Question

How do you solve `(x-2)/(x^2-7x+10)=1/(5x-10)-1/(x-5)`?

Answer 1

Simplify by getting everything on the same common denominator and then rearrange to find x

Explanation:

`(x-2)/((x-2)(x-5)) = ((x-5) - (5x - 10))/((5x - 10)(x-5))`
`(x-2)/((x-2)(x-5)) = (-4x + 5)/(5(x-2)(x-5)`
`5(x-2) = (5 - 4x)`
`5x + 4x = 5 + 10`
`x = 15/9 = 5/3`

Answer 2

You can simplify the left side.

Explanation:

Obtaining the roots for `x^2-7x+10=0`
`Delta = 49 -40= 9`
`x=(7+-sqrt(9))/2` => `x_1 = 2` and `x_2=5`
Then `x^2-7x+10 = (x-x_1)(x-x_2)=(x-2)(x-5)`

In the expression:
`cancel(x-2)/(cancel((x-2))(x-5))=1/(5x-10)-1/(x-5)`
`2/(x-5)=1/(5x-10)`
`10x-20=x-5`
`9x=15` => `x=5/3`

Answer 3

`x=5/3`

Explanation:

Factor each denominator.

`(x-2)/((x-5)(x-2))=1/(5(x-2))-1/(x-5)`

Simplify the first fraction by dividing `(x-2)/(x-2)=1`.

`1/(x-5)=1/(5(x-2))-1/(x-5)`

Add `1/(x-5)` to both sides.

`2/(x-5)=1/(5(x-2)`

Cross multiply.

`10(x-2)=x-5`

Simplify and solve.

`10x-20=x-5`

`9x-20=-5`

`9x=15`

`x=15/9`

`x=5/3`

When solving rational equations such as these, it's always a good idea to go back and check that this answer isn't an invalid answer. Invalidity would occur if `x=5/3` caused a denominator to equal `0`, but the only two values for `x` that would cause that here are `2` and `5`, so this is a valid answer.