Answer: (a) \(\displaystyle \left(\frac32,\frac13\right)\) and \(\displaystyle \left(-\frac12,-1\right)\); (b) \(\displaystyle x=3^{-5}\) or \(x=3^2\).

Answer: (a) \(\displaystyle \left(\frac32,\frac13\right)\) and \(\displaystyle \left(-\frac12,-1\right)\); (b) \(\displaystyle x=3^{-5}\) or \(x=3^2\).

(a) From

\(3y-2x+2=0,\)

we get

\(2x=3y+2,\)

so

\(x=\frac{3y+2}{2}.\)

Substitute this into \(xy=\frac12\):

\(y\left(\frac{3y+2}{2}\right)=\frac12.\)

Multiplying by \(2\),

\(3y^2+2y=1.\)

So

\(3y^2+2y-1=0.\)

Factorising,

\((3y-1)(y+1)=0.\)

Hence

\(y=\frac13\quad\text{or}\quad y=-1.\)

If \(y=\frac13\), then

\(x=\frac32.\)

If \(y=-1\), then

\(x=-\frac12.\)

So the solutions are

(b) Let

\(u=\log_3x.\)

Then

\(\log_x3=\frac1{\log_3x}=\frac1u.\)

The equation becomes

\(u+3=\frac{10}{u}.\)

Multiplying by \(u\),

\(u^2+3u=10.\)

So

\(u^2+3u-10=0.\)

Factorising,

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

Thus

\(u=-5\quad\text{or}\quad u=2.\)

Therefore

\(\log_3x=-5\quad\text{or}\quad \log_3x=2.\)

Hence

\(x=3^{-5}\quad\text{or}\quad x=3^2.\)