Answer: \(x=\log_{10}2\), \(y=\dfrac12\log_{10}\dfrac52\); \(x=-8\) or \(x=\dfrac{125}{8}\).

(a) Taking \(\log_{10}\) of both equations gives

\(x+2y=\log_{10}5\)

and

\(3x+4y=\log_{10}50.\)

Subtract twice the first equation from the second:

\(x=\log_{10}50-2\log_{10}5.\)

So

\(x=\log_{10}\left(\frac{50}{25}\right)=\log_{10}2.\)

Now substitute into \(x+2y=\log_{10}5\):

\(\log_{10}2+2y=\log_{10}5.\)

Therefore

\(2y=\log_{10}5-\log_{10}2 =\log_{10}\frac52.\)

Hence

\(y=\frac12\log_{10}\frac52.\)

(b) Let

\(u=x^{\frac13}.\)

Then \(x^{\frac23}=u^2\), so the equation becomes

\(2u^2-u-10=0.\)

Factorising,

\((2u-5)(u+2)=0.\)

So

\(u=\frac52\quad\text{or}\quad u=-2.\)

Since \(u=x^{\frac13}\),

\(x=\left(\frac52\right)^3 \quad\text{or}\quad x=(-2)^3.\)

Therefore

\(x=\frac{125}{8}\quad\text{or}\quad x=-8.\)