Approaching modular arithmetic problems I'm a little stumbled on two questions.
How do I approach a problem like $x*41 \equiv 1 \pmod{99}$. 
And given $2$ modulo, $7x+9y \equiv 0 \pmod{31}$ and $2x−5y \equiv 2 \pmod{31}$ (solve for $x$ only)?
When I solve for $x$ for the latter, I got a fraction as the answer and I'm not sure if I can have a fraction as an answer? I'm not sure how to approach the first problem either.
 A: Finding the solution to $$x \times 41 \equiv 1 \pmod {99}$$ is equivalent to asking for the multiplicative inverse of $41$ modulo $99$.  Since $\gcd(99,41)=1$, we know $41$ actually has an inverse, and it can be found using the Extended Euclidean Algorithm:
\begin{align*}
99-2 \times 41 &= 17 \\
41-2 \times 17 &= 7 \\
17-2 \times 7 &= 3 \\
7-2\times 3 &= 1 &=\gcd(99,41). \\
\end{align*}
Going back, we see that
\begin{align*}
1 &= 7-2\times 3 \\
  &= 7-2\times (17-2 \times 7) \\
  &= 5 \times 7-2\times 17 \\
  &= 5 \times (41-2 \times 17)-2\times 17 \\
  &= -12 \times 17+5 \times 41 \\
  &= -12 \times (99-2 \times 41)+5 \times 41 \\
  &= 29 \times 41-12 \times 99 \\
\end{align*}
Hence $29 \times 41 \equiv 1 \pmod {99}$ and thus $x=29$.

In the second case, we have $$7x+9y \equiv 0 \pmod {31}$$ and $$2x-5y\equiv 2 \pmod {31}.$$  Here we want to take $7x+9y=0 \pmod {31}$ and rearrange it to get $x \equiv ?? \pmod {31}$, then substitute it into the other equation and solve for $y$.  This requires finding the multiplicative inverse of $7$ modulo $31$ (which we can do as above).  It turns out $7 \times 9 \equiv 1 \pmod {31}$.  Hence
\begin{align*}
& 7x+9y=0 \pmod {31} \\
\iff & 7x \equiv -9y \pmod {31} \\
\iff & x \equiv -9y \times 9 \pmod {31} \\
\iff & x \equiv 12y \pmod {31}.
\end{align*}
We then substitute this into the equation $2x-5y\equiv 2 \pmod {31}$, which implies $$2 \times 12y-5y \equiv 2 \pmod {31}$$ or equivalently $$19y \equiv 2 \pmod {31}.$$  Yet again, we find a multiplicative inverse, this time of $19$ modulo $31$, which turns out to be $18$.  So
\begin{align*}
& 19y \equiv 2 \pmod {31} \\
\iff & y \equiv 2 \times 18 \pmod {31} \\
\iff & y \equiv 5 \pmod {31}.
\end{align*}
Hence $$x \equiv 12y \equiv 29 \pmod {31}.$$  Thus we have the solution $(x,y)=(29,5)$.
A: For the first one, you have to find a multiplicative inverse for $41$ mod $99$. Use Euclid's algorithm to find the solutions of $41\cdot x + 99 \cdot y = 1$ to find $x$.
Are you familiar with abstract algebra? If yes, do you know that $\mathbb{Z}_{31}$ is a field because $31$ is a prime number and you can use tools of linear algebra because it works overy any field?
You have the following system of equations in $\mathbb{Z}_{31}$:
$$7x+9y=0$$
$$2x-5y=2$$
You can use Gaussian elimination in $\mathbb{Z}_{31}$ to find $x$ and $y$. But you have to be careful that your coefficients are in $\mathbb{Z}_{31}$ not in $\mathbb{R}$.
EDIT(suggested by Git Gud):
Notice that if you get a fraction like $\displaystyle \frac{a}{b}$ then you should think of it as $a \cdot b^{-1}$ and then find $b^{-1}$ in $\mathbb{Z}_{31}$. As previously said, $\mathbb{Z}_{31}$ is a field, so talking about fractions in it makes sense. The notation $\frac{a}{b}$ is actually nothing but $a.b^{-1}$ where $.$ denotes the multiplication in the field and $b^{-1}$ denotes the multiplicative inverse of $b$ in the field.
A: For the first one, you could approach it as follows:
$41x=1$ mod $99$
$140x=1$ mod $99$ (because 41 mod 99 = (41+99) mod 99)
$140x=100$ mod $99$ (because 1 mod 99 = (1+99) mod 99)
$7x=5$ mod $99$ (divide both sides by 20)
$7x=203$ mod $99$ (because 5 mod 99 = (5+99+99) mod 99)
$x=29$ mod $99$ (divide both sides by 7)
