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The third side of the triangle must have a length greater than zero and less than the sum of the other two sides.
1. 4x+5 2. x+2 3. 6x+1 4. 5x+6 5. 2x+17
i know its 1, 3 and 5 but why is 3 an answer?
adding the 2 given sides gives 5x + 5
Note x>0 says 3x+4 > 2x+1. So, the possibilities are those which lie between the following sum and difference: (3x+4) + (2x+1) = 5x+5 (3x+4) - (2x+1) = x+3
If you add the length of the two sides together, you get \(5x+5\). Set that equal to \(6x+1\) and solve for \(x\). Is \(x > 0\) ?
so its A, B
but look at B if you use 10 then it also doesn't work out
it COULD BE if, for example, \(x=1\)
\[5x + 5 = x+2\]\[4x = -3\]Can that be?
hmmm i see wow so i'd have to be extra extra careful..... this question is a tricky one
i have to add that this is, in my opinion ,a really foolish question that confuses the issue
for arguments sake let x = 2 5x + 5 = 15 6x + 1 = 13 so 3rd side is less than sum of the other 2 6x +1 can be the third side.
Satellite i 1000000% agree and the only reason it's such a dumb question is because it's a GRE question.
yeah standardized tests strike again
okay got it thanks everyone =)
I would do it this way: from 2x+1 and 3x+4 , we must be > x+3 and < 5x+5 1. x+3 < 4x+5 < 5x+5 4x>x and 5>3 and 4x<5x and 5=5. this is OK 2. x+3 < x+2 < 5x+5 for all x>0, x+3 is bigger than x+2. NOT legal 3. x+3 <6x+1 < 5x+5 x+3< 6x+1 means 2 < 5x, 2/5 < x and 6x+1 < 5x + 5 means x< 4. so : 2/5 < x < 4 works. OK 4. x+3 <5x+6 < 5x+5 for all x>0 5x+6 > 5x+5. NOT legal 5. x+3 < 2x+17< 5x+5 2x>x and 17>3 . also 2x+17<5x+5 means 12<3x, 4
4 this is OK
when I checked (for example) x+3 < 4x+5 I said it is obviously true for x>0 but we could analyze it by simplifying the relation to -2 < 3x (add -x to both sides, add -5 to both sides) -2/3 < x and it is true for all x > -2/3. because x>0, this will always be true.
and if we wanted to analyze x+3 < x+2 add -x to both sides 3<2 and we see this is not true. thus we could simplify each relation and make sure it is true for some x