How do you solve the compound inequalities $9 + 3 n \geq 6$ $9 + 3n>=6$ or $5 n < 25$ $5n < 25$ ?

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Answer 1 · 2015-05-04T17:08:10

Handle each inequality separately and then recombine.

$9 + 3 n \geq 6$ $9+3n>=6$
$\to n \geq - 1$ $rarr n>= -1$

$5 n < 25$ $5n<25$
$\to n < 5$ $rarr n<5$

So $9 + 3 n \geq 6 or 5 n < 25$ $9+3n>=6 " or " 5n<25$
is equivalent to
$n \geq - 1 or n < 5$ $n>=-1 " or " n<5$

But one or the other is true for any Real value of $n$ $n$
So
$n ε (- \infty, + \infty)$ $n epsilon (-oo,+oo)$

How do you solve the compound inequalities 9+3n≥69 + 3n>=6 or 5n<255n < 25?