The previous equation can be calculated in two steps \mathbf{L} \mathbf{y} = \mathbf{w};\qquad \mathbf{Ux}=\mathbf{y}.
To show that this is correct we use to the LU decomposition to rewrite our system of linear equations as \mathbf{LUx}=\mathbf{w}, and since the determinat of \mathbf{L} is equal to 1 (by construction since the diagonals of \mathbf{L} equal 1) we can use the inverse of \mathbf{L} to obtain \mathbf{Ux}=\mathbf{L^{-1}w}=\mathbf{y}, which yields the intermediate step \mathbf{L^{-1}w}=\mathbf{y} and as soon as we have \mathbf{y} we can obtain \mathbf{x} through \mathbf{Ux}=\mathbf{y} .