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What is the solution range of (a - 3b)x + b - 2a > 0, if the solution

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Math Revolution GMAT Instructor
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What is the solution range of (a - 3b)x + b - 2a > 0, if the solution  [#permalink]

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New post 17 Jan 2020, 00:08
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[GMAT math practice question]

What is the solution range of \((a - 3b)x + b - 2a > 0\), if the solution range of \((a + b)x + 2a - 3b < 0\) is \(x < -13\)?

A. \(x < -3\)

B. \(-3 < x < 3\)

C. \(-1 < x < 0\)

D. \(-1 < x < 3\)

E. \(0 < x < 4\)

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Re: What is the solution range of (a - 3b)x + b - 2a > 0, if the solution  [#permalink]

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New post 19 Jan 2020, 06:29
=>

\((a + b)x + 2a - 3b < 0\)

=> \((a + b)x < 3b - 2a\)

=> \(x < \frac{(3b - 2a) }{ (a + b)} = -(\frac{1}{3})\) under the assumption \(a + b > 0.\)

Then we have \(\frac{(3b - 2a) }{ (a + b)} = -(\frac{1}{3})\) or \((–3)(3b - 2a) = (a + b).\)

We have \(-9b + 6a = a + b\) or \(a = 2b\).

Thus the inequality \((a - 3b)x + b - 2a > 0\) is equivalent to \((2b - 3b)x + b - 2 \)

\((2b) > 0, -bx + b - 4b > 0, -bx – 3b > 0\), or \(-b(x + 3) > 0.\)

Then we have \(x + 3 < 0\) or \(x < -3\) since \(b < 0\).

Therefore, A is the answer.
Answer: A
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What is the solution range of (a - 3b)x + b - 2a > 0, if the solution  [#permalink]

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New post 21 Jan 2020, 06:33
[quote="MathRevolution"]=>

\((a + b)x + 2a - 3b < 0\)

=> \((a + b)x < 3b - 2a\)

=> \(x < \frac{(3b - 2a) }{ (a + b)} = -(\frac{1}{3})\) under the assumption \(a + b > 0.\)

Could you please help me understand the above step?

And the following step too

Thus the inequality \((a - 3b)x + b - 2a > 0\) is equivalent to \((2b - 3b)x + b - 2 \)

\((2b) > 0, -bx + b - 4b > 0, -bx – 3b > 0\), or \(-b(x + 3) > 0.\)

Then we have \(x + 3 < 0\) or \(x < -3\) since \(b < 0\).
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What is the solution range of (a - 3b)x + b - 2a > 0, if the solution   [#permalink] 21 Jan 2020, 06:33
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