How do you solve 3x+3<3 and $- 8 x + 6 \geq 0$ $-8x+6>=0$ ?

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Answer 1 · 2015-07-09T12:44:07

$x < 0$ $x<0$

Simplify the inequalities:
$3 x + 3 < 3$ $3x+3<3$ => $x < 0$ $x<0$
and
$- 8 x + 6 \geq 0$ $-8x+6>=0$ => $x \leq \frac{6}{8}$ $x<=6/8$

Take a union of the in equalities we see the first equality is only true if the second is true (the second is therefore redundant).

How do you solve 3x+3<3 and −8x+6≥0-8x+6>=0?