How to solve this logarithm system?

I am new to logarithms and I am having trouble with this logarithm system. \begin{align*} \log_9(x) + \log_y(8) & = 2, \\ \log_x(9) + \log_8(y) & = 8/3. \end{align*} A step-by-step procedure would be highly appreciated.

Use the fact that $$\log_b(a) = \dfrac1{\log_a(b)}$$
Hence, if we denote $\log_9(x) = a$ and $\log_y(8) = b$, we get that \begin{align} a+b & = 2\\ \dfrac1a + \dfrac1b & = \dfrac83 \implies \dfrac{a+b}{ab} = \dfrac83 \implies ab = \dfrac34 \end{align} Now solve for $a$ and $b$ and hence $x$ and $y$.
If $\log(\cdot)$ denotes logarithm in base $10,$ then $$\log_y(x) = \frac{\log(x)}{\log(y)}.$$ Apply through & simplify to see what system you will get. Substitute $X$ for $\log(x),$ and $Y$ for $\log(y).$ You should have 2 equations in 2 unknowns.