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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
Joined: 16 Aug 2015
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What is the solution range of (a - 3b)x + b - 2a > 0, if the solution  [#permalink]

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17 Jan 2020, 00:08
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13% (02:08) correct 87% (02:59) wrong based on 15 sessions

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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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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8584 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: What is the solution range of (a - 3b)x + b - 2a > 0, if the solution [#permalink] ### Show Tags 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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What is the solution range of (a - 3b)x + b - 2a > 0, if the solution  [#permalink]

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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 $$a + b > 0.$$

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 $$b < 0$$.
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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