Algebra

We can use subsitution to our advantage. 

Now, first, let's put a terms of b. From the second equation, we get that

$a=\frac{4}{b^2}$

Subsituting this value of x into the first equation, we get

$\frac{4}{b^2} \cdot b^4 = 48\\$

Now, the top and the bottom cancel out, so we have

$4b^2 = 48\\ b^2=12\\ b=2\sqrt{3},\:b=-2\sqrt{3}$

Thus, we find two values of b. We have

$b=2\sqrt{3},\:b=-2\sqrt{3}$

Thanks! :)

We could do $ab^4/ab^2=48/4$

So the A will cancel out and we will be left with $b^2=12$

So the answer is $2 \sqrt{3}$ or$-2\sqrt{3}$