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anonymous
 one year ago
How would I go about finding x and y? x + y = x + y
anonymous
 one year ago
How would I go about finding x and y? x + y = x + y

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think you make your own example like 5+3=8

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1there are infinite solutions for example 1! + 2 = 1 + 2 and many ,many more

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think the thing that I am more confused with is what would x and y have to be (such as negative or positive) that would be right, sorry I didn't explain the question correctly I mean for x + y < x+y xD

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0This is always true, just to let you know...it may help \[a+b\le a+b\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What you have is less than or equal to. I am thinking more of just less than, and switched around

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0you will never find \(a+b<a+b\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Interesting, How would you interpret the word problem: "Write the statement where the sum of the absolute values of two numbers is less than the absolute value of the sum of the two numbers."

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1you interpreted it correctly

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So that would interpret to x+y<x+y? and if so, there must be some value for x and y that would make the equation true, correct?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0There is not. \[a^2+b^2+2ab\ge a^2+b^2+2ab\] Now \(a^2=a^2\) for any number \(a\). So from above \[a^2+b^2\ge a+b^2\implies a+b\ge a+b\]
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