Find a positive integer $x$ less than $105$ satisfying the following simultaneous congruence equations. 
$$x=2 mod 3$$
  $$x=3 mod 5$$
  $$x=4 mod 7$$

I have only learnt modulo for 2 weeks so far... really basic theorems. 

My attempt using definitions of modulo

From Equation 1, $3a=x-2 \rightarrow 15a=5x-10$
From Equation 2, $5b=x-3 \rightarrow 15b=3x-9$
Adding them together, $15(a+b)=8x-19$ which implies, $8x=19mod15$ or $8x=4mod15$
and hence i am stuck....
please help, i can't seem to understand this at all.

$3a+2=5b+3\iff 3a=5b+1$
$\iff3(a-2)=5(b-1)$ (Subtracting either sides by $6$)
$\displaystyle\implies \frac{5(b-1)}3=a-2\text{ which is an integer}$
$\implies 3$ divides $5(b-1)\implies 3$ divides $(b-1)$ as $(5,3)=1$
$\implies b-1=3d\iff b=3d+1$ where $d$ is some integer
Again we have  $7c+4=5b+3=5(3d+1)+3\iff 7c+4=15d+8$
$\iff 7c=15d+4\iff 7(c-7)=15(d-3)$(Subtracting either sides by $49$)
$\displaystyle\implies \frac{7(c-7)}{15}=d-3$ which is an integer
$\implies 15$ divides $7(c-7)\implies 15$ divides $c-7$
$\implies c-7=15e$ where $e$ is any integer
$\implies x=7c+4=7(15e+7)+4=105e+53$
