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x/a > 4 and y/a < –6 a2 = 9 ab2 = –8
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28 Mar 2019, 13:37
28 Mar 2019, 13:37
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\(\frac{x}{a} > 4\) and \(\frac{y}{a} < – 6\)
\(a^2 = 9\)
\(ab^2 = –8\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
\(a^2 = 9\)
\(ab^2 = –8\)
Quantity A |
Quantity B |
x |
y |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Re: x/a > 4 and y/a < –6 a2 = 9 ab2 = –8
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30 Mar 2019, 11:54
30 Mar 2019, 11:54
2
Expert Reply
First, if a² = 9, then a = 3 or a = –3. Those are the only options for a.
Next, if ab² = –8, then either a or b² must be negative (note that ab² = a(b²), not (ab)²). b² cannot be negative, so a must be negative.
This means that a = –3, not positive 3. Now we can plug that into our inequalities:
\(\frac{x}{-3}> 4\)
Now remember that when we multiply by a negative, we have to flip the direction of the inequality. So, multiplying both sides by –3:
\(x< -12\)
Doing the same with the other inequality:
\(\frac{y}{-3}< -6\)
\(y> 18\)
So y is positive while x is negative. Hence, Quantity B is definitely greater.
Next, if ab² = –8, then either a or b² must be negative (note that ab² = a(b²), not (ab)²). b² cannot be negative, so a must be negative.
This means that a = –3, not positive 3. Now we can plug that into our inequalities:
\(\frac{x}{-3}> 4\)
Now remember that when we multiply by a negative, we have to flip the direction of the inequality. So, multiplying both sides by –3:
\(x< -12\)
Doing the same with the other inequality:
\(\frac{y}{-3}< -6\)
\(y> 18\)
So y is positive while x is negative. Hence, Quantity B is definitely greater.
Re: x/a > 4 and y/a < –6 a2 = 9 ab2 = –8
[#permalink]
21 Apr 2021, 21:35
21 Apr 2021, 21:35
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