I tried to solve the following equation system (from a 7th grade exam) in WolframAlpha, however, the steps it takes to get to that solution aren't clear to me.
$x \mod 7 = 3 $
$x \mod 9 = 1 $
$x \mod 4 = 0 $
I also looked into other questions into this platform, but those seem too hard to explain to a person in 7th grade.
On
Hint $\,x \equiv -4 \pmod{\!4\ \&\ 7}\ \iff 4,7\mid x\!+\!4\iff 28\mid x\!+\!4\iff x\equiv \color{#0a0}{-4\pmod{\!28}}$
so $\bmod 9\!:\,\ 1\equiv \color{#0a0}{x\equiv -4+28\color{#c00}k}\equiv -4+k\iff k\equiv 5\iff \color{#c00}{k = 5\!+\!9n}$
so we conclude: $\,\ x = -4 + 28(\color{#c00}{5\!+\!9n}) = 136 + 252 n$
Remark $ $ The optimization in the first sentence is a special case of this general reduction to CCRT = Constant case of Chinese Remainder Theorem, viz. $\,a_1 - m_1 = 3-7 = -4 = 0-4 = a_3 - m_3.\,$ This proves quite handy for manual computations.
If $x=3\pmod 7,$ then it means that $x$ is of the form $7p+3.$ Similarly, $x$ must also be of the forms $9q+1$ and $4r$ by the last equations, where $p,q,r$ are arbitrary integers. Thus, we find that we want a multiple of $4$ simultaneously of the forms $7p+3$ and $9q+1.$ Reducing these modulo $4$ then yields $3p+3=3m$ and $q+1=n$ respectively. Again, $m,n$ are integers. Since we want $n=3m,$ it follows that this number is also a multiple of $3.$ Thus the number is a multiple of $12.$
The question now is, which multiples of $12$ satisfy the constraints? That is, for what $p,q$ are $12k=7p+3$ and $12k=9q+1$ true? Those form the solutions.