Question

How do you simplify and find the restrictions for `(3x-2)/(x+3)+7/(x^2-x-12)`?

Answer 1

`=((x-3)(3x-5))/((x-4)(x+3)`
Restrictions are
`x!=4`
`x!=-3`

Explanation:

`(3x-2)/(x+3)+7/(x^2-x-12)`
`=(3x-2)/(x+3)+7/(x^2-4x+3x-12)`
`=(3x-2)/(x+3)+7/(x(x-4)+3(x-4)`
`=(3x-2)/(x+3)+7/((x-4)(x+3))`
`=((3x-2)(x-4)+7)/((x-4)(x+3))`
`=(3x^2-12x-2x+8+7)/((x-4)(x+3))`
`=(3x^2-14x+15)/((x-4)(x+3))`
`=(3x^2-9x-5x+15)/((x-4)(x+3))`
`=(3x(x-3)-5(x-3))/((x-4)(x+3))`
`=((x-3)(3x-5))/((x-4)(x+3)`
Restrictions are
`x-4!=0`
or
`x!=4`
`x+3!=0`
or
`x!=-3`