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What is the value of y?

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shanmugamgsn
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What is the value of y? [#permalink] New post   04 Dec 2012, 23:23
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What is the value of y?

(1) y^3+2y=y+2y^2

(2) y^2=y

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shanmugamgsn
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Re: What is the value of y? [#permalink] New post   04 Dec 2012, 23:30
shanmugamgsn wrote:
What is the value of \(y\)?

(1)\(y^3+2y=y+2y^2\)

(2)\(y^2=y\)



My choice was A..
I did this way can any one correct me if im wrong??

(2) INSUFFICIENT :
\(y^2-y=0\)
\(y(y-1)\)
So \(y=0\) or \(y=1\)

(1) i felt this is SUFFICIENT:
\(y^3+2y=y+2y^2\)
divide by y on both sides
\(y^2+2=1+2y\)
\(y^2-2y+1=0\)
\((y-1)^2=0\)
So solution will be \(y=1\)

Is there any mistakes in my calculation?
If y this soln is wrong???
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Re: What is the value of y? [#permalink] New post   05 Dec 2012, 00:42
shanmugamgsn wrote:
shanmugamgsn wrote:
What is the value of \(y\)?

(1)\(y^3+2y=y+2y^2\)

(2)\(y^2=y\)



My choice was A..
I did this way can any one correct me if im wrong??

(2) INSUFFICIENT :
\(y^2-y=0\)
\(y(y-1)\)
So \(y=0\) or \(y=1\)

(1) i felt this is SUFFICIENT:
\(y^3+2y=y+2y^2\)
divide by y on both sides
\(y^2+2=1+2y\)
\(y^2-2y+1=0\)
\((y-1)^2=0\)
So solution will be \(y=1\)

Is there any mistakes in my calculation?
If y this soln is wrong???


At first glance of the equation given in Statement 1, you should be able to see that 0 is a solution to the equation. Every expression contains a y term and hence 0 will satisfy the equation. Your first step is dividing by y. Since y is an unknown variable, that step cannot be allowed. As the famous mems of the internet go No one except Chuck Norris can divide by 0. :P...
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Re: What is the value of y? [#permalink] New post   05 Dec 2012, 02:02
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What is the value of y?

(1) y^3+2y=y+2y^2 --> \(y^3-2y^2+y=0\) --> \(y(y^2-2y+1)=0\) --> \(y(y-1)^2=0\) --> \(y=0\) or \(y=1\). Not sufficient.

(2) y^2=y --> \(y(y-1)=0\) --> \(y=0\) or \(y=1\). Not sufficient.

(1)+(2) \(y=0\) or \(y=1\). Not sufficient.

Answer: E.
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Re: What is the value of y? [#permalink] New post   05 Dec 2012, 02:04
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shanmugamgsn wrote:
shanmugamgsn wrote:
What is the value of \(y\)?

(1)\(y^3+2y=y+2y^2\)

(2)\(y^2=y\)



My choice was A..
I did this way can any one correct me if im wrong??

(2) INSUFFICIENT :
\(y^2-y=0\)
\(y(y-1)\)
So \(y=0\) or \(y=1\)

(1) i felt this is SUFFICIENT:
\(y^3+2y=y+2y^2\)
divide by y on both sides
\(y^2+2=1+2y\)
\(y^2-2y+1=0\)
\((y-1)^2=0\)
So solution will be \(y=1\)

Is there any mistakes in my calculation?
If y this soln is wrong???


Never reduce an equation by a variable (or expression with a variable), if you are not certain that the variable (or the expression with a variable) doesn't equal to zero. We can not divide by zero.

So, if you divide (reduce) y^3+2y=y+2y^2 by y, you assume, with no ground for it, that y does not equal to zero thus exclude a possible solution (notice that both y=1 AND y=0 satisfy the equation).

Hope it's clear.
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Re: What is the value of y? [#permalink] New post   06 May 2014, 06:48
What is the value of y?
(1) y^3+2y=y+2y^2

(2) y2=y

official ans. is E. But I got A. Please someone explain issue in following logic

y^3+2y = y+2y^2

y(Y^2+2) = y(1+2y)

now cancel y on both sides

y^2+2 = 1+2y
Y^2-2y+1 = 0

solving this I will get (y-1) = 0 so y =1

or we can see b^2-4ac = 0 so it has one root.

Please clarify

Thanks
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Re: What is the value of y? [#permalink] New post   06 May 2014, 08:22
Expert Reply
PathFinder007 wrote:
What is the value of y?
(1) y^3+2y=y+2y^2

(2) y2=y

official ans. is E. But I got A. Please someone explain issue in following logic

y^3+2y = y+2y^2

y(Y^2+2) = y(1+2y)

now cancel y on both sides

y^2+2 = 1+2y
Y^2-2y+1 = 0

solving this I will get (y-1) = 0 so y =1

or we can see b^2-4ac = 0 so it has one root.

Please clarify

Thanks


Merging similar topics. Please refer to the discussion above.

Your doubt is addressed here: what-is-the-value-of-y-143619.html#p1150937

Hope it helps.
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Re: What is the value of y? [#permalink] New post   22 Oct 2022, 04:19
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Re: What is the value of y? [#permalink]
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