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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 < -1/3?

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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 < -1/3?

by Max@Math Revolution » Fri Jan 17, 2020 12:05 am
[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 < -1/3?

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 range of (a + b)x + 2a - 3b < 0 is x < -1/3?

by deloitte247 » Sun Jan 19, 2020 2:50 pm
(a - 3b)x + b - 2a > 0 -- eqn (1)
(a + b)x + 2a - 3b < 0 ---- eqn (2)
From eqn. 2, subtract (2a - 3b) from both sides
(a+b)x + 2a - 3b - (2a - 3b) < 0 - (2a - 3b)
(a+b)x + 2a - 3b - 2a + 3b < 0 -2a + 3b
(a+b)x < -2a + 3b
Divide through by (a+b)
$$\frac{\left(a+b\right)x}{a+b}<\frac{-2a+3b}{a+b}$$
$$x<\frac{-2a+3b}{a+b}$$
Given that x = -1/3, from the question in equation 2
$$-\frac{1}{3}<\frac{-2a+3b}{a+b}$$
-a - b = -6a + 9b
-a + 6a = 9b + b
5a = 10b
a = 10b/5
a = 2b

Substituting a=2b in equation (1)
(2b - 3b)x + b - 2(2b) > 0
-bx + b - 4b > 0
-bx - 3b > 0
-bx > 3b
$$\frac{-bx}{-b}>\ \frac{3b}{-b}$$
$$x<-3$$

Answer = option A
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Re: 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 < -1/3?

by Max@Math Revolution » Sun Jan 19, 2020 5:53 pm
=>

(a + b)x + 2a - 3b < 0
=> (a + b)x < 3b - 2a
=> x < (3b - 2a) / (a + b) = -(1/3) under the assumption a + b > 0.
Then we have (3b - 2a) / (a + b) = -(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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