Help me to understand question on Linear Congruence in simplest and elaborated way. I came across the following congruence in which I have to get value of $x$. They devide it by $3$ which I understand how and multiply it by $7$ on both sides and proceeds further as shown by photo please help me to understand very first line and further too, please, in most elementry way you can. I don't understand meaning of Residues in theory of Congruence please make it clear too by giving some example. 
I would be very grateful to you, Thanks.


 A: I do not know how far you are in learning modular arithmetic, so it may be possible that I'm talking about things that you haven't covered yet in the course. In this case: please ignore this solution.
Think of modular computation as if you're walking from left to right on a fininte number line. When you reach the end of the number line, you start again at point one.
In Mathematics, you should think of the numbers as modular residues. That is simply a set of numbers: $1 \equiv 4 \equiv 7 \equiv \dots \equiv 1 \mod{3}$. Basically: $1 + 3\lambda \equiv 1 \mod 3$ for all $\lambda \in \mathbb{Z}$.
Written down more general: $a + b \lambda \equiv a \mod{b}$ for all $\lambda\in\mathbb{Z}$.
The problem
The solution that's proposed in your article is very brief and does not really explain why steps are made as stated. I will solve it the way that I would and document why I make the steps as I state. I think this may help you understanding the parts you're having problems with. And if it doesn't help: please ignore that.
By convention I'll write $\equiv$ if I perform modular operations and $=$ if I perform standard arithmetic operations.
The problem $9 x \equiv 21 \mod{30}$ can be re-written as the equation $\exists \lambda\in\mathbb{Z}: 9x = 21 + 30\lambda$. If you want solutions to this problem, you can divide by three. 
If you divide by three and continue computating with $\frac{\lambda}{3}\in\mathbb{Z}$, then you're dropping solutions from the set of solutions. So you need to distinguish the cases where $\lambda \in 3\mathbb{Z}$, $\lambda\in3\mathbb{Z}+1$ and $\lambda\in3\mathbb{Z}+2$. This covers all cases, because $3\mathbb{Z} + 3 = 3\mathbb{Z}$.
For some $\lambda\in\mathbb{Z}$:
$$3x = 7 + 30\frac{\lambda}{3} = 7 + 10\lambda$$
So we've reduced the problem to the form $3x = 7 \mod{10}$. Because $\gcd(7,10) = 1$, the number 7 has an inverse modulo 10. The numbers are small, so 'guessing' the inverse is faster than carrying out the Extended Euclidean Algorithm. The inverse is 3, since $3\cdot 7 = 21 \equiv 1 \mod 10$.
Multiply both sides of $3x = 7\mod{10}$ by 3, and you find 
$$9x \equiv 21\equiv1\mod{10}$$
Now compute $9^{-1} \mod{10}$. Note that $9 \equiv -1 \mod{10}$ and the inverse of -1 is -1. Therefore 9 is its own inverse.
We find $9 \cdot 9 = 81 \equiv 1 \mod 10$.
So the solution to $9x \equiv 21\equiv1\mod{10}$ is $x = 9$.
Because we've dropped solutions from the set of solution before, we need to look for other solutions. The original problem is $9x \equiv 21\mod{30}$, and one solution is $x = 9$. This works for some $\lambda \in\mathbb{Z}$:
$$9x = 21 + 30\lambda$$
If $x^*$ is another solution, then there is a $\mu \in\mathbb{Z}$ such that:
$$9x^* = 21 + 30\lambda + 30 \mu$$
Subtracting these two equations yields $9(x^*-x) = 30\mu \equiv 0\mod{30}$, so apparantly it suffices to find zeroes $\mod{30}$.
Again 'guessing' is faster than executing the Euclidean Algorithm. You quicly see that if $y=10$ that then $9y = 90 \equiv 0\mod{30}$, and the same if $y = 20$ and $y = 0$. So apparantly $x - x^*$ can take the values 0, 10 and 20.
You find that the solutions are $x = 9$, $x = 9 + 10 = 19$ and $x = 29$.
A: For the first step, you can look at it as $9x=21+k30$ => $3x=7+k10$.
Now it comes that if $gcd(n,10)=1$ then you can multiply the two sides by $n$, keeping the congruence with ten. (k will vary, but remains an integer and we do not care in congruences). That is a basic theorem of modular arithmetics. Of all the possible numbers prime to $10$, $\ 7$ is chosen, as to be able to get for $x$ a coefficient $=1\ mod\ 10$.
 So $(20+1)x=(40+9)+j10$ $\ => \ x=9+m10$
