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505-555 (Easy)|   Exponents|      
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MarcoL90
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If y is not equal to 4, x is not equal to 0, and (y^2 – 16)/(3x) = (y Updated on: 16 Oct 2016, 06:28
Originally posted by MarcoL90 on 16 Oct 2016, 06:21.
Last edited by Bunuel on 16 Oct 2016, 06:28, edited 2 times in total.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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Question Stats:

82% (01:52) correct 18% (02:32) wrong based on 355 sessions

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If y is not equal to 4, x is not equal to 0, and (y^2 – 16)/(3x) = (y – 4)/6, then in terms of x, y equals:

A: (x + 8) / 2

B: (x – 8) / 2

C: (-3x) / 2

D: (-3x + 8) / 2

E: (3x – 8) / 2
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B
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Bunuel
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If y is not equal to 4, x is not equal to 0, and (y^2 – 16)/(3x) = (y 16 Oct 2016, 06:31
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MarcoL90
If y is not equal to 4, x is not equal to 0, and (y^2 – 16)/(3x) = (y – 4)/6, then in terms of x, y equals:

A: (x + 8) / 2

B: (x – 8) / 2

C: (-3x) / 2

D: (-3x + 8) / 2

E: (3x – 8) / 2
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\(\frac{(y^2 – 16)}{(3x)} = \frac{(y – 4)}{6}\);

\(\frac{(y – 4)(y+4)}{(3x)} = \frac{(y – 4)}{6}\);

\(\frac{(y+4)}{(3x)} = \frac{1}{6}\)

\(6(y+4)=3x\);

\(2(y+4)=x\);

\(y=\frac{x-8}{2}\).

Answer:B.
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If y is not equal to 4, x is not equal to 0, and (y^2 – 16)/(3x) = (y 28 Nov 2017, 13:50
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Hi All,

Many Test Takers would approach this question algebraically (which is fine); however, it's designed in such a way that you can also get to the solution by TESTing VALUES (although we are not allowed to use X=4 or Y=0 in our work).

Let's TEST Y = 5.... that makes the equation...

9/3X = 1/6
3X = 54
X = 18

Thus, we're looking for an answer that equals 5 when X=18. There's only one answer that matches...

Final Answer:
Show Spoiler
B


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If y is not equal to 4, x is not equal to 0, and (y^2 – 16)/(3x) = (y 25 Jun 2020, 16:01
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Bunuel
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If y is not equal to 4, x is not equal to 0, and (y^2 – 16)/(3x) = (y – 4)/6, then in terms of x, y equals:

A: (x + 8) / 2

B: (x – 8) / 2

C: (-3x) / 2

D: (-3x + 8) / 2

E: (3x – 8) / 2

\(\frac{(y^2 – 16)}{(3x)} = \frac{(y – 4)}{6}\);

\(\frac{(y – 4)(y+4)}{(3x)} = \frac{(y – 4)}{6}\);

\(\frac{(y+4)}{(3x)} = \frac{1}{6}\)

\(6(y+4)=3x\);

\(2(y+4)=x\);

\(y=\frac{x-8}{2}\).

Answer:B.
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Hi Bunuel :

Per the yellow highlight, are you allowed to cancel (y-4) on both sides if y = 4 ?

I am trying to understand the significance of the information per the red font specifically.

If the red bit was not given, would we be able to cancel (y-4) on both sides

Thank you
Jd
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If y is not equal to 4, x is not equal to 0, and (y^2 – 16)/(3x) = (y 25 Jun 2020, 21:16
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Bunuel
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If y is not equal to 4, x is not equal to 0, and (y^2 – 16)/(3x) = (y – 4)/6, then in terms of x, y equals:

A: (x + 8) / 2

B: (x – 8) / 2

C: (-3x) / 2

D: (-3x + 8) / 2

E: (3x – 8) / 2

\(\frac{(y^2 – 16)}{(3x)} = \frac{(y – 4)}{6}\);

\(\frac{(y – 4)(y+4)}{(3x)} = \frac{(y – 4)}{6}\);

\(\frac{(y+4)}{(3x)} = \frac{1}{6}\)

\(6(y+4)=3x\);

\(2(y+4)=x\);

\(y=\frac{x-8}{2}\).

Answer:B.

Hi Bunuel :

Per the yellow highlight, are you allowed to cancel (y-4) on both sides if y = 4 ?

I am trying to understand the significance of the information per the red font specifically.

If the red bit was not given, would we be able to cancel (y-4) on both sides

Thank you
Jd
Show more

If we were not told that y is not equal to 4, we would not be able to reduce by y - 4.
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If y is not equal to 4, x is not equal to 0, and (y^2 – 16)/(3x) = (y 25 Jun 2020, 21:20
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If y is not equal to 4, x is not equal to 0, and (y^2 – 16)/(3x) = (y – 4)/6, then in terms of x, y equals:

A: (x + 8) / 2

B: (x – 8) / 2

C: (-3x) / 2

D: (-3x + 8) / 2

E: (3x – 8) / 2

\(\frac{(y^2 – 16)}{(3x)} = \frac{(y – 4)}{6}\);

\(\frac{(y – 4)(y+4)}{(3x)} = \frac{(y – 4)}{6}\);

\(\frac{(y+4)}{(3x)} = \frac{1}{6}\)

\(6(y+4)=3x\);

\(2(y+4)=x\);

\(y=\frac{x-8}{2}\).

Answer:B.

Hi Bunuel :

Per the yellow highlight, are you allowed to cancel (y-4) on both sides if y = 4 ?

I am trying to understand the significance of the information per the red font specifically.

If the red bit was not given, would we be able to cancel (y-4) on both sides

Thank you
Jd

If we were not told that y is not equal to 4, we would not be able to reduce by y - 4.
Show more

That's what I thought too !

Just wondering, why can't we reduce by y-4 ?

Is it because we are not allowed to divide by zero ?

Posted from my mobile device
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If y is not equal to 4, x is not equal to 0, and (y^2 – 16)/(3x) = (y 25 Jun 2020, 21:26
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Bunuel
MarcoL90
If y is not equal to 4, x is not equal to 0, and (y^2 – 16)/(3x) = (y – 4)/6, then in terms of x, y equals:

A: (x + 8) / 2

B: (x – 8) / 2

C: (-3x) / 2

D: (-3x + 8) / 2

E: (3x – 8) / 2

\(\frac{(y^2 – 16)}{(3x)} = \frac{(y – 4)}{6}\);

\(\frac{(y – 4)(y+4)}{(3x)} = \frac{(y – 4)}{6}\);

\(\frac{(y+4)}{(3x)} = \frac{1}{6}\)

\(6(y+4)=3x\);

\(2(y+4)=x\);

\(y=\frac{x-8}{2}\).

Answer:B.

Hi Bunuel :

Per the yellow highlight, are you allowed to cancel (y-4) on both sides if y = 4 ?

I am trying to understand the significance of the information per the red font specifically.

If the red bit was not given, would we be able to cancel (y-4) on both sides

Thank you
Jd

If we were not told that y is not equal to 4, we would not be able to reduce by y - 4.

That's what I thought too !

Just wondering, why can't we reduce by y-4 ?

Is it because we are not allowed to divide by zero ?

Posted from my mobile device
Show more

Yes. If we were not told that y is not equal to 4, then we'd have that y = 4 OR y = (x – 8)/2.
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If y is not equal to 4, x is not equal to 0, and (y^2 – 16)/(3x) = (y 25 Jun 2020, 23:09
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MarcoL90
If y is not equal to 4, x is not equal to 0, and (y^2 – 16)/(3x) = (y – 4)/6, then in terms of x, y equals:

A: (x + 8) / 2

B: (x – 8) / 2

C: (-3x) / 2

D: (-3x + 8) / 2

E: (3x – 8) / 2
Show more

Asked: If y is not equal to 4, x is not equal to 0, and (y^2 – 16)/(3x) = (y – 4)/6, then in terms of x, y equals:

6(y-4)(y+4) = 3x(y-4)
2(y+4) = x

\(y = \frac{x}{2} - 4 = [m](x-8)/2\)[/m]

IMO B
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If y is not equal to 4, x is not equal to 0, and (y^2 – 16)/(3x) = (y 26 Jun 2020, 03:46
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y not equal to 4 means that RHS cannot be 0
and x not equal to 0 means LHS cannot be indefinite.

Since, it is all variables, you can TEST IT. I find the algebra way faster.

\(\frac{(y^2 – 16)}{(3x)} = \frac{(y – 4)}{6}\)

\(\frac{(y-4)(y+4)}{3x} = \frac{(y-4)}{6}\)

\(\frac{(y+4)}{3x } = \frac{1}{6}\)

\(y+4 = \frac{x}{2}\)

\(y = \frac{x}{2} - 4\)

\(y = \frac{(x-8)}{2}\)

OA, B
MarcoL90
If y is not equal to 4, x is not equal to 0, and (y^2 – 16)/(3x) = (y – 4)/6, then in terms of x, y equals:

A: (x + 8) / 2

B: (x – 8) / 2

C: (-3x) / 2

D: (-3x + 8) / 2

E: (3x – 8) / 2
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If y is not equal to 4, x is not equal to 0, and (y^2 – 16)/(3x) = (y 19 May 2022, 21:21
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