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[GMAT math practice question]
The solution set of an inequality \((a + b)x + 2a - 3b < 0 is x < \frac{-1}{3}\). What is the solution set of the inequality \((a - 3b)x + b - 2a > 0\), in terms of \(x\)?
A. \(x < -3\)
B. \(-3 < x < 3\)
C. \(-1 < x < 0\)
D. \(-1 < x < 3 \)
E. \(0 < x < 4\)
The solution set of an inequality \((a + b)x + 2a - 3b < 0 is x < \frac{-1}{3}\). What is the solution set of the inequality \((a - 3b)x + b - 2a > 0\), in terms of \(x\)?
A. \(x < -3\)
B. \(-3 < x < 3\)
C. \(-1 < x < 0\)
D. \(-1 < x < 3 \)
E. \(0 < x < 4\)
24 Dec 2019, 01:01
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Given that \((a+b)x+2a−3b<0\)
\(x<\frac{(-2a+3b)}{(a+b)}\)
Given \(x<\frac{-1}{3}\)
Therefore \(\frac{(-2a+3b)}{(a+b)}=\frac{-1}{3}\)
\(-6a+9b=-a-b\)
\(-5a=-10b\)
\(a=2b\)
Substitute \(a=2b\) in \((a−3b)x+b−2a>0(a−3b)x+b−2a>0\)
\(2bx-3bx+b-4b>0\)
\(-bx-3b>0\)
\(-bx>3b\)
\(x<-3\)
The answer is A.
\(x<\frac{(-2a+3b)}{(a+b)}\)
Given \(x<\frac{-1}{3}\)
Therefore \(\frac{(-2a+3b)}{(a+b)}=\frac{-1}{3}\)
\(-6a+9b=-a-b\)
\(-5a=-10b\)
\(a=2b\)
Substitute \(a=2b\) in \((a−3b)x+b−2a>0(a−3b)x+b−2a>0\)
\(2bx-3bx+b-4b>0\)
\(-bx-3b>0\)
\(-bx>3b\)
\(x<-3\)
The answer is A.
26 Dec 2019, 03:57
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=>
\((a + b)x + 2a - 3b < 0\)
=> \((a + b)x < 3b - 2a\)
=> \(x < \frac{(3b - 2a) }{ (a + b)}\) under the assumption \(a + b > 0\)
Since its solution set is \(x < -(\frac{1}{3})\), we have \(\frac{(3b - 2a) }{ (a + b)} = -(\frac{1}{3})\) or \((-3)(3b - 2a) = a + b.\) Then we have \(6a – 9b = a + b\) or \(a = 2b.\)
Since \(a = 2b\), the inequality \((a - 3b)x + b - 2a > 0\) is equivalent to \((2b - 3b)x + b – 4b > 0\) or \(-bx – 3b > 0\).
Then we have \(bx < -3b\) or \(x < -3.\)
Therefore, A is the answer.
Answer: A
\((a + b)x + 2a - 3b < 0\)
=> \((a + b)x < 3b - 2a\)
=> \(x < \frac{(3b - 2a) }{ (a + b)}\) under the assumption \(a + b > 0\)
Since its solution set is \(x < -(\frac{1}{3})\), we have \(\frac{(3b - 2a) }{ (a + b)} = -(\frac{1}{3})\) or \((-3)(3b - 2a) = a + b.\) Then we have \(6a – 9b = a + b\) or \(a = 2b.\)
Since \(a = 2b\), the inequality \((a - 3b)x + b - 2a > 0\) is equivalent to \((2b - 3b)x + b – 4b > 0\) or \(-bx – 3b > 0\).
Then we have \(bx < -3b\) or \(x < -3.\)
Therefore, A is the answer.
Answer: A
