Question

Solving Rational equations. LCD method. Show work. Image attached.


`(4)/(k^(2) -8k +12 ) = (k)/(k-2) + (1)/(k-6)`

Answer 1

`(k-2)(k-6)=k^2-2k-6k+12=k^2-8k+12`

So in order to get all the fractions to have a common denominator, we need to multiply
`\frac k{k-2}` by
`(k-6)/(k-6)`, and
`\frac1{k-6}` by
`(k-2)/(k-2)`:

`\frac k{k-2}\cdot(k-6)/(k-6)=(k(k-6))/((k-2)(k-6))=(k^2-6k)/(k^2-8k+12)`

`\frac1{k-6}\cdot(k-2)/(k-2)=(k-2)/((k-2)(k-6))=(k-2)/(k^2-8k+12)`

Now,

`\frac4{k^2-8k+12}=((k^2-6k)+(k-2))/(k^2-8k+12)`

As long as
`k\\eq2` and
`k\\eq6` (which we can't have because otherwise
`k^2-8k+12=0`), we can cancel
`k^2-8k+12` in the denominators on both sides:

`4=(k^2-6k)+(k-2)`

`4=k^2-5k-2`

`0=k^2-5k-6`

We can factorize the right side:

`0=(k-6)(k+1)`

which tells us that
`k=6` and
`k=-1` are solutions.