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If |y|≤ - 4x and |3x - 4|= 2x + 6

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If |y|≤ - 4x and |3x - 4|= 2x + 6  [#permalink]

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New post 13 Jul 2018, 06:35
2
1
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

57% (01:45) correct 43% (01:35) wrong based on 35 sessions

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If |y|≤ - 4x and |3x - 4|= 2x + 6. Which of the following could be the value of x?

A). -3
B). -1/2
C). -2/5
D). 1/3
E). 10
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Re: If |y|≤ - 4x and |3x - 4|= 2x + 6  [#permalink]

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New post 13 Jul 2018, 06:47
1
\(|y| \leq -4x\)

The LHS of above equation has to be either 0 or positive

=> \(-4x \geq 0\)

=> \(-x \geq 0\)

=> \(x \leq 0\)

options D and E are thus eliminated

Now for remaining options we can substitute values of x and see which one satisfies the given equation

option A

\(|3x-4| = |2x+6|\)

=> \(|3(-3)-4| = |2(-3)+6|\)

option A doesn't satisfy the give equation

option B

\(|3x-4| = |2x+6|\)

\(|3(\frac{-1}{2})-4| = |2(\frac{-1}{2})+6|\)

option B doesn't satisfy the give equation

option C

\(|3x-4| = |2x+6|\)

\(|3(\frac{-2}{5})-4| = |2(\frac{-2}{5})+6|\)

\(|\frac{-6}{5}-4| = |\frac{-4}{5}+6|\)

\(|\frac{-24}{5}| = |\frac{-24}{5}+6|\)

option C satisfies the give equation

Hence option C
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Re: If |y|≤ - 4x and |3x - 4|= 2x + 6  [#permalink]

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New post 13 Jul 2018, 10:33
CaptainGMAT wrote:
If |y|≤ - 4x and |3x - 4|= 2x + 6. Which of the following could be the value of x?

A). -3
B). -1/2
C). -2/5
D). 1/3
E). 10


Since |y|≤ - 4x, we have x≤ 0 (Absolute value is non-negative)
Given, |3x - 4|= 2x + 6 when 3x-4>0 or x>\(\frac{4}{3}\) (Not possible since x is not positive)

Again |3x - 4|=-(3x-4)=2x + 6
Or, 4-3x=2x+6
Or, 5x=-2
Or, \(x=\frac{-2}{5}\)
Ans. (C)
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If |y|≤ - 4x and |3x - 4|= 2x + 6  [#permalink]

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New post 13 Jul 2018, 12:29
CaptainGMAT wrote:
If |y|≤ - 4x and |3x - 4|= 2x + 6. Which of the following could be the value of x?

A). -3
B). -1/2
C). -2/5
D). 1/3
E). 10

\(y\) imposes conditions on \(x\). In absolute value equation and inequality questions, if we are given a second variable that relates to the main variable, the former often sets conditions on the latter.

Simply solving the second equation yields two satisfactory solutions that are both answer choices.
But the \(y\) statement makes one solution not a possible value of \(x\).

So we must consider what "|y|" indicates: Absolute value is always non-negative.
Hence \(|y|\) MUST be \(\geq\) to 0

Replace the whole expression |y| with \(y\)'s value range
Put 0 (lower limit) on LHS
\((0≤-4x)\) -> divide by -4, switch sign
\((0\geq{x})=> (x\leq0)\)

You can plug A, B, and C into original equation. Or solve.
\(|3x-4|=2x+6\)

Case 1
\(3x-4=2x+6\)
\(x=10\)

Case 2
\(3x-4=-2x-6\)
\(5x=-2\)
\(x=-\frac{2}{5}\)

Case 1 violates the condition that \(x\leq0\)
\(x=-\frac{2}{5}\)

Answer C

Checking the solution \(x=-\frac{2}{5}\) gives \(|-\frac{26}{5}|=\frac{26}{5}\). Valid.
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If |y|≤ - 4x and |3x - 4|= 2x + 6 &nbs [#permalink] 13 Jul 2018, 12:29
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