Question

`x/(x-3)` subtracted from `(x-2)/(x+3)`?

rational expressions

Answer 1

`-(8x-6)/((x+3)(x-3))`

Explanation:

`"before we can subtract the fractions we require "`
`"them to have a "color(blue)"common denominator"`

`"this can be achieved as follows"`

`"multiply numerator/denominator of "(x-2)/(x+3)" by "(x-3)`

`"multiply numerator/denominator of "x/(x-3)" by "(x+3)`

`rArr(x-2)/(x+3)-x/(x-3)`

`=((x-2)(x-3))/((x+3)(x-3))-(x(x+3))/((x+3)(x-3))`

`"now the denominators are common subtract the numerators"`
`"leaving the denominator as it is"`

`=(cancel(x^2)-5x+6cancel(-x^2)-3x)/((x+3)(x-3))`

`=(-8x+6)/((x+3)(x-3))=-(8x-6)/((x+3)(x-3))`

`"with restrictions on the denominator "x!=+-3`

Answer 2

`(-8x+6)/((x+3)(x-3))`

Explanation:

In order to subtract fractions, we have to make sure the denominators (i.e, the bottom part of the fractions) are the same. We are given:

`(x-2)/(x+3)-x/(x-3)`

Notice that the denominators are different. The goal is to find the Least Common Multiple. A common denominator of both `(x+3)` and `(x-3)` is some value that has both those numbers as a multiple. The fastest, easiest number that is a multiple of both `(x+3)` and `(x-3)` is the value:

`(x+3)(x-3)`

Next, convert both fractions by multiplying (both numerator and denominator) by the missing multiple. Here is what that looks like:

`(x-2)/(x+3)*color(red)(x-3)/color(red)(x-3)-(x)/(x-3)*color(red)(x+3)/color(red)(x+3)`

Rewriting gives

`((x-2)(x-3))/((x+3)(x-3))-(x(x+3))/((x+3)(x-3))`

Now that the denominators are the same value, we can subtract them

`((x-2)(x-3)-x(x+3))/((x+3)(x-3))`

Simplifying the numerator requires using FOIL and the distributive law.

`(x^2-3x-2x+6-x^2-3x)/((x+3)(x-3))`

Combining like terms, we get

`(-8x+6)/((x+3)(x-3))`