Solve: $3x≡1 \pmod 7, 4x≡1 \pmod 9$

Question

My attempt:

3x≡1 mod 7 (1)

4x≡1 mod 9 (2)

Multiply (1) by 5

Multiply (2) by 7

x≡5 mod 7

x≡7 mod 9

So x≡9k+7

9k+7=5(mod7)

k=5(mod7)

k=7j+5

x=9(7j+5)+7

=63j+52

x≡52(mod63)

Answer 1

From the first equation (note that $5$ is the multplicative inverse of $3 \pmod 5$),

$3(5)x \equiv 5 \pmod 7 \implies 15x \equiv 5 \pmod 7 \implies x \equiv 5 \pmod 7$. So $x = 7k + 5$

Substitute into the second to give $28k + 20 \equiv 1 \pmod 9 \implies k \equiv -19\pmod 9 \implies k \equiv -1 \pmod 9$. So $k = 9t -1$

So $x = 7(9t-1) + 5 = 63t -2$, giving the solution $x \equiv -2 \equiv 61 \pmod {63}$.

In asker's question, the error is here:

Going from "9k+7=5(mod7)"

to "k=5(mod7)"

is unjustifiable. From the first, we get $9k \equiv -2 \pmod 7$ or $2k \equiv -2 \pmod 7$. That gives $k \equiv -1 \equiv 6 \pmod 7$.

It is also equally unclear how "So x≡9k+7" became "x=9(k)+5" later on. The latter should be $x = 9(k) + 7$.

Correcting for those two errors gives the expected solution (identical to mine).

Answer 2

Summing two equations you get $7x=2$, or $0=2$ in mod 7. So no solutions exist.

Answer 3

Notice that we need $x+2$ divisible by $7$ and by $9$, hence by LCM$(7,9)$