y-4}$

2. 2

   Factor all four quadratics.

   $6y^{2}+y-1=(3y-1)(2y+1),\qquad 4y^{2}+4y+1=(2y+1)^{2}$

   $3y^{2}+2y-1=(3y-1)(y+1),\qquad 2y^{2}-7y-4=(2y+1)(y-4)$

   Note $4y^{2}+4y+1$ is a perfect square — spotting that saves a round of the ac method.

3. 3

   Invert the divisor and assemble.

   $\frac{(3y-1)(2y+1)}{(2y+1)^{2}}\cdot\frac{(2y+1)(y-4)}{(3y-1)(y+1)}$

4. 4

   Cancel systematically. Multiplying out the factored form, the numerator is $(3y-1)(2y+1)^{2}(y-4)$ and the denominator is $(2y+1)^{2}(3y-1)(y+1)$. Both $(3y-1)$ and $(2y+1)^{2}$ appear top and bottom, so they cancel completely:

   $\frac{(3y-1)(2y+1)^{2}(y-4)}{(2y+1)^{2}(3y-1)(y+1)}=\frac{y-4}{y+1}$

   $\boxed{\dfrac{y-4}{y+1}}\qquad\left(y\neq -\tfrac12,\ \tfrac13,\ -1,\ 4\right)$

5. 5

   Check with a test value. At $y=0$: the top fraction is $\tfrac{-1}{1}=-1$, the bottom fraction is $\tfrac{-1}{-4}=\tfrac14$, so the quotient is $-1\div\tfrac14=-4$; the answer gives $\tfrac{-4}{1}=-4$ ✓.

6. 6

   Note the excluded values. Beyond the visible $y=-1$, the original expression also requires $y\neq-\tfrac12$ (both denominators of the inner fractions), $y\neq\tfrac13$ and $y\neq 4$ (so the divisor is neither zero nor undefined).