Question

I need help on this edmentum problem

Answer 1

`\displaystyle (x+3)/(5x-2)`

Explanation:

`\displaystyle (x-1)/(x^2-4x+3)/(5x-2)/(x^2-9)\\\\=(x-1)/((x-3)(x-1))/(5x-2)/((x-3)(x+3))\\\\=(1)/((x-3))/(5x-2)/((x-3)(x+3))\\\\=(1)/((x-3))\cdot((x-3)(x+3))/(5x-2)\\\\=((x-3)(x+3))/((x-3)(5x-2))\\\\=(x+3)/(5x-2)`

Remember when dividing fractions to take the reciprocal of the second fraction and then multiply the two fractions.

Answer 2

Explanation:

To divide fractions ....flip the second one and MULTIPY.....

Factor the denominator in the first one and the now numerator in the second one :

(x-1)/ ((x-1)(x-3) ) * (x+3)(x-3) / 5x-2) <====note all of the factors that 'cancel out'

to leave

1 * (x+3) / 5x-2 = (x+3) / 5x-2