Question

How do you divide `(r+9)/(r^2 - 8r + 15) -: (r^2 + 8r - 9)/(r-3)`?

Answer 1

Factoring! After I did that, I ended up with `1/(r^2-6r+5)`

Explanation:

`(r+9)/(r^2-8r+15) -: (r^2+8r-9)/(r-3)`

the complex equations can be factored down:

`(r+9)/((r-3)(r-5)) -: ((r+9)(r-1))/(r-3)`

then we can eliminate the r-9 and r-3 terms to get:

`1/(r-5) -: (r-1)`

flipping over the divisor and multiplying gives me the answer:

`1/((r-5)(r-1)) = 1/(r^2-6r+5)`

Answer 2

`(r+9)/(r^2-8r+15)-:(r^2+8r-9)/(r-3)=color(blue)(1/(r^2-6r+5)`

Explanation:

Divide:

`(r+9)/(r^2-8r+15)-:(r^2+8r-9)/(r-3)`

When dividing fractions, invert the second fraction (the divisor) and multiply.

`(r+9)/(r^2-8r+15)xx(r-3)/(r^2+8r-9)`

Factor the denominators.

`(r+9)/((r-5)(r-3))xx(r-3)/((r+9)(r-1))`

Multiply the numerators and denominators.

`((r+9)(r-3))/((r-5)(r-3)(r+9)(r-1))`

Since `(r+9)` and `(r-3)` occur in both the numerator and denominator, they cancel out.

`(color(red)cancel(color(black)((r+9)))^1color(red)cancel(color(black)((r-3)))^1)/((r-5)color(red)cancel(color(black)((r-3)))^1color(red)cancel(color(black)((r+9)))^1(r-1))`

Simplify.

`1/((r-5)(r-1))`

Expand.

`1/(r^2-6r+5)`