Find all $x\in \mathbb{Z}$ with $x \equiv 2 \mod 4$ , $x \equiv 8\mod9$ and $x \equiv 1\mod5$ 
Find all $x\in \mathbb{Z}$ with $x \equiv 2 \mod 4$ , $x \equiv 8\mod9$ and $x \equiv 1\mod5$

My attempt:
$\gcd\left(9,5,4\right) = 1 = 9r+5s+4\times0$
$9 = 1\times5+4$
$5 = 1\times4+1$
so $1 = 5-(9-5) = 2\times5 - 1\times9$
thus $r=-1$ and $s = 2$
so $x = 2\times5\times8 -1\times9\times1=80 - 9=71\mod 180$
But the problem is that $71\equiv8\mod9$ and $71\equiv1\mod5$, but $71 \ne 2 \mod 4$. What is wrong?

From $x\equiv2\mod4$, let $x=4a+2$ that $a$ is an integer. Then sub it into the second equation,$$4a+2\equiv8\mod9\\4a\equiv6\mod9\\2a\equiv3\mod9\\2a\equiv12\mod9\\a\equiv6\mod9$$ After that, we do the same thing: let $a=9b+6$ that $b$ is an integer, $x=4\left(9b+6\right)+2=36b+26$. Then sub it into the third equation, $$36b+26\equiv1\mod5\\36b\equiv0\mod5\\b\equiv0\mod5$$ Then, let $b=5k$ that $k$ is an integer. $x=36\left(5k\right)+26=180k+26$, and that is the answer.
Conclusion: The answer is $180k+26$ for $k\in\mathbb{Z}$.
