How do you solve $abs(2x+9)-2x>=0$ ?

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Answer 1 · 2017-02-24T18:34:04

$abs(2x+9)-2x>=0$ is true for all $x in RR$

$abs(2x+9)-2x>=0 -> abs(2x+9)-(2x+9)+9 ge 0$

then, for $x ne -9/2$

$1-(2x+9)/abs(2x+9)+9/abs(2x+9) ge 0$

Calling $9/abs(2x+9) = epsilon ge 0$ we have

$1 pm 1 + epsilon ge 0$ . This is always true so

$abs(2x+9)-2x>=0$ is true for all $x in RR$

How do you solve abs(2x+9)-2x>=0abs(2x+9)-2x>=0?