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

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If |y| ≤ -4x and |3x - 4|= 2x + 6. Which of the following could be the  [#permalink]

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New post 12 Nov 2019, 03:49
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If |y| ≤ -4x and |3x - 4|= 2x + 6. Which of the following could be the  [#permalink]

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New post 12 Nov 2019, 06:12
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Quote:
If \(|y| ≤ -4x\) and \(|3x - 4|= 2x + 6\). Which of the following could be the value of x?


\(|y| ≤ -4x\) MEANS \(0 ≤ -4x.....x\leq{0}\), because the least value of |y| is 0 and, therefore, -4x will surely be greater than it or equal to that.
|y| ≤ -4x And 0 ≤ |y|..........0 ≤|y| ≤ -4x, so 0 ≤-4x


Solve \(|3x - 4|= 2x + 6\)
1. \(3x-4=2x+6....x=10\) but \(x\leq{0}\)...Neglect
2. \(-(3x-4)=2x+6....5x=-2....x=\frac{-2}{5}\)..Possible as \(x\leq{0}\).

C. -2/5



C
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Re: If |y| ≤ -4x and |3x - 4|= 2x + 6. Which of the following could be the  [#permalink]

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New post 12 Nov 2019, 07:18
1
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 --> correct
D. 1/3
E. 10

Solution:
\(|y| ≤ -4x\)
=> \( -4x >= |y| >=0\)
=> -x>=0
=> x ≤ 0

So for |3x - 4|= 2x + 6
3x-4 can't be > 0 i.e. x can't be > 4/3, because x can't be > 0
So 3x - 4 ≤ 0
-(3x-4)=2x+6
=> 5x = -2
=> x = -2/5
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Re: If |y| ≤ -4x and |3x - 4|= 2x + 6. Which of the following could be the  [#permalink]

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New post 18 Nov 2019, 09:45
chetan2u wrote:
Quote:
If \(|y| ≤ -4x\) and \(|3x - 4|= 2x + 6\). Which of the following could be the value of x?


\(|y| ≤ -4x\) MEANS \(0 ≤ -4x.....x\leq{0}\)

Also \(|3x - 4|= 2x + 6\)
1. \(3x-4=2x+6....x=10\) but \(x\leq{0}\)...Neglect
2. \(-(3x-4)=2x+6....5x=-2....x=\frac{-2}{5}\)..Possible as \(x\leq{0}\).

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


C



Why |y|=0?

Thanks,
V
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Re: If |y| ≤ -4x and |3x - 4|= 2x + 6. Which of the following could be the  [#permalink]

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New post 18 Nov 2019, 20:31
chetan2u wrote:
Quote:
If \(|y| ≤ -4x\) and \(|3x - 4|= 2x + 6\). Which of the following could be the value of x?


\(|y| ≤ -4x\) MEANS \(0 ≤ -4x.....x\leq{0}\)

Also \(|3x - 4|= 2x + 6\)
1. \(3x-4=2x+6....x=10\) but \(x\leq{0}\)...Neglect
2. \(-(3x-4)=2x+6....5x=-2....x=\frac{-2}{5}\)..Possible as \(x\leq{0}\).

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


C


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Re: If |y| ≤ -4x and |3x - 4|= 2x + 6. Which of the following could be the  [#permalink]

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New post 18 Nov 2019, 20:44
1
vikgupta07 wrote:
chetan2u wrote:
Quote:
If \(|y| ≤ -4x\) and \(|3x - 4|= 2x + 6\). Which of the following could be the value of x?


\(|y| ≤ -4x\) MEANS \(0 ≤ -4x.....x\leq{0}\)

Also \(|3x - 4|= 2x + 6\)
1. \(3x-4=2x+6....x=10\) but \(x\leq{0}\)...Neglect
2. \(-(3x-4)=2x+6....5x=-2....x=\frac{-2}{5}\)..Possible as \(x\leq{0}\).

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


C



Why |y|=0?

Thanks,
V


Hi I have not taken |y|=0, but that is the least value of |y| and -4x will surely be greater than it equal to that.
|y| ≤ -4x And 0 ≤ |y|..........0 ≤|y| ≤ -4x, so 0 ≤-4x
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Re: If |y| ≤ -4x and |3x - 4|= 2x + 6. Which of the following could be the  [#permalink]

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New post 18 Nov 2019, 22:34
\(|y|\) is an absolute value which means \(|y|\) cannot be negative

Therefore since \(|y|≤−4x\), \(x\) must be negative as \(|y|\) cannot be less than or equal to a negative value

Eliminate (D) and (E)

\(|3x−4|=2x+6\)

This means \(3x-4=2x+6\) or \(4-3x=2x+6\)

On solving we get, \(x=10\) or \(x=-\frac{2}{5}\)

Since we know that x must be negative, our answer is \(x=-\frac{2}{5}\)

Answer is (C)
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Re: If |y| ≤ -4x and |3x - 4|= 2x + 6. Which of the following could be the  [#permalink]

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New post 18 Nov 2019, 23:37
2
This is a very good question which tests your fundamental understanding of the concepts of Absolute Values.

The fact that the absolute value of a number cannot be negative is a very useful fact. If you do not leverage this fact, you can miss out on the most crucial piece of the puzzle.

|y| < -4x. We know that |y| can never be negative; or in other words, it is always greater than or equal to ZERO. But, in the inequality, we have a negative sign. This can only mean that x HAS to be non-positive so that -4x becomes non-negative.

Therefore, x CANNOT take any positive values. Basis this inference, we can eliminate answer options D and E straight away. We are left with options A, B and C.

Beyond this stage, the best and the most efficient method of solving this question would be to plug in each of the values from the options and check which value satisfies the equation. Also, from the expression inside the modulus, we can estimate that a fractional value of x is more probable in satisfying the equation. Therefore, I’d try the negative fractions first rather than trying the -3 in option A.
When we plug in the options, we see that option C satisfies the equation. The correct answer option is C.

The most important data in this question was the fact that |y| < -4x because it led us to the conclusion that x CANNOT be positive. Do not get confused as to what is the role of the variable ‘y’ in this question. It doesn’t serve any other purpose than telling us that |y| cannot be negative.

Hope that helps!
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Re: If |y| ≤ -4x and |3x - 4|= 2x + 6. Which of the following could be the  [#permalink]

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New post 19 Nov 2019, 00:32
|3x-4|= 2x+6
taking (3x-4) as positive
3x-4= 2x+6
=> x= 10
taking (3x-4) as negative
-(3x-4)= 2x+6
=> x= -(2/5)

as |y|<= -4x
therefore -4x must be positive, hence x must be negative.
so the ans is x= -(2/5)
Answer: C
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Re: If |y| ≤ -4x and |3x - 4|= 2x + 6. Which of the following could be the  [#permalink]

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New post 19 Nov 2019, 07:33
From Second Equation we know that either X is 10 or -2/5

Substitute these values in the First equation

X=10 yields |y| ≤ -40 which is not possible as the modulus is always positive

Hence X is -2/5
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Re: If |y| ≤ -4x and |3x - 4|= 2x + 6. Which of the following could be the  [#permalink]

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New post 19 Nov 2019, 07:34
From Second Equation we know that either X is 10 or -2/5

Substitute these values in the First equation

X=10 yields |y| ≤ -40 which is not possible as the modulus is always positive

Hence X is -2/5
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Re: If |y| ≤ -4x and |3x - 4|= 2x + 6. Which of the following could be the   [#permalink] 19 Nov 2019, 07:34
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