Solving systems of mods (Chinese Remainder Theorem)

Question

Solve the following system of mods for $x$.

$3x$ $\equiv$ $5\ (mod \ 7)$
$7x$ $\equiv$ $5\ (mod \ 15)$

This is my current work, but am stuck about 75% of the way through:

1) Find multiplicative inverse of $7\ mod\ 15 $

$7b$ ($mod$ 15) = 1 so $b$ = $13$.

$13$ * $7x$ $\equiv$ $5$ * $13$ ($mod \ 15$)
$91x$ $\equiv$ $65$ $ mod \ 15$
$x$ $\equiv$ $5$ $ mod\ 15$

2) Find multiplicative inverse of $3\ mod\ 7%$

$3b$ ($mod$ 7) = 1 so $b$ = $5$.

$5$ * $3x$ $\equiv$ $5$ * $5$ ($mod \ 7$)
$15x$ $\equiv$ $25$ $ mod \ 7$
$x$ $\equiv$ $4$ $ mod\ 7$

3) Substitution method to solve for $x$

$x$ $\equiv$ $4$ $ mod\ 7$

$x$ = $7a+4 $ $ \equiv$ $5 \ (mod \ 15)$

The answer to this problem, according to my textbook, is $x \equiv 95 (mod \ 105)$. The $105$ makes sense since its the LCM of 15 and 7. But what do I do now? Find the multiplicative inverse of 7 mod 15, so let the coefficient of $a$ be 1 and then solve for $x$?

Answer 1

The inverse of $3 \ (mod \ 7)$ is $5$ since $3\cdot5 = 15$ and $15 \ (mod \ 7) \equiv 1 \ (mod \ 7)$, therefore the first equation is equivalent to $$x \ \equiv \ 25 \ (mod \ 7) \ \equiv \ 4 \ (mod \ 7)$$ and this can be seen as $x$ is of the form $4 + 7k, \ k \in \mathbb{Z}$. With this, we go to the other equation.

Now, we have $$ 7x \ \equiv \ 5 \ (mod \ 15) \iff 28 + 49k \ \equiv \ 5 \ (mod \ 15)$$ Reducing the actual equation: $$k \ \equiv \ 13 \ (mod \ 15)$$ hence $k$ is of the form $13 + 15q, \ q \in \mathbb{Z}$. So, $$x = 4 + 7k = 95 + 105q$$ and this is equivalent to $x \ \equiv \ 95 \ (mod \ 105)$.

Answer 2

You're on the right track. Now just compute the value of $a$ modulo $15$ from $7a + 4 \equiv 5 \pmod{15}$ and the you would get that $a$ is of the form $15k + b$ and now just substitute in $x = 7a + 4$.