I am learning how to solve system of congruences and I am having some trouble with one exercise.
$5x=3mod7$
$2x=4mod8$
$3x=6mod9$
For
$x=2mod7$
$x=2mod8$ and $x=6mod8$
$x=2mod9$, $x=5mod9$, and $x=8mod9$
$M: M=7*8*9=504.$
Now to figure out M1, M2, and M3:
M1=72, M2=63, and M3=56
Here is were I am having some trouble.
72X1=1mod7----------->X1=4
63X2=1mod8----------->X2=7
56X3=1mod9----------->X3=5
Therefore $X0= 5(4)72+2(7)(63)+3(5)(56)$
$X0=3162$
So $X=138 mod504$
Is this correct? I feel like I am missing some steps.
First we can cancel factors from the final two congruences as follows.
$2x\equiv 4\pmod{8}\iff 2x = 4+8j\iff x = 2+4j\iff x=2\pmod{4}$
Similarly $\,3x \equiv 6\pmod{9}\iff x\equiv 2\pmod{3}$
Note $\,x\equiv 2\,$ is also the (unique!) solution of $\,5x\equiv 3\pmod{7}$
So the congruences are equivalent to $\,x\equiv 2\pmod{\!3,4,7}\!\iff x\equiv 2\pmod{\!84}\,$ by CCRT.
If you must use the CRT formula then you can make it easy as follows. Let $X = x-2.\,$ Solve the system $X\equiv \color{#c00}0$ for all three moduli. The CRT formula will have a factor of $\,\color{#c00}0\,$ in all $3$ summands, so the result is $0$. Thus $\,x-2 = X \equiv 0\,$ so $\,x \equiv 2\,$ is the solution by the CRT formula. Note that we don't need to compute inverses of moduli etc since all those terms are annihilated by the $0$ factors.