M27-18

Official Solution:


If \(x\) and \(y\) are negative numbers, is \(x < y\)?

(1) \(3x + 4 < 2y + 3\).

Re-arrange to: \(3x < 2y-1\).

Express \(2y\) as \(3y -y\) to get: \(3x < 3y - y - 1\), which can also be written as: \(3x < 3y - (y + 1)\).

Divide each term by 3: \(x < y -\frac{y+1}{3}\).

Now, \(\frac{y+1}{3}\) can be positive if \(y > -1\), and negative if \(y < -1\).

When \(\frac{y+1}{3}\) is positive, \(y -\frac{y+1}{3}\) will be less than \(y\), so we'd have: \(x < y -\frac{y+1}{3} < y\).

When \(\frac{y+1}{3}\) is negative, \(y -\frac{y+1}{3}\) will be greater than y, so we'd have: \(y < y -\frac{y+1}{3}\). In this case, \(x\) could be between \(y\) and \(y -\frac{y+1}{3}\), thus being more than \(y\): \(y < x < y -\frac{y+1}{3}\), as well as to the left of \(y\), thus being less than \(y\): \(x < y < y -\frac{y+1}{3}\).

Not sufficient.

Alternatively, after getting \(3x < 2y-1\), we could plug in numbers. If \(x\) is a very small number, for instance -100, and \(y\) is -1, then \(x < y\) and the answer is YES. However, if \(x=-2\) and \(y=-2\), then \(x=y\) and the answer is NO.

(2) \(2x - 3 < 3y - 4\).

Re-arrange to: \(2x < 3y - 1\).

Express \(3y\) as \(2y + y\) to get: \(2x < 2y + y - 1\).

Divide each term by 2: \(x < y + \frac{y-1}{2}\).

Since \(y\) is negative, \(\frac{y-1}{2}\) will also be negative, thus \(y + \frac{y-1}{2}\) will be less than \(y\). Therefore, we have \(x < y + \frac{y-1}{2} < y\).

Sufficient.


Answer: B