Answer
See below
Work Step by Step
Given any invertible $n \times n$ matrix has a factorization of the form $A = QR$,
where $Q$ and $R$ are invertible, $R$ is upper triangular,
and $Q$ satisfies $Q^T Q = I_n$.
Then the system $Ax=b$ is equivalent to the two problems
$$Qy=b\\Rx=y$$
Transfer the first equation to $Q^TQy=y=Q^Tb$
Then, we have $Rx=Q^Tb$. Since $R$ is upper triangular, as a result, we can find x by back substitution
