Question

How do you evaluate `\frac{k^{2}+2k-35}{3k^{2}-15k}\div \frac{k+7}{2k+1}`?

Answer 1

The simplified expression is `(2k+1)/(3k)`.

Explanation:

Factor the polynomials and cancel the like terms:

`(k^2+2k-35)/(3k^2-15k)-:(k+7)/(2k+1)`

`((k+7)(k-5))/(3k(k-5))-:(k+7)/(2k+1)`

`((k+7)cancel((k-5)))/(3kcancel((k-5)))-:(k+7)/(2k+1)`

`cancel(k+7)/(3k)*(2k+1)/cancel(k+7)`

`(2k+1)/(3k)`