Question

How do you evaluate `\frac { 2x } { x ^ { 2} - 25} - \frac { 1} { x - 5} = \frac { 1} { 6}`?

Answer 1

`x=1` is the only solution
`x=5` is an extraneous solution.

Explanation:

`(2x)/(x^2-25)-1/(x-5)=1/6`

multiply both sides by`(x^2-25)`

Note that `color(red)(x^2 - 25= (x+5)(x-5))`

`2x-color(red)((x+5)cancel((x-5)))/cancel((x-5))=(x^2-25)/6`

`2x-(x+5)=(x^2-25)/6`

multiply both sides by 6

`12x-6x-30=x^2-25`

`6x-30=x^2-25`

`6x-x^2+25-30 =0" "` (make a quadratic equal to 0)

`-x^2+6x-5=0`

`x^2-6x+5=0" "` make `+x^2`

`(x-5)(x-1)=0" "` find factors

`x=5 or x=1`

Check:
`x=5` will give : `(2x)/(x^2-25) rarr 10/0`

Division by 0 is not defined, so `x!=5`

`x = 1`

`(2(1))/(1^2-25)-1/(1-5)=1/6`

`(2(1))/(1-25)-1/(1-5)=1/6`

`cancel2^1/cancel(-24)^-12-1/(-4)=1/6`

`1/(-12)-1/(-4)=1/6`

`- 1/12+1/4=1/6`

`(-1+3)/12=1/6`

`cancel2^1/cancel12^6=1/6`

`1/6=1/6`