Question

How do you solve `(-1)/(x-3)=(x-4)/(x^2-27)`?

Answer 1

`x=-3/2` or `x=5`

Explanation:

In `(-1)/(x-3)=(x-4)/(x^2-27)`, let us multiply each side by `(x-3)(x^2-27)` to simplify we get

`(-1)/(x-3)xx(x-3)(x^2-27)=(x-4)/(x^2-27)xx(x-3)(x^2-27)` or

`(-1)/cancel((x-3))xxcancel(x-3)(x^2-27)=(x-4)/cancel(x^2-27)xx(x-3)cancel((x^2-27))` or

`-1xx(x^2-27)=(x-4)(x-3)` or

`-x^2+27=x(x-3)-4(x-3)` or

`-x^2+27=x^2-3x-4x+12` or

`-x^2+27-x^2+7x-12=0` or

`-2x^2+7x+15=0`

`2x^2-7x-15=0`

`2x^2-10x+3x-15=0` or

`2x(x-5)+3(x-5)=0` or

`(2x+3)(x-5)=0` and hence

ether `2x+3=0` i.e. `x=-3/2`

or `x-5=0` i.e. `x=5`

Answer 2

`x=5`
`x=-3/2`

Explanation:

`-1/(x-3)=(x-4)/(x^2-27`
or
`-x^2+27=(x-3)(x-4)`
or
`-x^2+27=x^2-3x-4x+12`
or
`-x^2+27=x^2-7x+12`
or
`2x^2-7x-15=0`
or
`2x^2-10x+3x-15=0`
or
`2x(x-5)+3(x-5)=0`
or
`(x-5)(2x+3)=0`
or
`x-5=0`
or
`x=5`=======Ans `1`
or
`2x+3=0`
or
`2x=-3`
or
`x=-3/2`=======Ans `2`