Write expression without using the symbol for absolute value

[seli]

x+y+|x-y| where x>y
2

 

[Deleted member 4993]

x+y+|x-y| where x>y
2

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[JeffM]

[MATH]\dfrac{x + y + \sqrt{(x - y)^2}}{2}[/MATH]
You should KNOW that [MATH]|a| = \sqrt{a^2}.[/MATH]

 

[seli]

Please share your thoughts/work with us, so that we know where to begin to help you.

Please follow the rules of posting in this forum, enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/

[MATH]\dfrac{x + y + \sqrt{(x - y)^2}}{2}[/MATH]
You should KNOW that [MATH]|a| = \sqrt{a^2}.[/MATH]

My answer key says the answer should be x and I have no clue how to get there.

 

[Deleted member 4993]

My answer key says the answer should be x and I have no clue how to get there.

Can you tell me if x = 5 and y=1 (x>y)

|x - y| = ?

 

[JeffM]

My answer key says the answer should be x and I have no clue how to get there.

[MATH]x > y \implies x - y > 0. \implies y - x < 0.[/MATH]
[MATH]\sqrt{a^2} \ge 0.[/MATH]
So which is true:

[MATH]\text {GIVEN } x > y, \text { either } x - y = \sqrt{(x - y)^2} \text { or else } y - x = \sqrt{(x - y)^2}?[/MATH]
What do you do with that information?

 

[HallsofIvy]

My answer key says the answer should be x and I have no clue how to get there.

Yes, since the problem explicitly says \(\displaystyle x\ge y\) then x- y is non-negative so |x- y|= x- y. In that case \(\displaystyle \frac{x+ y+ |x- y|}{2}= \frac{x+ y+ x- y}{2}= \frac{2x}{2}= x\).

Notice that if x< y then x- y is negative so |x- y|= -(x- y)= y- x. In that case \(\displaystyle \frac{x+ y+ |x- y|}{2}= \frac{x+ y+ y- x}{2}= \frac{2y}{y}= y\).