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If y = |3x - 5|/(-x^2 - 3) for what value of x will the value of y be

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Re: If y = |3x - 5|/(-x^2 - 3) for what value of x will the value of y be [#permalink]
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Video solution from Quant Reasoning starts at 22:23
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General Discussion
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Re: If y = |3x - 5|/(-x^2 - 3) for what value of x will the value of y be [#permalink]
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parkhydel wrote:
If $$y = \frac{|3x - 5|}{-x^2 - 3}$$ for what value of x will the value of y be greatest?

A. -5
B. -3/5
C. 0
D. 3/5
E. 5/3

PS29580.02

We have $$y = \frac{|3x - 5|}{-x^2 - 3}$$

The denominator, $$-x^2 - 3$$ MUST be Negative

The Numerator, $$|3x - 5|$$ can NOT be Negative

i.e. $$y = \frac{|3x - 5|}{-x^2 - 3}$$ MUST be Non-Positive

i.e. Maximum Value of a non-Positive function can be ZERO when Numerator is Zero[

i.e. 3x-5 = 0
i.e. y is maximum at $$x = 5/3$$ and $$y_{max}=0$$

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Re: If y = |3x - 5|/(-x^2 - 3) for what value of x will the value of y be [#permalink]
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parkhydel wrote:
If $$y = \frac{|3x - 5|}{-x^2 - 3}$$ for what value of x will the value of y be greatest?

A. -5
B. -3/5
C. 0
D. 3/5
E. 5/3

PS29580.02

Confused of signs ? just substitute X= options

arrive at Solution which gives greatest takes 1 to 2 mins. But helps if u are confused

E.5/3 gives greatest value when substituted

HOPE this helps.
THANKS
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Re: If y = |3x - 5|/(-x^2 - 3) for what value of x will the value of y be [#permalink]
parkhydel wrote:
If $$y = \frac{|3x - 5|}{-x^2 - 3}$$ for what value of x will the value of y be greatest?

A. -5
B. -3/5
C. 0
D. 3/5
E. 5/3

PS29580.02

Segment the numerator and denominator to look for extreme combination: |3x-5|>=0 and $$-x^2-3<=-3$$, and hence y<=0.

Thus the greatest value of y is 0 --> 3x-5 = 0 --> x = $$5/3$$
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Re: If y = |3x - 5|/(-x^2 - 3) for what value of x will the value of y be [#permalink]
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parkhydel wrote:
If $$y = \frac{|3x - 5|}{-x^2 - 3}$$ for what value of x will the value of y be greatest?

A. -5
B. -3/5
C. 0
D. 3/5
E. 5/3

PS29580.02

$$y = \frac{+ numerator }{ - denominator}$$
Hence y will always be negative except when the numerator is zero which is the largest value y can take.

$$\implies$$ |3x-5| = 0
$$\implies x = \frac{5}{3}$$

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Re: If y = |3x - 5|/(-x^2 - 3) for what value of x will the value of y be [#permalink]
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BrentGMATPrepNow wrote:
parkhydel wrote:
If $$y = \frac{|3x - 5|}{-x^2 - 3}$$ for what value of x will the value of y be greatest?

A. -5
B. -3/5
C. 0
D. 3/5
E. 5/3

Key idea: $$x^2$$ is always greater than or equal to 0
This means $$-x^2$$ is always less than or equal to 0
And this means $$-x^2 - 3$$ is always NEGATIVE

In other words, the denominator of $$y$$ is most definitely NEGATIVE

From here, there are two possible cases:
case i: If the numerator of $$y$$ is NEGATIVE, then y is POSITIVE
case ii: If the numerator of $$y$$ is POSITIVE, then y is NEGATIVE

Since we're trying to MAXIMIZE the value of y, we need the denominator to be negative (which would make y POSITIVE)
However, the denominator, |3x - 5|, will always be greater than or equal to 0.
Since we can't make the numerator NEGATIVE, the only way to maximize the value of y is make the numerator ZERO
In other words: $$|3x - 5|=0$$
Which means $$3x -5=0$$
So $$3x =5$$
Ergo $$x = \frac{5}{3}$$

Cheers,
Brent

Hi BrentGMATPrepNow Brent, the explanation is nice i just have a few questions:

When i saw |3x - 5| i thought i should consider two case

3x-5 or -3x+5 so I kinda confused...

3x=5 or -3x=-5

x = 5/3 or x = 5/3 (negative/negative =positive) both options give the same results thats strange to me Why ?

General question: if the absolute value by definition is distance from 0 and the distance is always positive or zero. why do we consider two cases positive and negative

p.s. i think you meant $$-x^2 - 3$$ in the highlighted part
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Re: If y = |3x - 5|/(-x^2 - 3) for what value of x will the value of y be [#permalink]
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dave13 wrote:

Hi BrentGMATPrepNow Brent, the explanation is nice i just have a few questions:

When i saw |3x - 5| i thought i should consider two case

3x-5 or -3x+5 so I kinda confused...

3x=5 or -3x=-5

x = 5/3 or x = 5/3 (negative/negative =positive) both options give the same results thats strange to me Why ?

General question: if the absolute value by definition is distance from 0 and the distance is always positive or zero. why do we consider two cases positive and negative

p.s. i think you meant $$-x^2 - 3$$ in the highlighted part

You have taken the equation |3x - 5| = 0 and concluded that EITHER 3x - 5 = 0 OR -(3x - 5) = 0
Noticed that both of these equations are the same.
If we take -(3x - 5) = 0 and multiply both sides by -1, we got the equivalent equation 3x - 5 = -0, which is identical to the equation 3x - 5 = 0

Thanks for the heads-up about $$-x^2 - 3$$
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If y = |3x - 5|/(-x^2 - 3) for what value of x will the value of y be [#permalink]
BrentGMATPrepNow wrote:
parkhydel wrote:
If $$y = \frac{|3x - 5|}{-x^2 - 3}$$ for what value of x will the value of y be greatest?

A. -5
B. -3/5
C. 0
D. 3/5
E. 5/3

Key idea: $$x^2$$ is always greater than or equal to 0
This means $$-x^2$$ is always less than or equal to 0
And this means $$-x^2 - 3$$ is always NEGATIVE

In other words, the denominator of $$y$$ is most definitely NEGATIVE

From here, there are two possible cases:
case i: If the numerator of $$y$$ is NEGATIVE, then y is POSITIVE
case ii: If the numerator of $$y$$ is POSITIVE, then y is NEGATIVE

Since we're trying to MAXIMIZE the value of y, we need the denominator to be negative (which would make y POSITIVE)
However, the numerator, |3x - 5|, will always be greater than or equal to 0.
Since we can't make the numerator NEGATIVE, the only way to maximize the value of y is make the numerator ZERO
In other words: $$|3x - 5|=0$$
Which means $$3x -5=0$$
So $$3x =5$$
Ergo $$x = \frac{5}{3}$$

Cheers,
Brent

BrentGMATPrepNow this time i picked A because i thought since Absolute value is always positive i can pick -5 so and in denominator since it it is --
-x^2 after plugging in -5, the sign will change into positive - *-5^2 = 25

So
$$y = \frac{|3 (-5) - 5|}{- -5^2 - 3}$$

= 10/22 greater than 0

why its wrong with it?
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Re: If y = |3x - 5|/(-x^2 - 3) for what value of x will the value of y be [#permalink]
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dave13 wrote:
BrentGMATPrepNow this time i picked A because i thought since Absolute value is always positive i can pick -5 so and in denominator since it it is --
-x^2 after plugging in -5, the sign will change into positive - *-5^2 = 25

So
$$y = \frac{|3 (-5) - 5|}{- -5^2 - 3}$$

= 10/22 greater than 0

why its wrong with it?

Think of -x² as -(x²)
So, -x² - 3 = -(x²) - 3
So, when we plug in x = -5, we get: -(x²) - 3 = -((-5)²) - 3 = -(25) - 3 = -28

Does that help?
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Re: If y = |3x - 5|/(-x^2 - 3) for what value of x will the value of y be [#permalink]
IMO - E
I did it by subsituting the given value of X in option and found the largest value for Y is obtained by option E (5/3)
also realized, is denominator is negative and numerator is positive,
to obain highest value Y when the numerator is zero
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Re: If y = |3x - 5|/(-x^2 - 3) for what value of x will the value of y be [#permalink]
Denominator is always negative -x^2 - 3

Max value can be 0

x = 5/3

E
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Re: If y = |3x - 5|/(-x^2 - 3) for what value of x will the value of y be [#permalink]
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BrentGMATPrepNow wrote:
parkhydel wrote:
If $$y = \frac{|3x - 5|}{-x^2 - 3}$$ for what value of x will the value of y be greatest?

A. -5
B. -3/5
C. 0
D. 3/5
E. 5/3

Key idea: $$x^2$$ is always greater than or equal to 0
This means $$-x^2$$ is always less than or equal to 0
And this means $$-x^2 - 3$$ is always NEGATIVE

In other words, the denominator of $$y$$ is most definitely NEGATIVE

From here, there are two possible cases:
case i: If the numerator of $$y$$ is NEGATIVE, then y is POSITIVE
case ii: If the numerator of $$y$$ is POSITIVE, then y is NEGATIVE

Since we're trying to MAXIMIZE the value of y, we need the denominator to be negative (which would make y POSITIVE)
However, the numerator, |3x - 5|, will always be greater than or equal to 0.
Since we can't make the numerator NEGATIVE, the only way to maximize the value of y is make the numerator ZERO
In other words: $$|3x - 5|=0$$
Which means $$3x -5=0$$
So $$3x =5$$
Ergo $$x = \frac{5}{3}$$

Cheers,
Brent

Hi Brent,

Thank you for the great explanation. I understood the concept based on your walk-through.

Although, I did catch one issue with your one statement and please correct me if I am wrong.

BrentGMATPrepNow wrote:
Since we're trying to MAXIMIZE the value of y, we need the denominator to be negative (which would make y POSITIVE)

Regards,
Sud
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Re: If y = |3x - 5|/(-x^2 - 3) for what value of x will the value of y be [#permalink]
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BrentGMATPrepNow wrote:
parkhydel wrote:
If $$y = \frac{|3x - 5|}{-x^2 - 3}$$ for what value of x will the value of y be greatest?

A. -5
B. -3/5
C. 0
D. 3/5
E. 5/3

Key idea: $$x^2$$ is always greater than or equal to 0
This means $$-x^2$$ is always less than or equal to 0
And this means $$-x^2 - 3$$ is always NEGATIVE

In other words, the denominator of $$y$$ is most definitely NEGATIVE

From here, there are two possible cases:
case i: If the numerator of $$y$$ is NEGATIVE, then y is POSITIVE
case ii: If the numerator of $$y$$ is POSITIVE, then y is NEGATIVE

Since we're trying to MAXIMIZE the value of y, we need the denominator to be negative (which would make y POSITIVE)
However, the numerator, |3x - 5|, will always be greater than or equal to 0.
Since we can't make the numerator NEGATIVE, the only way to maximize the value of y is make the numerator ZERO
In other words: $$|3x - 5|=0$$
Which means $$3x -5=0$$
So $$3x =5$$
Ergo $$x = \frac{5}{3}$$

Cheers,
Brent

Hi Brent,

I have a question regarding the modulus of the numerator.....since this is an absolute value function, the sign of the numerator will depend on the range of X. In this case, if X<5/3 the function should be written as -(3X-5), which makes the numerator negative for all range of values of X<5/3. So there can be values greater than 0 now.
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Re: If y = |3x - 5|/(-x^2 - 3) for what value of x will the value of y be [#permalink]
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$$y = \frac{|3x - 5|}{-x^2 - 3}$$

= $$\frac{|3x - 5|}{ -(x^2 + 3)}$$
= - $$\frac{|3x - 5|}{x^2 + 3}$$

Now, both |3x - 5| and $$x^2 + 3$$ are non-negative values and we have a negative sign outside
=> Value can be maximum only when the numerator is either zero or is equal to denominator so that we get a small negative value overall

|3x - 5|=0
=> x = $$\frac{5}{3}$$

So, y = 0 when $$\frac{5}{3}$$