Correct, if $x$ is a root of the congruences then $\, x= j\,m_1 + r_1 = -k\, m_2 + r_2\,$ has roots $\,j,k\in\Bbb Z$.
This argument reverses:$ $ if $\,j,k\,$ are integers with $\ j\, \color{#c00}{m_1}+ k\, \color{#0a0}{m_2} =\, \color{#0a0}{r_2} - \color{#c00}{r_1}\, $ then by $\rm\color{#c00}{re}\color{#0a0}{arranging}$ $\ x :=\ \color{#c00}{r_1} +\, j\ \color{#c00}{m_1}^{\phantom{|}}\ =\,\ \ \color{#0a0}{r_2}\: -\,\ k\ \color{#0a0}{m_2}\ $ is one solution of the given system of congruences because $\,x\equiv \color{#c00}{r_1}\!\!\pmod{\!\!\color{#c00}{m_1}}^{\phantom{|^|}}\!\!\!,$ $x\equiv \color{#0a0}{r_2}\!\pmod{\!\!\color{#0a0}{m_2}}.\,$ Since you've already discovered one such solution for $\,j,k,\,$ viz. $\,j=hu,\,k^{\phantom{|^|}}\!\! = hv,\,$ you need only substitute into the above rearranged CRT solution for $\,x.$
Remark $ $ Combining both directions above and adding a final gcd equivalence yields the following
CRT Solvability Criterion $ $ (which immediately generalizes to ideals)
Theorem $\ \ \left.\exists\, x\in\Bbb Z\!: \begin{align}x\equiv r_1\!\!\!\pmod{\!m_1}\\ x\equiv r_2\!\!\!\pmod{\!m_2}\end{align}\right\} \begin{array}{l}\!\iff \exists\,j,k\in\Bbb Z\!:\ j\,m_1\! + k\, m_2 =\, r_2\!-r_1 \\ \!\iff\, \gcd(m_1,\,m_2)\mid r_2 -r_1\end{array}$
Proof $ $ Clearly $\,d := \gcd(m_1,m_2)\mid r_2-r_1 \,$ is a necessary condition for the equation to have roots $\,j,k\in \Bbb Z,\,$ by $\,d\mid m_1,m_2\Rightarrow\, d^{\phantom{|}}_{\phantom{i}}\!\mid j m_1\! + km_2 = r_2 - r_1.\,$ Further this condition is also sufficient by Bezout (or, constructively, by the extended Euclidean algorithm), i.e. we can scale the Bezout equation $\, a m_1\! + b m_2 = d\,$ by $\, c = \large \frac{r_2\,-\,r_1^{\phantom{.}}}{d}\,$ to get $\,ca\,m_1\!+cb\,m_2 = r_2-r_1 \,$ so, as above, rearranging this yields a congruence system solution: $\ x\, :=\, r_1 + ca\,m_1 = r_2 - cb\,m_2$.
Thus the congruence system is solvable $\iff d=\gcd(m_1,m_2)\mid r_2-r_1, \,$ i.e. iff the pair of congruences is consistent mod their moduli gcd, and when true we can constructively read off a solution from the Bezout equation for the moduli by translating it into the equivalent system language as above, i.e. scale the Bezout equation to obtain the residue difference $\,r_1-r_2\,$ then rearrange it as above to obtain $\,x.\,$ Here is a worked example from this viewpoint, and here is some motivational examples. Summarizing the above method, we have derived the following very simple Bezout-based CRT (Chinese Remainder Theorem) method for solving congruence systems
$$\bbox[8px,border:2px solid #c00]{\text{$\color{#90f}{\text{scale}}$ the Bezout equation by the residue difference - then ${\rm \color{#c00}{re}\color{#0a0}{arrange}}$}}\qquad$$
$$\begin{align}
{\rm if}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\begin{array}{rr} &x\equiv\, r_1\pmod{\!m_1}\\ &x\equiv\, r_2\pmod{\! m_2} \end{array}\ \ \ \ {\rm and}\ \ \ \ \gcd(m_1,m_2) = 1\\[.4em]
{\rm then}\ \ \ r_1 - r_2\, &=:\ \delta\qquad\qquad\qquad \rm residue\ difference \\[.2em]
\times\qquad\quad\ \ \ 1\ &=\ \ a m_1\, +\, b m_2\quad\ \rm Bezout\ equation\ for \ \gcd(m_1,m_2) \\[.5em]\hline
\Longrightarrow\ \,r_1\ \color{#c00}{-\ r_2}\ \ \ \ &=\, \color{#0a0}{\delta am_1}\! + \delta bm_2\quad\ \rm product\ of \ above\ (= {\color{#90f}{scaled}}\ Bezout)\\[.2em]
\Longrightarrow\ \underbrace{r_1 \color{#0a0}{- \delta am_1}}_{\!\!\!\large \equiv\ r_1\! \pmod{\!m_1}}\! &=\,\ \ \underbrace{\color{#c00}{r_2}\ +\ \delta bm_2}_{\large\!\! \equiv\ r_2\! \pmod{\!m_2}}\ \ \ \
\underset{\large {\rm has\ sought\ residues}\phantom{1^{1^{1^{1^1}}}}}{\rm \color{#c00}{\!re}\color{#0a0}{arranged}\ product}\rm\! = {\small CRT}\ solution\end{align}\qquad$$
This generalizes: a system of $n$ congruences is solvable $\iff$ each pair of congruences is solvable as above, and we can solve the system by successively replacing a pair of congruences by a single equivalent congruence. By induction we eventually obtain a single congruence equivalent to the entire system.
Notice the above-linked ideal version can be expressed succinctly in terms of cosets, viz.
$$ \bbox[9px,border:2px solid #c00]{r_1\! +\! m_1\Bbb Z\,\cap\, r_2\! +\! m_2\Bbb Z \neq \phi \iff r_1-r_2 \in m_1\Bbb Z+m_2\Bbb Z}\qquad\qquad $$
Generally it is easier to solve them a pair at a time, as follows.
${\rm mod}\ 25\!:\,\ 7\equiv x\iff x = 7+25j\,$ for some integer $j$
${\rm mod}\ 21\!:\,\ 5 \equiv x\equiv 7+25j\equiv 7+4j\!\iff\! 4j\equiv -2\!\iff\! 2j \equiv -1\equiv 20 \!\iff\! j \equiv \color{#c00}{10}$
$\qquad\quad\, \iff x = 7+25(\color{#c00}{10}+21k) = 257 + 525k$
${\rm mod}\,\ 4\!:\,\ 3 \equiv x = 257+525k\equiv 1+k\iff k \equiv \color{#0a0}2$
$\qquad\ \ \iff x = 257 + 525(\color{#0a0}2 + 4n) \equiv 1307\pmod{\!2100}$
Remark $ $ Only very small numbers were involved because we solved the congruences with largest moduli first, so that, in the end, when the numbers are bigger, this is compensated by computing at smaller moduli.
This method also works even if the moduli are not pairwise coprime, though in this general case there may be zero or multiple solutions. Furthermore it leads to the well-known criterion for existence of a solution in the general case - which characterizes the ubiquitous Prüfer domains. They generalize many common number systems and are characterized by a remarkably huge number of interesting properties, e.g. Gauss's Lemma for contents, lcm distributes over gcd (and reversely), contains = divides for fin. gen. ideals, etc. Thus it is well-worth investing the small effort needed to master this viewpoint of CRT.
Best Answer
Welcome to Math SX! You have to use Euler's theorem as $\varphi(4)=2$, $\;\varphi(25)=20$ we have $$ 7^{30}\equiv7^{30\bmod2}=1\mod 4,\qquad 7^{30}\equiv7^{30\bmod20}=7^{10}\mod 25$$ To find the latter power, you can use the modular fast exponentiation algorithm, but here, it will be simpler: modulo $25$, $$7^2\equiv -1\enspace\text{so}\enspace 7^4=1,\enspace\text{hence } \;7^{30}\equiv 7^{30\bmod 4}=7^2\equiv -1.$$ Finally, since $\;25-6\cdot 4=1$ (Bézout's identity), $$7\equiv \begin{cases}\phantom{-}1\mod4\\-1\mod 25\end{cases}\iff 7\equiv 1\cdot 25-(-1)\cdot 6\cdot 4=49\mod 100.$$