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

Solve the system of congruences
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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
$x\equiv 1\left(\text{mod}3\right)$
$x\equiv 4\left(\text{mod}5\right)$
$x\equiv 6\left(\text{mod}7\right)$
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
$1\equiv 35c\equiv 2c\equiv -c\left(\text{mod}3\right)$
$4\equiv 21b\equiv b\left(\text{mod}5\right)$
$6\equiv 15a\equiv a\left(\text{mod}7\right)$
$⇒a=6,b=4,c=-1$
One solution is therefore
$x=3\cdot 5\cdot 6+3\cdot 7\cdot 4+5\cdot 7\cdot \left(-1\right)=139\left(\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}105\right)=34$