Proving $93x + 47 \equiv 61 \pmod {101}$ I am preparing for an exam. I am dealing with this right now:
$$93x + 47 \equiv 61\pmod{101}$$
However, I can't figure it out. Can someone describe steps for this example, or provide a link to any free pdf, website describing this problem and the way it is solved on attached photo, please? I repeat it needs to be solved like on the photo.
Thanks in advance
Notes 
 A: The equation is equivalent to
$$
93x\equiv 61-47\pmod{101}
$$
so to
$$
93x\equiv14\pmod{101}
$$
You just need to find the inverse of $93$ modulo $101$:
\begin{align}
101&=93\cdot 1+8\\
93&=8\cdot 11+5\\
8&=5\cdot1+3\\
5&=3\cdot1+2\\
3&=2\cdot1+1
\end{align}
so
\begin{align}
1&=3+2\cdot(-1)\\
&=3+(5-3\cdot1)\cdot(-1)\\
&=3\cdot2+5\cdot(-1)\\
&=(8-5)\cdot2+5\cdot(-1)\\
&=8\cdot2+5\cdot(-3)\\
&=8\cdot2+(93-8\cdot11)\cdot(-3)\\
&=93\cdot(-3)+8\cdot35\\
&=93\cdot(-3)+(101-93)\cdot35\\
&=93\cdot(-38)+101\cdot35
\end{align}
so the inverse is $-38\equiv63\pmod{101}$.
Thus
$$
x\equiv14\cdot63\equiv74\pmod{101}
$$
Yes, your computation seems right.
A: Rewrite our congruence as $93x\equiv 14\pmod{101}$.
Now we can use a general procedure, by finding the inverse $b$ of $93$ modulo $101$, and multiplying through by $b$.
Or else note that $93x\equiv -8x\pmod{101}$, and $14\equiv 14+202=216\pmod{101}$. That gives $x\equiv -27\pmod{101}$, or equivalently $x\equiv 74\pmod{101}$. 
A: By Gauss's algorithm $\ {\rm mod}\ 101\!:\ x\equiv \dfrac{67\!-\!41}{93}\equiv \dfrac{14}{-8}\equiv \dfrac{7}{-4}\equiv \dfrac{108}{-4}\equiv -27\equiv 74$
