Question

Anyone know the answer to this algebra 2 problem?

Answer 1

`\large\boxed{(x+3)/(x)}`

Explanation:

`((x)/(x-3))/((x^2)/(x^2-9))=(x)/(x-3)/(x^2)/(x^2-9)=(x)/(x-3)\cdot(x^2-9)/(x^2)\\\\\text{cancel one x}\\\\=(1)/(x-3)\cdot(x^2-3^2)/(x)\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=(1)/(x-3)\cdot((x-3)(x+3))/(x)\\\\\text{cancel (x-3)}\\\\=(1)/(1)\cdot(x+3)/(x)\\\\=(x+3)/(x)`