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Please help :( The lengths of two of the sides of a certain triangle are (x - 3) and (x+3) where x > 3. Which of the following ranges represent all of the possible values of the third side, s?

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The sum of any two side of a triangle have to be greater than the third side.
So, \[\large(x-3)+(x+3) >s\]\[\large(x-3)+s >(x+3)\]\[\large\cancel{(x+3)+s >(x-3)}\] use the first two inequalities to find the limit for s
why not the last one? does it cancel out or something?

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we already know that \(x+3>x−3\) .... so the last one happens to be true for all s ... it doesn't help
ohh, okay. So would the answer be 6 < s < 2x ?
yes \(\Large\checkmark\)
Thank you so much :) I really didn't know what to do for it so you helped me SO much :)

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