Find the reciprocal of the fraction 1/(5y²).

Step-by-step solution

1. 1

   Recall the definition rather than a rule. The reciprocal (multiplicative inverse) of a quantity $q$ is the quantity $q^{-1}$ with $q\cdot q^{-1}=1$. For a fraction this means swapping numerator and denominator:

   $\left(\frac{u}{v}\right)^{-1}=\frac{v}{u}$

2. 2

   Apply it to the given fraction. Here $u=1$ and $v=5y^2$, so

   $\left(\frac{1}{5y^2}\right)^{-1}=\frac{5y^2}{1}=5y^2$

   The whole denominator flips up as one block — the coefficient $5$ and the power $y^2$ travel together.

3. 3

   Check by multiplying.

   $\frac{1}{5y^2}\cdot 5y^2=\frac{5y^2}{5y^2}=1\ \checkmark$

   This is the only verification that matters, and it works for every $y$ where both sides are defined.

4. 4

   State the restriction. The original fraction needs $5y^2\neq 0$, so $y\neq 0$. The reciprocal $5y^2$ looks harmless at $y=0$, but it is only the reciprocal of the given expression on the domain the original had, namely $y\neq 0$ — at $y=0$ there is nothing to invert, since $0$ has no reciprocal.

5. 5

   Read the notation carefully — the answer depends on it. Typed without brackets, 1/5y^2 has two common readings:

   $\frac{1}{5y^2}\ \longrightarrow\ \text{reciprocal } 5y^2, \qquad \frac{1}{5}y^2\ \longrightarrow\ \text{reciprocal } \frac{5}{y^2}$

   If the $y^2$ sits in the denominator you get $5y^2$; if only the $5$ is a denominator you get $\frac{5}{y^2}$. Always bracket the denominator when typing.