How do you solve \frac { 1} { x - 3} = \frac { 3} { x } - \frac { 3} { x - 3}1x−3=3x−3x−3?

1 Answer
graces
May 6, 2017

First you need to get a common denominator. In this case it would be x^2 -3xx2−3x.

Explanation:

To get the term 1/(x-3) to have the common denominator, you need to multiply the numerator and the denominator by x. To get the term 3/x to have the common denominator, multiply the numerator and the denominator by x-3. To get the term 3/(x-3) to have the common denominator, multiply the numerator and the denominator by x.
Then you end up with:
x/(x^2 -3x) = [(3x-9)/(x^2 -3x)] - [(3x)/(x^2 -3x)]xx2−3x=[3x−9x2−3x]−[3xx2−3x]

Next, combine the two terms on the right of the equals sign.
You end up with:
x/(x^2 -3x) = -9/(x^2 -3x)xx2−3x=−9x2−3x

Then you have to multiply the entire equation by x^2 -3xx2−3x, so you can get rid of the denominator by cancelling.
x^2 -3x*[x/(x^2 -3x) = -9/(x^2 -3x)]x2−3x⋅[xx2−3x=−9x2−3x]

x=-9x=−9
Hope this helps :)

Related questions
Impact of this question
952 views around the world
You can reuse this answer
Creative Commons License