Solve the system of congruence's x-=1(mod 3), x-=4(mod 5), x-=6(mod 7)

Step 1
Systems of linear congruences may be solved using methods from linear algebra: Matrix inversion, Cramer's rule. In case the modulus is prime, everything we know from linear algebra goes over to systems of linear congruences
Step 2
Consider the system of congruence
${\displaystyle x\equiv 1\left(\text{mod}3\right)}$
${\displaystyle x\equiv 4\left(\text{mod}5\right)}$
${\displaystyle x\equiv 6\left(\text{mod}7\right)}$
Try a solution of the form
$x=3\cdot 5\cdot a+3\cdot 7\cdot b+5\cdot 7\cdot c$
Taking the remainders mod 3,5, and 7. It gives the three equations
${\displaystyle 1\equiv 35c\equiv 2c\equiv -c\left(\text{mod}3\right)}$
${\displaystyle 4\equiv 21b\equiv b\left(\text{mod}5\right)}$
${\displaystyle 6\equiv 15a\equiv a\left(\text{mod}7\right)}$
${\displaystyle \Rightarrow a=6,b=4,c=-1}$
One solution is therefore
$x=3\cdot 5\cdot 6+3\cdot 7\cdot 4+5\cdot 7\cdot (-1)=139(\mathrm{mod}105)=34$