How do I solve these questions using Diophantine equations? I have been told that it is easier to solve the below 2 questions using Diophantine equations instead of simply trial and error. 

1) Find the smallest positive integer which, when divided by 6, gives a       remainder of 1 and when divided by 11, gives a remainder
  of 6.
2) What is the least positive integer which has remainders of 1, 1 and 5 when divided by 3, 5 and 7 respectively?

In terms of my progress, I have attempted to generalize these questions by for example the first question:
Since the integer has a remainder of 1 when divided by 6, it must be of the form 6m + 1 
By the way, I have no prior knowledge of Diophantine equations and would appreciate both a brief introduction as well as an explanation of how to apply this knowledge to the above questions.
Thank you :)
 A: One should maybe say that using this approach replaces the guesswork by some modeling and applying a solution algorithm, so it is more systematic.
Problem 1):
From the equations for the divisions
$$
u = 6x+1 =11y+6 
$$
we get a Diophantine linear equation
$$
6x-11y=5
$$
which is the name for a linear equation with integer coefficients
$$
a x + b y = c
$$ 
where one looks for integer solutions $x$, $y$. It has either no solution or infinite many solutions, depending on if 
$$
\gcd(a,b) \,\rvert\, c
$$
where $\gcd(a,b)$ is the greatest common denominator of $a$ and $b$.
Here $\gcd(6,-11)=1$ and $1$ divides $5$, so we have solutions.
A particular solution can be obtained by applying the 
extended Euclidean algorithm. 
$$
\begin{array}{|c|c|c|c|c|}
\hline
i & q_{i-1} & r_i & s_i & t_i \\
\hline
0 &           & 6                       & 1 & 0 \\
\hline
1 &           & -11                     & 0 & 1 \\
\hline
2 & 6:-11 = 0 & 6- 0\cdot (-11) = 6    & 1-0\cdot 0=1 & 0-0\cdot 1=0 \\
\hline
3 & -11:6=-1  & -11 - (-1)\cdot 6 = -5 & 0 -(-1)\cdot 1=1 & 1-(-1)\cdot 0=1\\
\hline
4 & 6:-5=-1   & 6 - (-1) \cdot (-5) = 1 & 1-(-1)\cdot 1=2 & 0-(-1)\cdot 1=1\\
\hline
5 & -5:1=-5   & -5 - 1 \cdot (-5) = 0   & 1-(-5)\cdot 2=11 & 1-(-5)\cdot 1=6\\
\hline
\end{array}
$$
The row for $i=4$ gives $\gcd(6,-11) = 1$, $s = 2$, $t = 1$.
Checking: $6\cdot 2 + (-11) \cdot 1 = 1$, OK.
This solves
$$
  a s + b t = \gcd(a,b) = 1 \Rightarrow a (5s) + b (5t) = 5
$$
Thus $(x,y)=(5\cdot 2, 5\cdot 1) = (10,5)$ is a particular solution.
The general solution is all solutions of the homogenous equation
$$
a x + b y = 0 = a x - t ab + t ab + b y = a(x-tb) + b (y + at) 
$$
for $t \in \mathbb{Z}$ plus one particular solution.
Here this gives $(10+11t,5+6t)$ for $t \in \mathbb{Z}$ and
$u = 6(10+11t)+1=61-66t$, the smallest positive one is $61$.
Problem 2): The second example gives
$$
u = 3x+1 = 5 y + 1 = 7 z + 5
$$
So we could try 
$$
3x-5y=0 \wedge 3x+1=7z+5
$$
which gives
$$
y=(3/5)x \wedge 3x-7z=4
$$
This leads to $(x,z)=(6+7t,2+3t)$ and $y = (3/5)(6+7t) = (18+21t)/5$ which is integer for $t=2$, giving $(x,y,z)=(20,12,8)$ and $u=61$ again.
WA seems to think so as well (link).
