## simplifying (16^{3/4}+9^{-1/2}) / (16^{3/4}-9^{-1/2})

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.

### simplifying (16^{3/4}+9^{-1/2}) / (16^{3/4}-9^{-1/2})

$\frac{16^{3/4}+9^{-1/2}}{16^{3/4}-9^{-1/2}}$

I think $16^{3/4}$ is 8 and $9^{-1/2}$ is 1/3, so I am getting $\frac{8\frac{1}{3}}{7\frac{2}{3}}$ as my answer.

The answer is $\frac{25}{23}$ so apparently I am an idiot.
EdSewersPark

Posts: 22
Joined: Mon May 23, 2011 8:30 pm

EdSewersPark wrote:...I am getting $\frac{8\frac{1}{3}}{7\frac{2}{3}}$ as my answer.

The answer is $\frac{25}{23}$....

What are you getting when you convert the "mixed number" form to improper fractions, and do the division (that is, flip and multiply)?

stapel_eliz

Posts: 1715
Joined: Mon Dec 08, 2008 4:22 pm

### Re: simplifying (16^{3/4}+9^{-1/2}) / (16^{3/4}-9^{-1/2})

So if I take $\frac{8\frac{1}{3}}{7\frac{2}{3}}$ and convert it to: $\left(\frac{\frac{25}{3}}{\frac{23}{3}}\right)$ and multiply by $(\frac{3}{3})$ My answer will be: $\frac{25}{23}$ and that was the correct way to get there?
EdSewersPark

Posts: 22
Joined: Mon May 23, 2011 8:30 pm

EdSewersPark wrote:...convert it to: $\left(\frac{\frac{25}{3}}{\frac{23}{3}}\right)$ and multiply by $(\frac{3}{3})$...

Or just flip-n-multiply, as usual when dividing by a fraction:

. . . . .$\frac{\left(\frac{25}{3}\right)}{\left(\frac{23}{3}\right)}\, =\, \left(\frac{25}{3}\right)\left(\frac{3}{23}\right)\, =\, \frac{25}{23}$

stapel_eliz

Posts: 1715
Joined: Mon Dec 08, 2008 4:22 pm

### Re: simplifying (16^{3/4}+9^{-1/2}) / (16^{3/4}-9^{-1/2})

Aha. Thank you.
EdSewersPark

Posts: 22
Joined: Mon May 23, 2011 8:30 pm