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Mathematics 9 Online
OpenStudy (anonymous):

How would I go about finding x and y? |x| + |y| = |x + y|

OpenStudy (anonymous):

i think you make your own example like 5+3=8

OpenStudy (welshfella):

there are infinite solutions for example |1! + |2| = |1 + 2| and many ,many more

OpenStudy (anonymous):

I 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

OpenStudy (zzr0ck3r):

This is always true, just to let you know...it may help \[|a+b|\le |a|+|b|\]

OpenStudy (anonymous):

What you have is less than or equal to. I am thinking more of just less than, and switched around

OpenStudy (zzr0ck3r):

you will never find \(|a|+|b|<|a+b|\)

OpenStudy (anonymous):

Interesting, 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."

OpenStudy (welshfella):

you interpreted it correctly

OpenStudy (anonymous):

So 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?

OpenStudy (zzr0ck3r):

There is not. \[a^2+b^2+2|a||b|\ge a^2+b^2+2ab\] 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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