Question

What is the difference 2x-3/3x^2-x+2/6x

Answer 1

Answer: D

Step-by-step explanation:

To find the difference, we want to make sure both denominators are equal.

$\frac{(6x)(2x-3)}{(6x)(3x^2)} -\frac{(3x^2)(x+2)}{(3x^2)(6x)}$

Now that the denominators are equal, we can distribute and multiply out.

$\frac{12x^2-18x}{18x^3} -\frac{3x^3+6x^2}{18x^3}$

With the fractions multiplied out, we can actually put it into one large fraction.

$\frac{12x^2-18x-3x^3-6x^2}{18x^3}$

Let's subtract the top expression.

$\frac{-3x^3+6x^2-18x}{18x^3}$

This may seem like our final answer, but we can actually factor out an 3x in the numerator and denominator.

$\frac{(3x)(-x^2+2x-6)}{(3x)(6x^2)}$

With the 3x factored out, they cancel out because 3x/3x=1.

Our final answer is:

$\frac{-x^2+2x-6}{6x^2}$