Question

How do you multiply `(1/(r-2)) + (1/(r^2-7r+10))=6/(r-2)`?

Answer 1

In order to sum (and not multiply) the first two fractions, we must find the lowest common denominator among them. In order to simplify things for us, we can factor the second denominator, finding its roots, as follows:

`r^2-7r+10`

Using Bhaskara:

`(7+-sqrt(49-4(1)(10)))/2`
Solving: `r_1=5` and `r_2=2`
Rewriting these roots, we can say that `r-5=0` and `r-2=0`, so

`r^2-7r+10 = (r-5)(r-2)`

Rewriting your sum, we get

`1/(r-2)+1/((r-2)(r-5)` where `(r-2)(r-5)` is the lowest common denominator.

Summing them: `((r-5)+1)/((r-2)(r-5)) = (r-4)/((r-2)(r-5))`

Now, we have the equation

`(r-4)/((r-2)(r-5)) = 6/(r-2)`

Cross multiplying them, we get

`(cancel((r-2))(r-4))/(cancel((r-2))(r-5)) = 6`

Now, moving `(r-5)` to he right side, we aim at isolating `r`:

`r-4=6(r-5)`
`r-4=6r-30`
`26=5r`
`color(Green)(r=5.2)`