Question

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?

Answer 1

The third side of the triangle must have a length greater than zero and less than the sum of the other two sides.

Answer 2

1. 4x+5
2. x+2
3. 6x+1
4. 5x+6
5. 2x+17

Answer 3

i know its 1, 3 and 5 but why is 3 an answer?

Answer 4

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

Answer 5

adding the 2 given sides gives 5x + 5

Answer 6

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

Answer 7

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 \(x > 0\) ?

Answer 8

so its A, B

Answer 9

but look at B if you use 10 then it also doesn't work out

Answer 10

"COULD BE"

Answer 11

13<61<55

Answer 12

it COULD BE if, for example, \(x=1\)

Answer 13

\[5x + 5 = x+2\]\[4x = -3\]Can that be?

Answer 14

hmmm i see wow so i'd have to be extra extra careful..... this question is a tricky one

Answer 15

i have to add that this is, in my opinion ,a really foolish question that confuses the issue

Answer 16

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.

Answer 17

Satellite i 1000000% agree and the only reason it's such a dumb question is because it's a GRE question.

Answer 18

yeah standardized tests strike again

Answer 19

okay got it thanks everyone =)

Answer 20

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<x
so for x>4 this is OK

Answer 21

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.

Answer 22

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