anonymous
  • anonymous
Please help.... If x>0, and two sides of a certain triangle have lengths 2x+1 and 3x+4 respectively, which of the following could be the length of the third side of the triangle?
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
The third side of the triangle must have a length greater than zero and less than the sum of the other two sides.
anonymous
  • anonymous
1. 4x+5 2. x+2 3. 6x+1 4. 5x+6 5. 2x+17
anonymous
  • anonymous
i know its 1, 3 and 5 but why is 3 an answer?

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ali2x2
  • ali2x2
http://lmgtfy.com/?q=If+x%3E0%2C+and+two+sides+of+a+certain+triangle+have+lengths+2x%2B1+and+3x%2B4+respectively%2C+which+of+the+following+could+be+the+length+of+the+third+side+of+the+triangle%3F
welshfella
  • welshfella
adding the 2 given sides gives 5x + 5
ali2x2
  • ali2x2
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
anonymous
  • anonymous
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 \(x > 0\) ?
ali2x2
  • ali2x2
so its A, B
anonymous
  • anonymous
but look at B if you use 10 then it also doesn't work out
anonymous
  • anonymous
"COULD BE"
anonymous
  • anonymous
13<61<55
anonymous
  • anonymous
it COULD BE if, for example, \(x=1\)
anonymous
  • anonymous
\[5x + 5 = x+2\]\[4x = -3\]Can that be?
anonymous
  • anonymous
hmmm i see wow so i'd have to be extra extra careful..... this question is a tricky one
anonymous
  • anonymous
i have to add that this is, in my opinion ,a really foolish question that confuses the issue
welshfella
  • welshfella
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.
anonymous
  • anonymous
Satellite i 1000000% agree and the only reason it's such a dumb question is because it's a GRE question.
anonymous
  • anonymous
yeah standardized tests strike again
anonymous
  • anonymous
okay got it thanks everyone =)
phi
  • phi
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, 44 this is OK
phi
  • phi
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.
phi
  • phi
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

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