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16 Sep 2014, 01:27
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If \(x\) and \(y\) are negative numbers, is \(x < y\)?
(1) \(3x + 4 < 2y + 3\)
(2) \(2x - 3 < 3y - 4\)
(1) \(3x + 4 < 2y + 3\)
(2) \(2x - 3 < 3y - 4\)
16 Sep 2014, 01:27
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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
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
07 Mar 2017, 17:37
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Graphically
One easy way to do this is to see then the lines intersect by graphing them. In the attached, red is (1), blue is (2), and yellow is the prompt. You can see that (1) intersects with y>x when x, y negative, and thus you can pick options that allow the red line to be true, but can literally go on either side of y > x. For the blue line, that is also possible, but only when y and x are positive. Thus, we know that (2) is sufficient for all x,y negative.
Algebraically
Another easy way to do this is to see when the lines intersect, but by using algebra.
(1) 2y > 3x+1
Pretend this is 2y=3x+1.
Set this equal to what we are comparing it to, y=x, or, 2y=2x.
2y=3x+1=2x
They intersect at x = -1. Either side of this, you will a Y or N answer when seeing if (1) is sufficient.
(2) 3y > 2x+1
Pretend this is 3y=2x+1.
Set this equal to what we are comparing it to, y=x, or, 3y=3x.
3y=2x+1=3x
They intersect at x=1. If less than 1, you will always get the same answer when seeing if (2) is sufficient. Since x < 0 < 1, (2) is sufficient.
One easy way to do this is to see then the lines intersect by graphing them. In the attached, red is (1), blue is (2), and yellow is the prompt. You can see that (1) intersects with y>x when x, y negative, and thus you can pick options that allow the red line to be true, but can literally go on either side of y > x. For the blue line, that is also possible, but only when y and x are positive. Thus, we know that (2) is sufficient for all x,y negative.
Algebraically
Another easy way to do this is to see when the lines intersect, but by using algebra.
(1) 2y > 3x+1
Pretend this is 2y=3x+1.
Set this equal to what we are comparing it to, y=x, or, 2y=2x.
2y=3x+1=2x
They intersect at x = -1. Either side of this, you will a Y or N answer when seeing if (1) is sufficient.
(2) 3y > 2x+1
Pretend this is 3y=2x+1.
Set this equal to what we are comparing it to, y=x, or, 3y=3x.
3y=2x+1=3x
They intersect at x=1. If less than 1, you will always get the same answer when seeing if (2) is sufficient. Since x < 0 < 1, (2) is sufficient.
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