which expression is equivalent to the expression below?\n$\frac{\frac{6c^{2}+3c}{-4c + 2}}{\frac{2c+1}{4c-2}}…

which expression is equivalent to the expression below?\n$\frac{\frac{6c^{2}+3c}{-4c + 2}}{\frac{2c+1}{4c-2}}$\n$\frac{3c(2c - 1)}{2c+1}$\n$\frac{-3c(2c + 1)^{2}}{4(2c - 1)^{2}}$\n$3c$\n$-3c$

which expression is equivalent to the expression below?\n$\frac{\frac{6c^{2}+3c}{-4c + 2}}{\frac{2c+1}{4c-2}}$\n$\frac{3c(2c - 1)}{2c+1}$\n$\frac{-3c(2c + 1)^{2}}{4(2c - 1)^{2}}$\n$3c$\n$-3c$

Answer

Explanation:

Step1: Simplify the complex - fraction

A complex - fraction $\frac{\frac{a}{b}}{\frac{c}{d}}$ is equivalent to $\frac{a}{b}\times\frac{d}{c}$. So, $\frac{\frac{6c^{2}+3c}{-4c + 2}}{\frac{2c + 1}{4c-2}}=\frac{6c^{2}+3c}{-4c + 2}\times\frac{4c-2}{2c + 1}$.

Step2: Factor the expressions

Factor out the greatest - common factor from $6c^{2}+3c$ and $-4c + 2$ and $4c-2$. $6c^{2}+3c=3c(2c + 1)$, $-4c + 2=-2(2c - 1)$, and $4c-2=2(2c - 1)$. Then the expression becomes $\frac{3c(2c + 1)}{-2(2c - 1)}\times\frac{2(2c - 1)}{2c + 1}$.

Step3: Cancel out the common factors

Cancel out the common factors $(2c + 1)$ and $(2c - 1)$ and the common factor of 2 in the numerator and denominator. $\frac{3c(2c + 1)}{-2(2c - 1)}\times\frac{2(2c - 1)}{2c + 1}=-3c$.

Answer:

$-3c$