First, notice that \(\left(14,19\right)=1\). We can find a solution to \(14x+19y=1\).
By playing around with different numbers, we get our solutions to be \(x=-4\) and \(y=3\).
We now have all our values: \(m_1=14,m_2=19,a_1=3,a_2=7,x=-3,y=7\) and can now solve \(N=m_1a_2x+m_2a_1y\).
So \(N=105\), however, our unique solution is up to our desired modulus which is \(14\cdot19=266\).
Thus, \(N=105\) (mod \(266\)).