Simplification of a rational expression

Question

I have been trying to simplify a rational expression; however, I am not able to reach the simplest form of the expression, here is my work:

$$\frac{5g+2}{5g-2}\div\frac{3g-1}{1-3g}$$

Then I try to simplify the equation by first multiplying by its reciprocal

$$\frac{5g+2}{5g-2}\times\frac{1-3g}{3g-1}$$

So once I have multiplication I find a common denominator

$$\frac{(5g+2)(3g-1)}{(5g-2)(3g-1)}\times\frac{(1-3g)(5g-2)}{(3g-1)(5g-2)}$$

However, after I cross out the terms due to division I end up with the same expression at the start

$$\frac{5g+2}{5g-2}\div\frac{3g-1}{1-3g}$$

What step am I missing to simplify it even more?

Answer 1

One might wish to note, as mentioned in the comments by @avs, that

$$\frac{3g-1}{1-3g}=\frac{-(1-3g)}{1-3g}=-1$$

Thus, we have

$$\frac{5g+2}{5g-2}\times(-1)$$

Answer 2

.The step you are missing is that $-1(1-3g) = -1 + 3g = 3g-1$, therefore $\frac{3g-1}{1-3g} = -1$! Thank you to avs for the point out in the comments.

Therefore, $$ \frac{5g+2}{5g-2} \div \frac{3g-1}{1-3g} = \frac{5g+2}{5g-2} \div -1 = \frac{-1(5g+2)}{5g-2} = \frac{-5g-2}{5g-2} $$

You can check that: $$ \frac{-5g-2}{5g-2} = \frac{-4}{5g-2} -1 $$

This simplification puts the expression in "partial sum" form, where only constants are on the numerators.