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.
 A: Existence: $\, x\!+\!ja = z = y\!+\!kb\, \Rightarrow\, (a,b)\mid ja\!-\!kb = y-x,\,$ so $\, y-x = e(a,b)\,$ for some $\,e.$
Bezout $\Rightarrow\exists\, c,d\!:\,  ac\!-\!bd = (a,b),\,$ so  $\,ace\!-\!bde = e(a,b) = y\!-\!x\,$
so  $\, x \!+\! (ce)a = z = y\!+\!(de) b$ 
Uniqueness: $ $  mod $\, a,b\!:\ z'\equiv z\,\Rightarrow\, a,b\mid z'-z\,\Rightarrow\ {\rm lcm}(a,b)\mid z'-z$
A: 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
$$
z_1\equiv z_2\pmod{\mathrm{lcm}(a,b)}
$$
A: Put $d=\gcd(a,b)$ and $\delta=x\bmod d=y\bmod d$ (here "mod" is the remainder operation). Then the numbers $x'=x-\delta$, $y'=y-\delta$ are both divisible by$~d$. In terms of a new variable $z'=z-\delta$ we need to solve the system
$$
  \begin{align}z'&\equiv x'\pmod a,\\z'&\equiv y'\pmod b.\end{align}
$$
Since $x',y',a,b$ are all divisible by $d$, any solution $z'$ will have to be as well; therefore we can divide everything by$~d$, and the system is equivalent to
$$
  \begin{align}\frac{z'}d&\equiv \frac{x'}d\pmod{\frac ad},\\
               \frac{z'}d&\equiv \frac{y'}d\pmod{\frac bd}.\end{align}
$$
Here the moduli $\frac ad,\frac bd$ are relatively prime, so by the Chinese remainder theorem there is a solution $\frac{z'}d\in\mathbf Z$, and it is unique modulo $\frac ad\times\frac bd$. Then the solutions for $z'$ will then form a single class modulo $\frac ad\times\frac bd\times d=\frac{ab}d=\operatorname{lcm}(a,b)$, and so will the solutions for $z=z'+\delta$.
