The equivalent resistance of two resistors connected in series is \( 6 \Omega \) and their parallel
equivalent resistance is \( \frac{4}{3} \Omega \) What are the values of resistances?

Let the two resistances be \(R_{1}\) and \(R_{2}\)
When they are connected in series, equivalent resistance, \(R_{s}=6 \Omega\)
\(R_{1}+R_{2}=6 \rightarrow(1)\)
When they are connected in parallel, equivalent resistance,
\(R_{p}=\frac{4}{3} \Omega\)
\(\frac{1}{R_{p}}=\frac{1}{R_{1}}+\frac{1}{R_{2}} \Rightarrow \frac{1}{R_{1}}+\frac{1}{R_{2}}=\frac{3}{4} \rightarrow(2)\)
Solving Eq. (2), we get
\(\frac{R_{1}+R_{2}}{R_{1} R_{2}}=\frac{3}{4} \Rightarrow R_{1} R_{2}=\left(R_{1}+R_{2}\right) \times \frac{4}{3}\)
\(\Rightarrow R_{1} R_{2}=6 \times \frac{4}{3}\)
\(\Rightarrow R, R_{2}=2 \times 4=8 \Omega(\) Using Eq. \((1))\)
\(\Rightarrow R_{1} R_{2}=2 \times 4=8 \Omega \rightarrow(3)\)
From Eq. (1), we have \(R_{2}=6-R_{1}\)
\(\Rightarrow R_{1}\left(6-R_{1}\right)=8 \Rightarrow 6 R_{1}-R_{1}^{2}=8 \Rightarrow R_{1}^{2}-6 R_{1}+8=0\)
\(R_{1}=\frac{6 \pm \sqrt{36-32}}{2}=\frac{6 \pm \sqrt{4}}{2}=\frac{6 \pm 2}{2}=4\) or 2
Using Eq. (3), we get \(R_{2}=2\) or 4
Therefore, the two resistances are \(4 \Omega\) and \(2 \Omega\).