The original question is about the perturbation in matrix equation. Assume we have $Ax=b$ where $A$ is invertible and $(A+E)(X+\delta)=b$ where $|\cdot|$ is one of those matrix norm derived from vector norm, compatible under mutiplication and $|E|\leq 10^{-8}|A|$. Now the question asks me to prove $|\delta|\leq 10^{-7}|x|$
It can be derived that $x=(A+E)^{-1}E\delta$. And here comes the question in title. I have no idea where to begin.
The hypothesis $|E|\le10^{-8}|A|$ is not sufficient to ensure $|\delta|\le10^{-7}|x|.$ Here is a counterexample: $A=\begin{pmatrix}1&0\\0&10^{-2}\end{pmatrix},$ $E=\begin{pmatrix}0&0\\0&10^{-8}\end{pmatrix},$ $x=\begin{pmatrix}0\\1\end{pmatrix},$ $b=\begin{pmatrix}0\\10^{-2}\end{pmatrix},$ $\delta=\begin{pmatrix}0\\-\frac1{1+10^6}\end{pmatrix}.$
A stronger and sufficient hypothesis is $|A^{-1}E|\le10^{-8},$ which gives $$\begin{align}|(A+E)^{-1}E|&=|(I+A^{-1}E)^{-1}A^{-1}E|\\&\le\sum_{n=1}^\infty|A^{-1}E|^n\\&\le\sum_{n=1}^\infty10^{-8n}\\&=\frac{10^{-8}}{1-10^{-8}}\\&<10^{-7},\end{align}$$ hence $$|\delta|=|(A+E)^{-1}Ex|\le10^{-7}|x|.$$