How do you solve the compound inequality $k+2>12$ and $k+2<=18$ ?

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Answer 1 · 2017-03-02T04:26:49

$10 < k <= 16$

Isolate the variable $k$ by subtracting $2$ on both sides of the equation:
$k + 2 > 12$
$k + 2 - 2 > 12 - 2$
$k > 10$ or $10 < k$

$k + 2 <= 18$
$k + 2 - 2 <= 18 - 2$
$k <= 16$

In a combined statement: $10 < k <= 16$

How do you solve the compound inequality k+2>12k+2>12 and k+2<=18k+2<=18?