Let $R$ be a ring with $1_R$, $u = a + x $ is a unit and $x \in J(R)$, prove $a$ is a unit. 
Definition: For a ring with unit $R$, Jacobson Radical is defined as the ideal $J(R) = \{r \in R: rM = 0 \}$ where $M$ is simple.

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*Let $R$ be a ring with $1_R$, $u = a + x $ is a unit and $x \in J(R)$, prove $a$ is a unit.



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*So I wrote that if $M$ is simple, let $m \in M$. Then $um = (a + x)m = am + xm = am$. So therefore $um = am \iff (u-a)m = 0 \iff u - a \in J(R).$ But this is where I am stuck. To show $a$ is a unit, I have to show $a \not\in J(R)$ at least. Now I think here I have to assume $J(R) \neq R$, then this gives me $u \not\in J(R)$ since $J(R)$ is an ideal of $R$, so therefore $a \not\in J(R)$. But I am still stuck. I tried also using $u$ being a unit to generate $1$, but I can't get any leads.

EDIT: I just realized $J(R)$ can be defined as the intersection of right maximal ideals in $R$ and I can turn the problem into showing $x\not\in J(R) $, but I was only able to show $1-sx$ is non-unit. where $s\in R.$
 A: 
EDIT: I just realized $J(R)$ can be defined as the intersection of right maximal ideals in $R$ and I can turn the problem into showing $x\not\in J(R) $, but I was only able to show $1-sx$ is non-unit. where $s\in R.$

If you believe the bolded characterization above, AND that its left-hand analogue is the same set, then there is a fast solution (not along the lines you were pursuing.)
Suppose $u=a+x$ as in your hypotheses, and that $a$ is not a unit. Then $a$ is contained in some maximal left ideal or some maximal right ideal, call it $A$.  In that case, $x\in A$ also, by the characterizations mentioned above.
But since $A$ is additively closed, that would mean $u\in A$ as well, but when an ideal contains a unit, it is the entire ring.  This would contradict the fact that $A$ is proper.
Therefore $a$ must be a unit.
A: If $a$ is not a unit then $aR$ is a proper right ideal of $R$, so there is a maximal right ideal $m$ with $aR \subset m$. But then $a \in m$, and $x \in J(R) \subset m$ since $J(R)$ is the intersection of all the maximal right ideals, so $u \in m$, but $u$ is a unit, which contradicts $m$ being a maximal ideal.
