To determine values for □, ○, △ such that the system has no solution, we use the property of linear systems: parallel lines (same slope) but non-coincident (different constants) have no solution.

Key Insight:

For equations (a_1x + b_1y = c_1) and (a_2x + b_2y = c_2) to have no solution:
(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2})

Application to the Problem:

The second equation is (6x + 8y = 4). Let the first equation be (\square x + \bigcirc y = \triangle).

Choose (\square = 6) and (\bigcirc =8) (so (\frac{6}{6}=\frac{8}{8}=1)), then (\triangle) must not equal (4) (since (\frac{c_1}{4}\neq1)).

Example Solution:

(\square =6), (\bigcirc=8), (\triangle=5) (any (\triangle\neq4) works).

Answer: □=6, ○=8, △=5 (or similar valid combinations).

(\boxed{6}) (□), (\boxed{8}) (○), (\boxed{5}) (△)

(Note: △ can be any number except 4; 5 is just an example.)

Final Answer
The values are (\boxed{6}), (\boxed{8}), and (\boxed{5}) (△ can be any non-4 value). For simplicity, we present this combination:
(\boxed{6})
(\boxed{8})
(\boxed{5})

But since the problem might expect a single line of answer (depending on format), but given the placeholders, the most straightforward is:

(\boxed{6}) (□), (\boxed{8}) (○), (\boxed{5}) (△)

However, if we need to write as per the original input's boxes, the answer could be:

□=6, ○=8, △=5 → so the filled boxes are 6,8,5.

(\boxed{6})
(\boxed{8})
(\boxed{5})

(This is a valid solution.)

(\boxed{6})
(\boxed{8})
(\boxed{5})

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