# Write the set of solutions of x -= 5 (mod 24) x -= 17(mod 18)

Write the set of solutions of
$x\equiv 5\left(\text{mod}24\right)$
$x\equiv 17\left(\text{mod}18\right)$
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Step 1
We have a set of congruence as
$x\equiv 5\left(\text{mod}24\right)$
$x\equiv 17\left(\text{mod}18\right)$
From the first equation we can write the set of solutions as $x={\left[5\right]}_{24}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}x=5t+24,t\in \mathbb{Z}$.
Now, we substitute for x in the second congruence $x\equiv 17\left(\text{mod}18\right)$, we get
$5t+24\equiv 17\left(\text{mod}18\right)$
$⇒5t\equiv 11\left(\text{mod}18\right)$
Step 2
Multiplying both sides by ${5}^{-1}\text{mod}\left(18\right)=11$, we get,
$t\equiv 13\left(\text{mod}18\right)$
So, $t={\left[13\right]}_{18}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}t=13+18s,s\in \mathbb{Z}$.
Substituting t back into x, we get $x=5t+24\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}x=5\left(13+18s\right)+24=65+90s+24=90s+89,s\in \mathbb{Z}$.
Which gives us the solution $x={\left[84\right]}_{90}⇒x\equiv 89\left(\text{mod}90\right)$