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I need help on this edmentum problem

I need help on this edmentum problem-example-1
User Adavo
by
8.3k points

2 Answers

3 votes

Answer:


\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.

User Lucas Huang
by
8.2k points
3 votes

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

User PongBongoSaurus
by
7.6k points

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