<s>[INST] If $x\equiv y \pmod{\gcd(a,b)}$, show that there is a unique $z\pmod{\text{lcm}(a,b)}$ with $z\equiv x\pmod a$ and $z\equiv y\pmod b$ 
If $x\equiv y \pmod{\gcd(a,b)}$, show that there is a unique $z\pmod{\text{lcm}(a,b)}$ with $z\equiv x\pmod{a}$ and $z\equiv y\pmod{b}$

What I have so far:
Let $z \equiv x\pmod{\frac{a}{\gcd(a,b)}}$ and $
z \equiv y\pmod b
 $. Then by the chinese remainder theorem there is a unique $z\pmod{\text{lcm}(a,b)}$ which satisfies this...
Is this the right approach here? I can't figure out how to get from $$z \equiv x\pmod{\frac{a}{\gcd(a,b)}}$$ what I need.
 [/INST] Existence
Bezout's Identity says that we have $g_a,g_b$ so that
$$
g_aa+g_bb=\gcd(a,b)\tag{1}
$$
Thus,
$$
\begin{align}
g_bb&\equiv\gcd(a,b)&\pmod{a}\\
g_bb&\equiv0&\pmod{b}
\end{align}\tag{2}
$$
Since $x\equiv y\pmod{\gcd(a,b)}$, we have $\gcd(a,b)\mid x{-}y$. Multiplying $(2)$ by $\frac{x-y}{\gcd(a,b)}$ gives
$$
\begin{align}
\frac{x-y}{\gcd(a,b)}g_bb&\equiv\frac{x-y}{\gcd(a,b)}\gcd(a,b)=x-y&\pmod{a}\\[6pt]
\frac{x-y}{\gcd(a,b)}g_bb&\equiv\frac{x-y}{\gcd(a,b)}0=0&\pmod{b}
\end{align}\tag{3}
$$
Adding $y$ to $(3)$ and setting $z=\frac{x-y}{\gcd(a,b)}g_bb+y$, we have
$$
\begin{align}
z&\equiv x&\pmod{a}\\
z&\equiv y&\pmod{b}
\end{align}
$$

Uniquenes
Suppose that $z_1\equiv x\pmod{a}$ and $z_2\equiv x\pmod{a}$. Then $z_1\equiv z_2\pmod{a}$.
Suppose that $z_1\equiv y\pmod{b}$ and $z_2\equiv y\pmod{b}$. Then $z_1\equiv z_2\pmod{b}$.
Therefore, we have $z_a,z_b$ so that
$$
z_1-z_2=z_aa=z_bb\tag{4}
$$
Multiplying $(1)$ by $(4)$ yields
$$
\begin{align}
g_aa\color{#C00000}{z_bb}+g_bb\color{#C00000}{z_aa}&=\gcd(a,b)\color{#C00000}{(z_1-z_2)}\\
(g_az_b+g_bz_a)\frac{ab}{\gcd(a,b)}&=z_1-z_2
\end{align}
$$
Therefore, since $\gcd(a,b)\,\mathrm{lcm}(a,b)=ab$, we have
$$
z_1\equiv z_2\pmod{\mathrm{lcm}(a,b)}
$$
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