Super Duper Tedious Casework

Algebra Level 4

− x + 8 , 2 x − 7 , − 3 x + 6 , 4 x − 5 -x + 8, \quad 2x-7, \quad -3x + 6, \quad 4x-5

The above are 4 distinct real numbers. Among these 4 numbers, which of these cannot be the smallest number?

− x + 8 -x + 8 4 x − 5 4x - 5 2 x − 7 2x - 7 − 3 x + 6 -3x + 6

Pi Han Goh by Pi Han Goh , 30, Malaysia

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1 solution

Jesse Nieminen
Oct 16, 2016

− x + 8 < − 3 x + 6 ⟹ x < − 1 -x + 8 < -3x + 6 \implies x < -1 and − x + 8 < 2 x − 7 ⟹ x > 5 -x + 8 < 2x - 7 \implies x > 5 , hence − x + 8 \boxed{-x + 8} cannot be the smallest.

(After checking the other numbers, we notice that we do not find a contradiction for their minimality.)

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