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 \( \mathbf{Ux}=\mathbf{y} \).