Solution of a congruence

Question

Where is my mistake please tell ,I also tried $x \equiv 1 \pmod 7$ instead of $25x \equiv 4 \pmod 7$ and $7x \equiv 4 \pmod 9$ instead of $25x \equiv 4 \pmod 9$

Answer 1

The answer you give (i.e. $x\equiv 4\pmod{63}$) is incorrect; try putting in $x=4$ to the original congruence equation.

I don't really see why you need CRT. All you need to do is find the inverse of $25$ modulo $63$. This can be done by the Euclidean algorithm to find $a,b\in\Bbb Z$ such that $25a+63b=1$. Then $a$ is the inverse of $25$ modulo $63$, and we get $$25x\equiv 4\pmod{63}\iff x\equiv 4a\pmod {63}$$

Spoiler:

You should find that $a=-5$ and $b=2$ (I did this simply by inspection). So $$x\equiv 4\cdot (-5)\equiv -20\equiv 43\pmod{63}$$

Answer 2

But $x=1$ works, $$25\equiv 4 \mod 7$$

Answer 3

You solved for $x \equiv 4 \mod 63$.

You need to solve for $25 x \equiv 4 \mod 63$

$25 x\equiv 4x \equiv 4 \mod 7$ so $x \equiv 1 \mod 7$

And $25 x \equiv 7x \mod 4 \mod 9$ so a bit of finaggling... $7y \equiv 1 \mod 9 \implies y \equiv 4 \mod 9$ so $4*7*x \equiv x \equiv 4*4 \equiv 7 \mod 9$.

Now use the CRT to solve

$x \equiv 1 \mod 7$

$x \equiv 7 \mod 9$

This time you will get the correct answer