SOLUTION: What are all the possible values for x using the pythagorean theorum of a right triangle when the hypotenuse is 27 and one leg is (2x+5)

Question 1016498: What are all the possible values for x using the pythagorean theorum of a right triangle when the hypotenuse is 27 and one leg is (2x+5)
Found 2 solutions by ikleyn, macston:
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
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What are all the possible values for x using the pythagorean theorum of a right triangle when the hypotenuse is 27 and one leg is (2x+5)
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The second leg is . The only restriction for x is this inequality: 2x + 5 < 27, which says that the leg is shorter than the hypotenuse. The last inequality means that x < 11.

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Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
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0 < 2x+5 < 27
-5 < 2x < 22
-2.5 < x < 11
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