ax + by=10, ay + bx=12. which one is greater here, x or y ?

After replying to a PM, I thought I have to make a few comments on this question.

This question is about comparing two numbers \(x\) and \(y\).
The best way to do this, is to look at the difference of the two:
if \(x - y = 0\), the two numbers are equal
if \(x - y > 0\), then \(x\) is greater than \(y\)
if \(x - y < 0\), then \(x\) is less than \(y\)

The "trap" here is the given system of the two equations. We don't need to solve it.
The two numbers are totally irrelevant, the only important thing is that 10 < 12.
Therefore, we can write that
\(ax + by < ay + bx\), or \(ax - ay - bx + by < 0\).

After factorisation we get
\(a(x - y) - b(x - y) = (x - y)(a - b) < 0\).

This means that \(x - y\) and \(a - b\) have opposite signs and that neither one can be 0.

From (1), \(a > b\) or \(a - b > 0\), so, necessarily \(x - y < 0\) and \(y\) is greater than \(x\).
Sufficient.

From (2), if \(b > 0\), then \(a > b\) (multiply the given inequality by \(b > 0\), direction preserved).
Then \(a - b > 0\), so \(x - y\) must be negative and \(x\) is smaller than \(y\).
If \(b < 0\), then \(a < b\) (multiply the given inequality by \(b < 0\), direction is reversed).
Then \(a - b < 0\), so \(x - y\) must be positive and \(x\) is greater than \(y\).
Not sufficient.

Hence, Answer A.

Remark:
When dealing with the inequality \(a/b > 1\), since \(b\) cannot be 0, we can multiply by \(b^2\), which is for sure positive and get
\(ab > b^2\)
This leads to \(ab - b^2\) > 0 or \(b(a - b) > 0\).
Now, we can easily see that if \(b > 0\), then \(a > b\).
When \(b < 0\), then \(a < b\).

Suggestion:
An inequality involving a fraction with unknown sign of the denominator can always be multiplied by the square of that
denominator, and the direction of the inequality is preserved.