English: Now we can solve the two equations and I get $2A = \left( 1 + \frac{k_{2}}{k_{1}} \right)C$ or $C = \frac{2k_{1}}{k_{1} + k_{2}}A$ and I get B = C − A which is $\left( \frac{2k_{1}}{k_{1} + k_{2}} - 1 \right)A$ which is $2A = \left( 1 + \frac{k_{2}}{k_{1}} \right)C$ ಅಥವಾ So, I get C which is $\frac{2k_{1}}{k_{1} + k_{2}}A$ implies $\frac{C}{A} = \frac{2k_{1}}{k_{1} + k_{2}}$ and B which is $\left( \frac{k_{1} - k_{2}}{k_{1} + k_{2}}\ \right)A$ which implies that Now, reflection coefficient $R = \frac{current\ going\ to\ the\ left}{current\ going\ to\ the\ right\ }$ in region x < 0, this is going to be the speed which will be given by k times B² divided by speed A².
Kannada: ಈಗ ನಾವು ಎರಡು ಸಮೀಕರಣಗಳನ್ನು ಪರಿಹರಿಸಬಹುದು ಮತ್ತು ನಾನು $2A = \left (1 + \frac{k_{2}}{k_{1}} \right)C$ ಅಥವಾ $C = \frac{2k_{1}}{k_{1} + k_{2}}A$ ಮತ್ತು ನಾನು B = C − A ಅನ್ನು ಪಡೆಯುತ್ತೇನೆ, ಇದು $\left (\frac{2k_{1}}{k_{1} + k_{2} k_{1} k_{2}} - 1 \right)A$  ನಾನು C ಅನ್ನು ಪಡೆಯುತ್ತೇನೆ, ಅದು $\frac{2k_{1}}{k_{1} + k_{2}}A$ ಎಂದರ $R ೆ $\frac{C}{A} = \frac{2k_{1}}{k_{1} + k_{2}}$ ಮತ್ತು B ಅನ್ನು $\left (\frac{k_{1} - k_{2}}{k_{1} + k_{2}}\ \right)A$ ಎಂದು <.