|x|=|2y|, what is the value of x-2y?

Dear Bunuel,

whenever absolute value is analysed, we take two scenarios of <0 and >0.
So, why the same is not considered for |x| ?

Tricky question.... I gave 2 much time to evaluate stmt 1 and went with A.

Bunuel: can we rewrite |x|=|2y| as x^2-4x^2=0 ?
I have solved the problem doing so, but not sure if it algebraically correct.
Below what i did:

(x-2y)(x+2y)=0

Using statement 1:
(x-2y)*6=0
so, (x-2y)=0. Sufficient

Using statement 2:
x=2y [same sign]
(x-2y)=0. Sufficient

D

Yes, you can square \(|x|=|2y|\) and write \(x^2=4y^2\) --> \((x-2y)(x+2y)=0\) --> either \(x=2y\) or \(x=-2y\) the same two options as in my solution above.

Hi Bunuel,

I had a query regarding an official statement in the solution to this problem.
Actually, the book says that , as, x+2y=6 , so a positive sum indicates that both x and 2y must be positive.
However, -4+10= 10+(-4) = 6 =positive sum [both x and 2y are not positive] 10+4=14= positive sum [both x & 2y are positive] isn't it?
Please clarify the confusion here..

x + 2y = 6
Hence we know that x is not equal to -2y, but |x| = |2y|
So, x = 2y

Hello, I am a bit confused regarding absolute value.

If \(|x|=|2y|\), then why why aren't x and 2y both positive? If the abs. value of something (i.e. it's positive value) is equal to something else, doesn't that imply that they are both positive? For example, if x=|2y| doesn't that mean that x is positive?

Also, for #2, xy both have the same signs. If x and y are negative, why would |x|=|2y| become -x = -2y? I get that it's equal to |x|=|2y| but why even take that step? |-x| = |-2y| will always be positive, right?

The absolute value cannot be negative \(|some \ expression|\geq{0}\), or \(|x|\geq{0}\) (absolute value of x, |x|, is the distance between point x on a number line and zero, and the distance cannot be negative).

So, if given that \(x=|2y|\) then \(x\) must be more than or equal to zero (RHS is non-negative thus LHS must also be non-negative).

But in our case we have that \(|x|=|2y|\). In this case \(x\) and/or \(y\) could be negative. For, example \(x=-2\) and \(y=-1\) --> \(|x|=2=|2y|\).

As for (2):
When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

So, if \(x<0\) and \(y<0\), then \(|x|=-x\) and \(|2y|=-2y\) --> \(-x=-2y\) --> \(x=2y\). If \(x>0\) and \(y>0\), then \(|x|=x\) and \(|2y|=2y\) --> \(x=2y\), the same as in the first case.

For more check Absolute Value chapter of Math Book: math-absolute-value-modulus-86462.html

DS questions on absolute value to practice: search.php?search_id=tag&tag_id=37
PS questions on absolute value to practice: search.php?search_id=tag&tag_id=58

Tough absolute value and inequity questions with detailed solutions: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope it helps.

Ok, so I get that Abs. value cannot be negative...distance cannot have a negative value.

We are trying to solve for x-2y, so naturally we are trying to determine x-2y. So,

If x=2y then the value of x-2y = 2y-2y = 0
OR
If x=-2y (the absolute value of 2y) then the value of x-2y = -2y-2y = -4y, correct?

I guess what throws me off is when you write

When x\leq{0} then |x|=-x. What you're saying is that, for example, |-4| = -(-4) or |-4| = 4. What is the point of writing |-4| = -(-4)

One final thing...In the stem you derived x=2y, x=-2y. Okay, but in #2. one of the cases is xy>0 so we could have -x and -y. If x and y are negative, doesn't that mean that you would substitute -x and y in to get -x=-2(-y) = -x=2y?

I'm sorry for being such a dolt. Sometimes, concepts that I know are very simple are extremely difficult to understand.

Yes, that's correct: if x=2y, then x-2y=0 and if x=-2y, then x-2y=-4y.

As for the red part: it's just an example of the statement that if \(x\leq{0}\) then \(|x|=-x\).

|x|=|2y|, what is the value of x-2y?

(1) x+2y = 6
(2) xy>0

1) that means that x=3 and 2y=3, so difference is only 0
2) that means that x and y is not 0 and both positive or negative and x=2y, so 2y-2y=0

D

|x|=|2y|, what is the value of x-2y?

x=2y
OR
x=-2y

(1) x+2y = 6

2y+2y = 6
4y = 6
y=3/2

x+2(3/2) = 6
x+3 = 6
x=3

OR
-2y+2y = 6
0=6 (Invalid...6 cannot equal 0)
With only one valid solution for x and y we can solve for x-2y.
SUFFICIENT

(2) xy>0

xy>0 means that BOTH x and y are positive or BOTH x and y are negative.
We can choose numbers to make this easier:
x=2, y=1

If x=2y, then 2=2(1)
OR
Id x=-2y, then -2 = 2(-1)

If x and y are both positive: x-2y ===> 2-2(1) = 0
If x and y are both negative: x-2y ===> -2 - 2(-1) ===> -2+2 = 0

SUFFICIENT

If |x|=|2y|
what is the value of x-2y?

1. x+2y=6
2. xy>0

Am stuck with solving the statement 1 with case scenarios.
Somebody please explain your entire solutions especially the statement one positive negative scenarios.

Thanks.

Hello mpingo
This topic discussed here:
x-2y-what-is-the-value-of-x-2y-133307.html

Please, use search before posting

Also, please, read rule of posting #3 about naming of topic
rules-for-posting-please-read-this-before-posting-133935.html

If after reading discussions above, you will have questions, than write them here and I will be glad to help.

ok, x= 2y, or x= -2y---> x-2y =0 or x+2y =0
1. x+2y =6 so x-2y =0
2. x,y >0--> x+2y can't be 0 , so x-2y =0
D

Hi Bunuel, thanks for that. Just have 1 qs. In case of statement2 when we consider the case when X and y are both negative wouldn't we consider the -(minus) inside the absolute |-x|=|-2y| instead of outside.

|something| = |something| ----> for constructs like these square both sides and remove the Mod sign.

Stem:
\(|x|=|2y|\) ----> \((x)^2= (2y)^2\)--->\( x^2-(2y)^2=0\)--- > is of the form \(a^2-b^2\)= \((a-b) * (a+b)\)

stm1>\(x+2y = 6\)

from the equation we formed in the stem: \(x^2-(2y)^2=0 \)----> \((x-2y)*(x+2y)=0\)

put \(x+2y=6\) in the above equation, and we get \(x-2y=0\)

Suff.

(2) xy>0 (means both x and y are of the same sign, both +ve or -ve)

\(x^2-(2y)^2=0-\)---> \((x+2y)* (x-2y)=0\) ----> \(x-2y= 0\). Sufficient.