If $7b-\frac{1}{4b}=7$, then what is the value of $16b^2+\frac{1}{49b^2}$?

The correct answer is $\frac{120}{7}$.

We are given:

$7b-\frac{1}{4b}=7$

We need to find the value of $16b^2+\frac{1}{49b^2}$.

Step 1: Transform the given equation into a useful form.

Notice that the expression we want to find contains $16b^2$ and $\frac{1}{49b^2}$, which are perfect squares of $4b$ and $\frac{1}{7b}$ respectively. This hints that we should try to form the expression $4b-\frac{1}{7b}$.

Multiply both sides of the given equation by $\frac{4}{7}$:

$\frac{4}{7}\times \left(7b-\frac{1}{4b}\right)=\frac{4}{7}\times 7$

$\frac{4}{7}\times 7b-\frac{4}{7}\times \frac{1}{4b}=4$

$4b-\frac{1}{7b}=4$

Step 2: Square both sides of this new equation.

Using the identity ${\left(x-y\right)}^2=x^2-2xy+y^2$, square the left side with $x=4b$ and $y=\frac{1}{7b}$:

${\left(4b-\frac{1}{7b}\right)}^2=4^2$

${\left(4b\right)}^2-2\times 4b\times \frac{1}{7b}+{\left(\frac{1}{7b}\right)}^2=16$

$16b^2-\frac{8}{7}+\frac{1}{49b^2}=16$

Step 3: Isolate the target expression.

Add $\frac{8}{7}$ to both sides:

$16b^2+\frac{1}{49b^2}=16+\frac{8}{7}$

Step 4: Simplify the right side.

Convert 16 to a fraction with denominator 7:

$16+\frac{8}{7}=\frac{112}{7}+\frac{8}{7}=\frac{120}{7}$

Therefore:

$16b^2+\frac{1}{49b^2}=\frac{120}{7}$