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\[2\left| x+3 \right| = 4\] and \[(\left| y + 1 \right|) \div 3 = 2\] then \[\left| x + y \right|\] could equal all of the following EXCEPT: (A) 0 (B) 4 (C) 8 (D) 10 (E) 12 I know the answer is 10, but why?
@pooja195
There are 2 answers to both the absolute value equations. If you add mix the 4 answers and add them in all different ways, there would be 4 different values from absolute value of (x+y)
i will do x \(2|x+3|=4\) \(|x+3|=2\) \(x+3=\pm 2\) \(x = -3 \pm 2 = -5, -1\) if you do y, you should get 2 poss values of y |dw:1439324981027:dw| pop them in here, then do | | to them, thus you will have the 4 possible outcomes
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