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\(-2\sqrt{3}\)