Systems of Absolute Inequalities | 3 − 2 x | ≥ 1 x 2 / ( x + 2 ) ≤ 0Please advise on this with the answer to this.

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Systems of Absolute Inequalities  |  3  −  2  x  |  ≥  1      x    2    /  (  x  +  2  )  ≤  0Please advise on this with the answer to this.

jugf5 · Accepted Answer

x  =  0 is also a possible solution, the solution set is therefore   x  &amp;lt;  −  2   or   x  =  0 or   x  &amp;lt;  −  2   or   x  =  0Here&#039;s the proof:We first focus on the second inequality:             x      2              x      +      2        ≤  0 implies that either the nominator or the denominator are negative (not both). Or that the nominator is equal to 0. The denominator is never negative. Hence we are left with   x  =  0 (nominator is zero). Or   x  &amp;lt;  −  2 (denominator negative). We can plug in all these into the first inequality and see that they all hold.

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