# 2016_Question_2

Solution:
We first set bounds on y and solve for y within those bounds.
0 < y \leq \frac{1}{25}

For these values of y, {{25y}} = \frac{1}{25y} and {{8y}} = \frac{1}{8y} as well.
Thus our equation simplifies to \frac{1}{40y}=1 and y=\frac{1}{40}.

This value of y fits inside our previous interval and it is therefore a solution.

\frac{1}{25} \leq y \leq \frac{1}{8}

For these values of y, {{25y}} = 25y and {{8y}} = \frac{1}{8y} and thus \frac{125y}{8}=1 and y = \frac{8}{125}.
This value also fits within our bounds so it is also a solution.

\frac{1}{8} \leq y < \infty
Similarly, {{25y}} = 25y and {{8y}} = 8y.

Solving, we get y = \frac{1}{10} but this value thus not fit the required bounds and so we only have two solutions, y=\frac{1}{40} and y=\frac{8}{125}.

Team Blue!!!

Brilliant!!!