By Kirchoff's voltage law, the sum of the voltage drops across the left and right loop must be zero. This leads to the equations
We solve this system using row reduction. This system is represented by the augmented matrix
![{\displaystyle \left[{\begin{array}{cc|c}3&-1&9\\-1&4&-14\end{array}}\right]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/2c92d898ca06114c3731e2f32ec9a4ca5b8653be)
.
We add 
of the first row to the second row:
![{\displaystyle \left[{\begin{array}{cc|c}3&-1&9\\0&11/3&-11\end{array}}\right]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/fd5abd2a97c96bc6e969a8afb8a9db0ff52f82f0)
.
Now, we multiply the second row by 
:
![{\displaystyle \left[{\begin{array}{cc|c}3&-1&9\\0&1&-3\end{array}}\right]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/5517a3caf6a994fdf326f98a5eb73e368ecef71d)
.
We add the second row to the first row:
![{\displaystyle \left[{\begin{array}{cc|c}3&0&6\\0&1&-3\end{array}}\right]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/76d0d76bd5dcbec31bada8d90e7b33c8d93e4ea6)
.
Finally, we divide the first row by 3:
![{\displaystyle \left[{\begin{array}{cc|c}1&0&2\\0&1&-3\end{array}}\right]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/2339b4227485cfd269ddeb2b47e00126fa5dcfba)
.
Therrefore, we get that 
and that 
.
We now verify that this answer makes sense. The direction of 
is opposite the direction of the current, because the current is running from high voltage to low voltage. Therefore, we expect that should be negative and 
should be positive. This is confirmed by our calculations.
Answer: 
