Question

collin left the party one half hour before gabriel. terrence left an hour and a half later than gabriel. together the three of them stayed more than four hours. how long did they each stay at the party?

Answer 1

Are you sure you type the question correctly?

It is too ambiguous to have one answer.

Answer 2

oh i forgot a part to translate into inequality?

Answer 3

c + g + t = 4
g + .5 = c
t + 1.5 = g

you can solve those with substitution

Answer 4

It says they all stayed together more than 4 hours.

What if Colin stayed for 20 hours, terrance Stayed for 22 hours, and gabriel stayed for 20.5 hours?

That is a valid answer, the way this question is typed.

Did you mean to say that they all stayed PRECISELY four hours at the party?

Answer 5

No, it can be at least 4 hours. You can find the minimum times with the equations i posted before. The only precise thing you would have to change is
\[c + g + t \ge 4\]
You can still solve with inequalities.

Answer 6

So you're saying:

c + g + t >=4
c = g - .5
t = g + 1.5

So we have

g - .5 + g + g + 1.5 >= 4
3g + 1 >= 4
3g >= 3
g >= 1

Right?

Answer 7

The idea is correct, but you substituted t incorrectly. t + 1.5 = g not -1.5.
Other than that, once you have g, you can solve c and t.

I was going to let salvarican do that math.

Answer 8

terrance left an hour and a half later than Gabriel,
this implies T = G + 1.5 ?

Answer 9

Oh yeah, my bad, copied the equation down wrong. Good catch!

Answer 10

I think whoever formulated the problem runs a scabby operation anyway

Answer 11

I've actually seen this question on here multiple times. It must be from a common algebra book.

Answer 12

scabby operation i tells ya