Balancing reaction between copper and nitric acid given the ratio of formed nitrogen oxides
The final arbiter of formal correctness of chemical reaction enumeration  are laws of mass and charge conservation. If total counts of charge and atoms of every element are not the same on each side, the equation is wrong. Reaction enumeration by following these laws may be troublesome, as general solution leads to resolving a set of linear equations. That can be useful for educational purposes, or rather to implant terror and despair in eyes of young chemistry students.
Chemists do not go that way. They usually use various heuristics to speed up the process, enumerating equations in several steps. For redox reactions, the heuristic is inventory of exchanged electrons, which is tracked by formal oxidation numbers (ON). These numbers are rather just arbitrary convention, but extremely useful.
For enumeration of your reaction ( with expressed ONs of relevant atoms):
$$\ce{Cu^{0} + HN^{+V}O3 -&gt; Cu^{+II}(N^{+V}O3)2 + N^{+II}O + N^{+IV}O2 + H2O}$$
the first step is matching components undergoing redox change.

Copper changes ON from 0 to +II.i.e. by 2.
Nitrogen changes ON from +V to +IV ( $\ce{NO2}$), or from +V to +II($\ce{NO}$), i.e. by 1 resp. 3.
There is given $\ce{NO/NO2}$ ratio 2:3, so there would be as products $\ce{2 NO + 3 NO2}$, or their multiples.
$\ce{2 NO}$ exchange $2 \cdot 3 = 6$ electrons.
$\ce{3 NO2}$ exchange $3 \cdot 1 = 3$ electrons.

For simplicity, we can take formally $\ce{2 NO + 3 NO2}$ as a single entity, created from $\ce{HNO3}$ by exchanging totally 9 electrons.
Now comes the &quot;magical&quot; formula $9 \cdot 2 = 2 \cdot 9$.
9 Cu exchanging 2 electrons each gives 18 electrons totally.
2 ( $\ce{2 NO + 3 NO2}$  ) exchanging 9 electrons each has accepted 18 electrons.
So the intermediate result is:
$$\ce{9 Cu + HNO3 -&gt; 9 Cu(NO3)2 + 4 NO + 6 NO2 + H2O}$$
This is enough for the question answer, but we have to finish the equation. It may sometimes happen some coefficients would get fractional, so all would get multiplied to have all integer.
We need supply enough nitrogen for the salt and oxides: $2\cdot 9 + 4 + 6 = 28$
$$\ce{9 Cu + 28 HNO3 -&gt; 9 Cu(NO3)2 + 4 NO + 6 NO2 + H2O}$$
We need to create water from hydrogens:
$$\ce{9 Cu + 28 HNO3 -&gt; 9 Cu(NO3)2 + 4 NO + 6 NO2 + 14 H2O}$$
And we are done. We can formally check the item conservation:
$\ce{Cu}$: $9=9$
$\ce{H}$: $28=28$
$\ce{N}$: $28=28$
$\ce{O}$: $3\cdot 28=84=54+4+12+14=84$
charge: $0=0$
For a quick check,we may just check items we were not enumerating before, like just oxygen and charge.