System of parametric inequalitiesI'm doing some systems of parametric inequalities, but I can not understand how to proceed.This is the system: x − 2 a &lt; 1 + a x 2 − 2 a &lt; 2 x − ( a − 3 ) 1 3 − x &lt; a per a &gt; 0These are the solutions I've found: x &lt; 3 a + 1 x &gt; − 2 a − 6 3 x &gt; − 3 a + 1 3 But now I do not know how to proceed. I thought I could compare the solutions like: 3 a + 1 &gt; − 2 a − 6 3 &gt; − 3 a + 1 3

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System of parametric inequalitiesI&#039;m doing some systems of parametric inequalities, but I can not understand how to proceed.This is the system:  x  −  2  a  &amp;lt;  1  +  a      x    2    −  2  a  &amp;lt;  2  x  −  (  a  −  3  )      1    3    −  x  &amp;lt;  a per   a  &amp;gt;  0These are the solutions I&#039;ve found:  x  &amp;lt;  3  a  +  1  x  &amp;gt;            −      2      a      −      6        3    x  &amp;gt;            −      3      a      +      1        3  But now I do not know how to proceed. I thought I could compare the solutions like:  3  a  +  1  &amp;gt;            −      2      a      −      6        3    &amp;gt;            −      3      a      +      1        3

trajeronls · Accepted Answer

The strategy when you have two inequalities, like   x  &amp;gt;            6      −      2      a        3   and   x  &amp;gt;            1      −      3      a        3  , where the signs are the same (i.e. the both say   x  &amp;gt;  expression) is to compare the things   x is compared to. So we will determine whether             −      6      −      2      a        3    &amp;gt;            1      −      3      a        3   or not.For that inequality, one should solve the corresponding equation             −      6      −      2      a        3    =            1      −      3      a        3  . Its solution is   a  =  7. Now, one can solve the inequality. Pick a test value   a  =  0, and plug it in both sides. On the left hand side it is   −  2 and on the right,   1  /  3. Thus             −      6      −      2      a        3    &amp;lt;            1      −      3      a        3   holds for all   a  &amp;lt;  7, while             −      6      −      2      a        3    &amp;gt;            1      −      3      a        3   holds for   a  &amp;gt;  7.Therefore, the assertion that   x is greater than both             −      6      −      2      a        3   and             1      −      3      a        3   is the same as saying   x  &amp;gt;            1      −      3      a        3   when   a  ≤  7 and   x  &amp;gt;            −      6      −      2      a        3   when   a  ≥  7Since we only have one inequality of the form   expression  &amp;gt;  x, we do not have to do anything else, as the system of inequalities now splits into two pieces:  1  +  3  a  &amp;gt;  x  &amp;gt;            1      −      3      a        3     if   a  ≤  7    1  +  3  a  &amp;gt;  x  &amp;gt;            −      6      −      2      a        3     if   a  ≥  7which, without any further information, counts as a solution.

System of parametric inequalities I'm doing some systems of parametric inequalities, but I can not

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